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Calculus and Above

Calculus Questions? Ask a Mathematician for Answers ASAP

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This answer was rated:

★★★★★

The population mean annual...

The population mean annual salary for environmental compliance specialists is about

$66 comma 00066,000.

A random sample of

4040

specialists is drawn from this population. What is the probability that the mean salary of the sample is less than

$62 comma 50062,500?

Assume

sigmaσequals=$6 comma 4006,400.

**Tutor's Assistant:** The Math Tutor can help you get an A on your homework or ace your next test. Is there anything else important you think the Math Tutor should know?

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Customer reply replied 3 years ago

Posted by JustAnswer at customer's request) Hello. I would like to request the following Expert Service(s) from you: Live Phone Call.Let me know if you need more information, or send me the service offer(s) so we can proceed.how can I contact you

Answered in 8 minutes by:

7/21/2017

Ryan, Engineer

Category: Calculus and Above

Satisfied Customers: 9,218

Experience: B.S. in Civil Engineering

Verified

Hi,

Thank you for using the site. I'll be happy to help you with this problem.

I assume that the repeated digits is because the original text has those values in bold-faced type.

Just to be sure though, can you please confirm the following values:

$66,000

40

$62,500

sigma =$6,400

Thanks,

Ryan

Customer reply replied 3 years ago

it is right

Ok, thanks. I'll have the solution posted for you very shortly.

Ryan, Engineer

Category: Calculus and Above

Satisfied Customers: 9,218

Experience: B.S. in Civil Engineering

Verified

Ryan and 87 other Calculus and Above Specialists are ready to help you

Ask your own question now

Phone call session started

Customer reply replied 3 years ago

$34 for how long

There is no specified limit to the length of the phone call, but it would need to be within reason. I couldn't spend 3 hours on the phone for that rate, obviously.

Please note that the phone call offer was generated by the computer system, and not by my personally.

Customer reply replied 3 years ago

Well the questions is on computer it is easier on email just copy and pest

Here is a link to a Word document with the solution for the problem:

Please feel free to ask if you have any questions about this solution.

If you would still like to receive a phone call, please let me know.

Thanks,

Ryan

Customer reply replied 3 years ago

Is it available all day

Are you asking about the link?

Customer reply replied 3 years ago

I sign up before a week a go but i am not able to login to get service

I'm not sure what you are specifically referring to.

If you are asking about the website, then, yes, in general the site is available 24 hours a day. Of course, it depends on whether there is anybody online who works in the category that your question is place in. If there are no available experts, then there would be some delay in getting an answer.

If you are asking about the link that I posted, then the answer is that the link will be available for about a week.

Is there anything else that I can help you with?

Customer reply replied 3 years ago

What is the specific sight to log in

This site can be reached at www.justanswer.com.

If you are having trouble accessing your account, you should contact a Customer Service representative to get assistance. There should be a link on the page that you can use to contact them, or you can call them at(###) ###-####

Customer reply replied 3 years ago

I still dont get my questions

Did you click on the link that I posted? The one labeled "Probability"?

If needed, I can post a new link. Or I can repost the solution as text here.

If you would prefer to use email, please let me know and we can make arrangements to do that.

Customer reply replied 3 years ago

I dont see any link

Here is the solution in text:

Calculate the z value:

z = (x-bar - mu) / (sigma / √n) = (62,500 - 66,000) / (6400 / √40) = **-3.4587**

Then, since the question asks for the probability that the average is "less than 62,500", we want the area to the left of z = -3.4587:

P(x-bar < 62,500) = P(z < -3.4587) =** 0.0003**

You can get this value from tables, from a statistical calculator, or by entering the following in a cell in Excel:

=normsdist(-3.4587)

Solution: The probability is 0.0003 (rounded to four decimal places).

Customer reply replied 3 years ago

That was correctIf i would like to use for 20 days how much it cost

Great!

Is there anything else I can help you with?

Customer reply replied 3 years ago

i have so many questionsOnly

0.030.03%

of samples of

4040

specialists will have a mean salary less than

$62 comma 50062,500.

This is an

extremely unusualextremely unusual

event.B.About

0.030.03%

of samples of

4040

specialists will have a mean salary less than

$62 comma 50062,500.

This is not an unusual event.C.Only

33%

of samples of

4040

specialists will have a mean salary less than

$62 comma 50062,500.

This is an

extremely unusualextremely unusual

event.D.About

33%

of samples of

4040

specialists will have a mean salary less than

$62 comma 50062,500.

This is not an unusual event.Find the probability and interpret the results. If convenient, use technology to find the probability. During a certain week the mean price of gasoline was

$2.7062.706

per gallon. A random sample of

3333

gas stations is drawn from this population. What is the probability that the mean price for the sample was between

$2.6992.699

and

$2.7272.727

that week? Assume

sigmaσequals=$0.0450.045.

LOADING...

Click the icon to view page 1 of the standard normal table.

LOADING...

Click the icon to view page 2 of the standard normal table.The probability that the sample mean was between

$2.6992.699

and

$2.7272.727

is

nothingm.

(Round to four decimal places as needed.)

0.030.03%

of samples of

4040

specialists will have a mean salary less than

$62 comma 50062,500.

This is an

extremely unusualextremely unusual

event.B.About

0.030.03%

of samples of

4040

specialists will have a mean salary less than

$62 comma 50062,500.

This is not an unusual event.C.Only

33%

of samples of

4040

specialists will have a mean salary less than

$62 comma 50062,500.

This is an

extremely unusualextremely unusual

event.D.About

33%

of samples of

4040

specialists will have a mean salary less than

$62 comma 50062,500.

This is not an unusual event.Find the probability and interpret the results. If convenient, use technology to find the probability. During a certain week the mean price of gasoline was

$2.7062.706

per gallon. A random sample of

3333

gas stations is drawn from this population. What is the probability that the mean price for the sample was between

$2.6992.699

and

$2.7272.727

that week? Assume

sigmaσequals=$0.0450.045.

LOADING...

Click the icon to view page 1 of the standard normal table.

LOADING...

Click the icon to view page 2 of the standard normal table.The probability that the sample mean was between

$2.6992.699

and

$2.7272.727

is

nothingm.

(Round to four decimal places as needed.)

The site doesn't charge on a "per day" basis like that. Also, please be forewarned that if you are offered some kind of a "subscription" for the site, that their policies do not allow subscriptions to be used in the homework categories.

The best approach for your situation is to post a set of questions, and adjust the price to a level that you are comfortable with for that amount of work involved. If one of the experts is willing to take on the assignment for the offered price, then they will contact you.

How many more questions do you need help with tonight?

The interpretation of the result from the previous question is the first choice:

**Only 0.03% of samples of 40 specialists will have a mean salary less than$62,500. This is an extremely unusual event.**

For the problem about the gasoline prices, do you just need the final answer, or do you need the intermediate steps as well?

Customer reply replied 3 years ago

Just answer

Ok.

Customer reply replied 3 years ago

as many as you can

The answer the gasoline price problem is:

Probability = **0.8106**

If you have additional problems that you want help with, can you please include a problem number? That will make it easier to refer to specific problems, and to match up the answers with the correct problem.

Thanks,

Ryan

Customer reply replied 3 years ago

are you with me

Yes.

Customer reply replied 3 years ago

Interpret the results. Choose the correct answer below.A.About

1919%

of samples of

3333

gas stations that week will have a mean price between

$ 2.699$2.699

and

$2.7272.727.B.About

1919%

of the population of

3333

gas stations that week will have a mean price between

$ 2.699$2.699

and

$2.7272.727.C.About

8181%

of the population of

3333

gas stations that week will have a mean price between

$ 2.699$2.699

and

$2.7272.727.D.About

8181%

of samples of

3333

gas stations that week will have a mean price between

$ 2.699$2.699

and

$2.7272.727.

1919%

of samples of

3333

gas stations that week will have a mean price between

$ 2.699$2.699

and

$2.7272.727.B.About

1919%

of the population of

3333

gas stations that week will have a mean price between

$ 2.699$2.699

and

$2.7272.727.C.About

8181%

of the population of

3333

gas stations that week will have a mean price between

$ 2.699$2.699

and

$2.7272.727.D.About

8181%

of samples of

3333

gas stations that week will have a mean price between

$ 2.699$2.699

and

$2.7272.727.

**D. About 81% of samples of 33 gas stations that week will have a mean price between$ 2.699 and $2.727.**

Customer reply replied 3 years ago

The height of women ages 20-29 is normally distributed, with a mean of

64.464.4

inches. Assume

sigmaσequals=2.72.7

inches. Are you more likely to randomly select 1 woman with a height less than

65.165.1

inches or are you more likely to select a sample of

2727

women with a mean height less than

65.165.1

inches? Explain.

LOADING...

Click the icon to view page 1 of the standard normal table.

LOADING...

Click the icon to view page 2 of the standard normal table.What is the probability of randomly selecting 1 woman with a height less than

65.165.1

inches?nothingm (Round to four decimal places as needed.)can you show me some step

64.464.4

inches. Assume

sigmaσequals=2.72.7

inches. Are you more likely to randomly select 1 woman with a height less than

65.165.1

inches or are you more likely to select a sample of

2727

women with a mean height less than

65.165.1

inches? Explain.

LOADING...

Click the icon to view page 1 of the standard normal table.

LOADING...

Click the icon to view page 2 of the standard normal table.What is the probability of randomly selecting 1 woman with a height less than

65.165.1

inches?nothingm (Round to four decimal places as needed.)can you show me some step

Ok.

What they want you to do here is compare the probability of a single individual from the population having a height that is less than 65.1 inches, with the probability of a sample of 27 individuals having an **mean **(average) height that is less than 65.1 inches.

For the single individual, calculate the z-value as:

z = (x - mu) / sigma = (65.1 - 64.4) / 2.7 = 0.2593

The corresponding probability is:

P(x < 65.1) = P(z < 0.2593) = **0.6023**

For the sample of 27 individuals selected from this population, calculate the z-value as:

z = (x-bar - mu) / (sigma / √n) = (65.1 - 64.4) / (2.7 / √27) = 1.3472

The corresponding probability is then:

P(x-bar < 65.1) = P(z < 1.3472) = **0.9110**

The probability is much higher in the second case, so you are far more likely to select a sample of 27 individuals with a mean height that is less than 65.1 inches than you are to select a single individual who is less than 65.1 inches tall.

Customer reply replied 3 years ago

sorry 0.9110 is incorrect

Did you enter the 0.6023 answer first?

Customer reply replied 3 years ago

no

Ok, the answer for the first part:

*What is the probability of randomly selecting 1 woman with a height less than 65.1 inches?*

is **0.6023**

Customer reply replied 3 years ago

A news reporter reports the results of a survey and states that 45% of those surveyed responded "yes" with a margin of error of "plus or minus 5%." Explain what this means.Choose the correct answer below.A.The percent of the population who fall under the "no" category is 5% less than the "yes" category.B.It means that 5% of the surveys are invalid.C.The true percent of the population who fall under the "yes" category is either 40% or 50%.D.The percent of the population who fall under the "yes" category probably lies between 40% and 50%.Use the standard normal table to find the z-score that corresponds to the given percentile. If the area is not in the table, use the entry closest to the area. If the area is halfway between two entries, use the z-score halfway between the corresponding z-scores. If convenient, use technology to find the z-score.Upper P 15P15Click to view page 1 of the table.

LOADING...

Click to view page 2 of the table.

LOADING...The z-score that corresponds to ***** 15P15

is

nothingm.

(Round to

twotwo

decimal places as needed

LOADING...

Click to view page 2 of the table.

LOADING...The z-score that corresponds to ***** 15P15

is

nothingm.

(Round to

twotwo

decimal places as needed

**D. The percent of the population who fall under the "yes" category probably lies between 40% and 50%.**

The z-core that corresponds to the upper 15th percent is **1.04 **(rounded to two decimal places).

Customer reply replied 3 years ago

sorry 1.04 is incorrect

I'm not sure what to say about that. The only thing that I can think of is that they have some kind of wonky definition of "Upper P15". I interpreted it to mean the top 15%. Normally though, this would be referred to as "P85" instead of "Upper P15".

If we ignore the "Upper" part, and just compute P15, then the answer is **-1.04. (Note the negative sign.)**

Customer reply replied 3 years ago

Find the critical value

z Subscript czc

necessary to form a confidence interval at the level of confidence shown below.cequals=0.830.83z Subscript czcequals=nothingmmeaning zero

z Subscript czc

necessary to form a confidence interval at the level of confidence shown below.cequals=0.830.83z Subscript czcequals=nothingmmeaning zero

zc = **1.372**

round off as directed

Customer reply replied 3 years ago

The site doesn't charge on a "per day" basis like that. Also, please be forewarned that if you are offered some kind of a "subscription" for the site, that their policies do not allow subscriptions to be used in the homework categories.The best approach for your situation is to post a set of questions, and adjust the price to a level that you are comfortable with for that amount of work involved. If one of the experts is willing to take on the assignment for the offered price, then they will contact you.How many more questions do you need help with tonight?The interpretation of the result from the previous question is the first choicewhat do you mean

Do you have more questions to work on tonight?

Customer reply replied 3 years ago

I accidentaly exite and not able to log in computer

I see. If you are able to get back in, please feel free to post the next problem here.

Customer reply replied 3 years ago

U see me on my phone but the qution is on my computerHow can i login so i can copy and past the qusion

So you can't get to this site on your computer?

Customer reply replied 3 years ago

I ued to copy and past the questions but am exite the pageYe i got it

Ok, good.

Customer reply replied 3 years ago

Find the margin of error for the given values of c, s, and n.cequals=0.900.90,sequals=3.63.6,nequals=100100Eequals=nothingm

E = **0.5922**

Customer reply replied 3 years ago

You are given the sample mean and the sample standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Which interval is wider? If convenient, use technology to construct the confidence intervals.A random sample of

3434

gas grills has a mean price of

$632.20632.20

and a standard deviation of

$55.4055.40.

3434

gas grills has a mean price of

$632.20632.20

and a standard deviation of

$55.4055.40.

The 90% confidence interval is (616.5722,(###) ###-####

The 95% confidence interval is (613.5783,(###) ###-####

Please let me know if the above intervals do not match the answer form that you are expected to use. Sometimes the intervals are expressed as "mean ± margin of error".

Customer reply replied 3 years ago

In a random sample of

5050

refrigerators, the mean repair cost was

$134.00134.00

and the population standard deviation is

$16.5016.50.

A

9090%

confidence interval for the population mean repair cost is

left parenthesis 130.16 comma 137.84 right parenthesis(130.16,137.84).Change the sample size to

nequals=100100.

Construct a

9090%

confidence interval for the population mean repair cost. Which confidence interval is wider? Explain.Construct a

9090%

confidence interval for the population mean repair cost.The

9090%

confidence interval is

(nothingm,nothingm).

(Round to two decimal places as needed.)can you show me how u do it

5050

refrigerators, the mean repair cost was

$134.00134.00

and the population standard deviation is

$16.5016.50.

A

9090%

confidence interval for the population mean repair cost is

left parenthesis 130.16 comma 137.84 right parenthesis(130.16,137.84).Change the sample size to

nequals=100100.

Construct a

9090%

confidence interval for the population mean repair cost. Which confidence interval is wider? Explain.Construct a

9090%

confidence interval for the population mean repair cost.The

9090%

confidence interval is

(nothingm,nothingm).

(Round to two decimal places as needed.)can you show me how u do it

For a 90% confidence interval, use z = 1.645.

The margin of error is then:

E = z * s / √n = (1.645)(16.5) / √100 = 2.7143

Then, the limits of the confidence interval are:

Lower limit = mean - E = 134 - 2.71 = 131.29

Upper limit = mean + E = 143 + 2.71 = 136.71

The new confidence interval is thus **(131.29, 136.71)**

The original confidence interval (with n = 50) is wider than the new confidence interval. This is because the larger sample size narrows the confidence interval.

Customer reply replied 3 years ago

The

nequals=5050

confidence interval is wider because a smaller sample is taken, giving less information about the population.B.The

nequals=100100

confidence interval is wider because a larger sample is taken, giving more information about the population.C.The two intervals are the same size because the confidence interval is based on the level of confidence and sample standard deviation.

nequals=5050

confidence interval is wider because a smaller sample is taken, giving less information about the population.B.The

nequals=100100

confidence interval is wider because a larger sample is taken, giving more information about the population.C.The two intervals are the same size because the confidence interval is based on the level of confidence and sample standard deviation.

**A. The n = 50 confidence interval is wider because a smaller sample is taken, giving less information about the population.**

Customer reply replied 3 years ago

A random sample of

fifty dash sevenfifty-seven

200-meter swims has a mean time of

3.5043.504

minutes and a standard deviation of

0.0700.070

minutes. A

9090%

confidence interval for the population mean time is

left parenthesis 3.489 comma 3.519 right parenthesis(3.489,3.519).Construct a

9090%

confidence interval for the population mean time using a standard deviation of

0.040.04

minutes. Which confidence interval is wider? Explain.The

9090%

confidence interval

fifty dash sevenfifty-seven

200-meter swims has a mean time of

3.5043.504

minutes and a standard deviation of

0.0700.070

minutes. A

9090%

confidence interval for the population mean time is

left parenthesis 3.489 comma 3.519 right parenthesis(3.489,3.519).Construct a

9090%

confidence interval for the population mean time using a standard deviation of

0.040.04

minutes. Which confidence interval is wider? Explain.The

9090%

confidence interval

Mean = 3.504

New standard deviation = 0.040

Sample size = 57

Margin of error = z * s / √n = (1.645)(0.040) / √57 = 0.0087

Lower limit = mean - E = 3.504 - 0.0087 = 3.495

Upper limit = mean + E = 3.504 + 0.0087 = 3.513

The new confidence interval is **(3.495, 3.513).**

The original confidence interval is wider.

Customer reply replied 3 years ago

A.The

sequals=0.0700.070

confidence interval is wider because of the increased variability within the sample.B.The

sequals=0.040.04

confidence interval is wider because of the decreased variability within the sample.C.The two intervals are the same size because the confidence interval is based on the level of confidence and sample size.

sequals=0.0700.070

confidence interval is wider because of the increased variability within the sample.B.The

sequals=0.040.04

confidence interval is wider because of the decreased variability within the sample.C.The two intervals are the same size because the confidence interval is based on the level of confidence and sample size.

**A. The s = 0.070 confidence interval is wider because of the increased variability within the sample.**

Customer reply replied 3 years ago

A doctor wants to estimate the HDL cholesterol of all 20- to 29-year-old females. How many subjects are needed to estimate the HDL cholesterol within

44

points with

99 %99%

confidence assuming

sigma equals 11.7 question mark σ=11.7?

Suppose the doctor would be content with

90 %90%

confidence. How does the decrease in confidence affect the sample size required?A 99% confidence level requires

nothingm subjects.

(Round up to the nearest whole number as needed.)

44

points with

99 %99%

confidence assuming

sigma equals 11.7 question mark σ=11.7?

Suppose the doctor would be content with

90 %90%

confidence. How does the decrease in confidence affect the sample size required?A 99% confidence level requires

nothingm subjects.

(Round up to the nearest whole number as needed.)

For the 99% confidence level:

Required sample size = (z * s / E)^2 = (2.576 * 11.7 / 0.44)^2 = **4693 **(when rounded up to the next whole number).

Customer reply replied 3 years ago

for how long/ day my account will be active

I don't know. I don't have any access to the details of your account. A customer service representative can help you with that inquiry.

Customer reply replied 3 years ago

meaning I already pay 37 dollar so how long survicei will getA doctor wants to estimate the HDL cholesterol of all 20- to 29-year-old females. How many subjects are needed to estimate the HDL cholesterol within

44

points with

99 %99%

confidence assuming

sigma equals 11.7 question mark σ=11.7?

Suppose the doctor would be content with

90 %90%

confidence. How does the decrease in confidence affect the sample size required?A 99% confidence level requires

nothingm subjects.

(Round up to the nearest whole number as needed.)A beverage company uses a machine to fill one-liter bottles with water (see figure). Assume that the population of volumes is normally distributed.(a) The company wants to estimate the mean volume of water the machine is putting in the bottles within 1 milliliter. Determine the minimum sample size required to construct a

9090%

confidence interval for the population mean. Assume the population standard deviation is

55

milliliters.

(b) Repeat part (a) using an error tolerance of

22

milliliters. Which error tolerance requires a larger sample size? Explain.A machine cuts plastic into sheets that are

3030

feet

(360360

inches) long. Assume that the population of lengths is normally distributed. Complete parts (a) and (b).(a)The company wants to estimate the mean length the machine is cutting the plastic within

0.1250.125

inch. Determine the minimum sample size required to construct a

9090%

confidence interval for the population mean. Assume the population standard deviation is

0.250.25

inch.nequals=nothingm

(Round up to the nearest whole number as needed.)

44

points with

99 %99%

confidence assuming

sigma equals 11.7 question mark σ=11.7?

Suppose the doctor would be content with

90 %90%

confidence. How does the decrease in confidence affect the sample size required?A 99% confidence level requires

nothingm subjects.

(Round up to the nearest whole number as needed.)A beverage company uses a machine to fill one-liter bottles with water (see figure). Assume that the population of volumes is normally distributed.(a) The company wants to estimate the mean volume of water the machine is putting in the bottles within 1 milliliter. Determine the minimum sample size required to construct a

9090%

confidence interval for the population mean. Assume the population standard deviation is

55

milliliters.

(b) Repeat part (a) using an error tolerance of

22

milliliters. Which error tolerance requires a larger sample size? Explain.A machine cuts plastic into sheets that are

3030

feet

(360360

inches) long. Assume that the population of lengths is normally distributed. Complete parts (a) and (b).(a)The company wants to estimate the mean length the machine is cutting the plastic within

0.1250.125

inch. Determine the minimum sample size required to construct a

9090%

confidence interval for the population mean. Assume the population standard deviation is

0.250.25

inch.nequals=nothingm

(Round up to the nearest whole number as needed.)

I'm afraid that I can't answer your question about billing because:

1) I am not authorized by the website to discuss those types of issues, and

2) I don't have any information about your account, what you paid, the terms of the agreement, etc

This one:

*A doctor wants to estimate the HDL cholesterol of all 20- to 29-year-old females. How many subjects are needed to estimate the HDL cholesterol within 44 points with 99% confidence assumingσ=11.7?Suppose the doctor would be content with 90% confidence. How does the decrease in confidence affect the sample size required?*

*A 99% confidence level requires ____ subjects.*

was already answered. The answer was **4693.**

Customer reply replied 3 years ago

I will see u after a few minutgot u

For this one:

A beverage company uses a machine to fill one-liter bottles with water (see figure). Assume that the population of volumes is normally distributed.

(a) The company wants to estimate the mean volume of water the machine is putting in the bottles within 1 milliliter. Determine the minimum sample size required to construct a

90% confidence interval for the population mean. Assume the population standard deviation is

55 milliliters.

(b) Repeat part (a) using an error tolerance of 22 milliliters. Which error tolerance requires a larger sample size? Explain.

a) n = (z * s / E)^2 = (1.645 * 55 / 1)^2 = 8186 (rounded up to the next whole number)

b) n = (z * s / E)^2 = (1.645 * 55 / 22)^2 = 17 (rounded up to the next whole number)

The smaller error tolerance requires the larger sample.

And for this one:

A machine cuts plastic into sheets that are 30 feet (360 inches) long. Assume that the population of lengths is normally distributed. Complete parts (a) and (b).

(a) The company wants to estimate the mean length the machine is cutting the plastic within

0.125 inch. Determine the minimum sample size required to construct a 90%

confidence interval for the population mean. Assume the population standard deviation is

0.25 inch.

n = (z * s / E) = (1.645 * 0.25 / 0.125)^2 = **11** (rounded up to the next whole number)

Customer reply replied 3 years ago

are u still there

Yes

Customer reply replied 3 years ago

can you check

For the 99% confidence level: it say incorect

Required sample size = (z * s / E)^2 = (2.576 * 11.7 / 0.44)^2 = 4693 (when rounded up to the next whole number).

For the 99% confidence level: it say incorect

Required sample size = (z * s / E)^2 = (2.576 * 11.7 / 0.44)^2 = 4693 (when rounded up to the next whole number).

Part of the trouble there is how one should interpret "44 points".

Taking 44 as a whole number, the required sample size would be:

n = (2.576 * 11.7 / 44)^2 = 1 (when rounded up to the next highest whole number)

This seemed unreasonable, which is why I went with the previous answer, but try it and see.

Customer reply replied 3 years ago

Why I am not able to send you

I don't know. It seems to be working now though.

Customer reply replied 3 years ago

Do you get new questionsThis is from my phoneA machine cuts plastic into sheets that are 3030 feet (360360 inches) long. Assume that the population of lengths is normally distributed. Complete parts (a) and (b).

(a)

The company wants to estimate the mean length the machine is cutting the plastic within 0.1250.125 inch. Determine the minimum sample size required to construct a 9090% confidence interval for the population mean. Assume the population standard deviation is 0.250.25 inch.

nequals=

1111

(Round up to the nearest whole number as needed.)

(b)

Repeat part (a) using an error tolerance of 0.06250.0625 inch.

nequals=

nothing

(Round up to the nearest whole number as needed.)

(a)

The company wants to estimate the mean length the machine is cutting the plastic within 0.1250.125 inch. Determine the minimum sample size required to construct a 9090% confidence interval for the population mean. Assume the population standard deviation is 0.250.25 inch.

nequals=

1111

(Round up to the nearest whole number as needed.)

(b)

Repeat part (a) using an error tolerance of 0.06250.0625 inch.

nequals=

nothing

(Round up to the nearest whole number as needed.)

No. The only thing I received recently was the bit about the sample size answer being wrong. I didn't get anything else.

Customer reply replied 3 years ago

360, 90,0.125,11 must be like this

*A machine cuts plastic into sheets that are 30 feet (360 inches) long. Assume that the population of lengths is normally distributed. Complete parts (a) and (b).*

*(a) The company wants to estimate the mean length the machine is cutting the plastic within 0.125 inch. Determine the minimum sample size required to construct a 90% confidence interval for the population mean. Assume the population standard deviation is 0.25 inch.*

*n = 11 (Round up to the nearest whole number as needed.)*

*(b) Repeat part (a) using an error tolerance of 0.0625 inch.*

n = (z * s / E) = (1.645 * 0.25 / 0.0625)^2 = **44** (rounded up to the next whole number)

Customer reply replied 3 years ago

The tolerance Eequals=0.1250.125 inch requires a larger sample size. As error size increases, a larger sample must be taken to ensure the desired accuracy.

B.

The tolerance Eequals=0.06250.0625 inch requires a larger sample size. As error size decreases, a larger sample must be taken to ensure the desired accuracy.

C.

The tolerance Eequals=0.1250.125 inch requires a larger sample size. As error size decreases, a larger sample must be taken to ensure the desired accuracy.

D.

The tolerance Eequals=0.06250.0625 inch requires a larger sample size. As error size increases, a larger sample must be taken to ensure the desired accuracy

B.

The tolerance Eequals=0.06250.0625 inch requires a larger sample size. As error size decreases, a larger sample must be taken to ensure the desired accuracy.

C.

The tolerance Eequals=0.1250.125 inch requires a larger sample size. As error size decreases, a larger sample must be taken to ensure the desired accuracy.

D.

The tolerance Eequals=0.06250.0625 inch requires a larger sample size. As error size increases, a larger sample must be taken to ensure the desired accuracy

**B. The tolerance E =0.0625 inch requires a larger sample size. As error size decreases, a larger sample must be taken to ensure the desired accuracy.**

Customer reply replied 3 years ago

U got itA soccer ball manufacturer wants to estimate the mean circumference of soccer balls within 0.050.05 in.

(a) Determine the minimum sample size required to construct a 9090% confidence interval for the population mean. Assume the population standard deviation is 0.250.25 in.

(b) Repeat part (a) using a standard deviation of 0.400.40 in. Which standard deviation requires a larger sample size? Explain.

(a) Determine the minimum sample size required to construct a 9090% confidence interval for the population mean. Assume the population standard deviation is 0.250.25 in.

(b) Repeat part (a) using a standard deviation of 0.400.40 in. Which standard deviation requires a larger sample size? Explain.

a) n = (1.645 * 0.25 / 0.05)^2 = **68**

b) n = (1.645 * 0.40 / 0.05)^2 = **174**

Customer reply replied 3 years ago

A population standard deviation of

▼

in. requires a larger sample size. Due to the increased variability in the population, a

▼

sample size is needed to ensure the desired accuracy.

▼

in. requires a larger sample size. Due to the increased variability in the population, a

▼

sample size is needed to ensure the desired accuracy.

A population standard deviation of

**0.40**

in. requires a larger sample size. Due to the increased variability in the population, a

**larger**

sample size is needed to ensure the desired accuracy.

Customer reply replied 3 years ago

for the first 0.40 or 0.25 for the next small or large

**0.40** for the first

**large** for the second

Customer reply replied 3 years ago

6.1.64 Question Help

Use a technology tool to construct a 9090% confidence interval for the population mean. Interpret your answer.

The table below is a random sample of the closing stock prices for a company for a recent year.

18.4318.43

16.8716.87

16.8316.83

17.6817.68

15.5115.51

15.5315.53

18.0518.05

19.1519.15

19.8419.84

18.2618.26

18.6218.62

20.7320.73

20.6820.68

21.0321.03

21.7221.72

22.1422.14

21.9421.94

22.1722.17

22.8622.86

20.8220.82

20.7620.76

22.0922.09

21.4321.43

22.3822.38

22.8322.83

24.3224.32

17.9717.97

14.4414.44

19.1019.10

18.4318.43

20.8320.83

21.4421.44

21.9121.91

21.8421.84

The 9090% confidence interval for the population mean is left parenthesis nothing comma nothing right parenthesis

,

.

(Round to two decimal places as needed.)

Enter your answer in the edit fields and then click Check Answer.

1

part remaining

Clear All Check Answer

Use a technology tool to construct a 9090% confidence interval for the population mean. Interpret your answer.

The table below is a random sample of the closing stock prices for a company for a recent year.

18.4318.43

16.8716.87

16.8316.83

17.6817.68

15.5115.51

15.5315.53

18.0518.05

19.1519.15

19.8419.84

18.2618.26

18.6218.62

20.7320.73

20.6820.68

21.0321.03

21.7221.72

22.1422.14

21.9421.94

22.1722.17

22.8622.86

20.8220.82

20.7620.76

22.0922.09

21.4321.43

22.3822.38

22.8322.83

24.3224.32

17.9717.97

14.4414.44

19.1019.10

18.4318.43

20.8320.83

21.4421.44

21.9121.91

21.8421.84

The 9090% confidence interval for the population mean is left parenthesis nothing comma nothing right parenthesis

,

.

(Round to two decimal places as needed.)

Enter your answer in the edit fields and then click Check Answer.

1

part remaining

Clear All Check Answer

**(19.28, 20.64)**

Customer reply replied 3 years ago

With 9090% confidence, it can be said that the confidence interval contains the sample mean closing stock price.

B.

The confidence interval contains 9090% of the closing stock prices.

C.

With 9090% confidence, it can be said that the confidence interval contains the true mean closing stock price.

B.

The confidence interval contains 9090% of the closing stock prices.

C.

With 9090% confidence, it can be said that the confidence interval contains the true mean closing stock price.

**C. With 90% confidence, it can be said that the confidence interval contains the true mean closing stock price. **

Customer reply replied 3 years ago

Find the critical value tc for the confidence level cequals=0.9999 and sample size nequals=2525.

LOADING... Click the icon to view the t-distribution table.

tcequals=

nothing (Round to the nearest thousandth as needed.)

LOADING... Click the icon to view the t-distribution table.

tcequals=

nothing (Round to the nearest thousandth as needed.)

**tc = 2.797**

Customer reply replied 3 years ago

Find the critical value tc for the confidence level cequals=0.9090 and sample size nequals=2020.

LOADING... Click the icon to view the t-distribution table.

tcequals=

nothing (Round to the nearest thousandth as needed.)

LOADING... Click the icon to view the t-distribution table.

tcequals=

nothing (Round to the nearest thousandth as needed.)

**tc = 1.729**

Customer reply replied 3 years ago

Find the margin of error for the given values of c, s, and n.

cequals=0.9898, sequals=44, nequals=66

LOADING... Click the icon to view the t-distribution table.

The margin of error is

nothing. (Round to one decimal place as needed.)

cequals=0.9898, sequals=44, nequals=66

LOADING... Click the icon to view the t-distribution table.

The margin of error is

nothing. (Round to one decimal place as needed.)

**E = 5.5**

Customer reply replied 3 years ago

please show me some step, no need of explanation just calculation

ind the margin of error for the given values of c, s, and n.

cequals=0.9999, sequals=3.83.8, nequals=1414

LOADING... Click the icon to view the t-distribution table.

The margin of error is

nothing. (Round to one decimal place as needed.)

ind the margin of error for the given values of c, s, and n.

cequals=0.9999, sequals=3.83.8, nequals=1414

LOADING... Click the icon to view the t-distribution table.

The margin of error is

nothing. (Round to one decimal place as needed.)

For this you need to first calculate the number of degrees of freedom:

df = n - 1 = 14 - 1 = 13

Then, get the critical t value from tables or a statistical calculator. The critical value with c = 0.99 and 13 degrees of freedom is tc = 3.012.

The margin of error is then E = tc * s / √n = 3.012 * 3.8 / √14 = **3.1**

Customer reply replied 3 years ago

sorry say incorrect

Is the value of c 0.99 or 0.9999?

Customer reply replied 3 years ago

0.99

Does it give you any clues as to why? I can't find any error there.

Customer reply replied 3 years ago

E= tc times S divided by square root of nsorry you did right that was my faul

Ok, no worries.

Customer reply replied 3 years ago

Find the margin of error for the given values of c, s, and n.

cequals=0.8080, sequals=2.52.5, nequals=1616

LOADING... Click the icon to view the t-distribution tab

cequals=0.8080, sequals=2.52.5, nequals=1616

LOADING... Click the icon to view the t-distribution tab

For this one

df = n - 1 = 16 - 1 = 15

The critical t-value with c = 0.80 and 15 degrees of freedom is tc = 1.341

E = tc * s / √n = 1.341 * 2.5 / √16 = **0.8381** (round off as directed)

Customer reply replied 3 years ago

Construct the indicated confidence interval for the population mean muμ using (a) a t-distribution. (b) If you had incorrectly used a normal distribution, which interval would be wider?

cequals=0.950.95, x overbarxequals=13.813.8, sequals=4.04.0, nequals=66

(a) The 9595% confidence interval using a t-distribution is left parenthesis nothing comma nothing right parenthesis .

,

.

(Round to one decimal place as needed.)

cequals=0.950.95, x overbarxequals=13.813.8, sequals=4.04.0, nequals=66

(a) The 9595% confidence interval using a t-distribution is left parenthesis nothing comma nothing right parenthesis .

,

.

(Round to one decimal place as needed.)

(a) (9.6, 18.0)

(b) (10.6, 17.00)

The interval using the t-distribution is wider.

Customer reply replied 3 years ago

Construct the indicated confidence interval for the population mean muμ using a t-distribution.

cequals=0.9999, x overbarxequals=113113, sequals=1010, nequals=2222

The confidence interval is left parenthesis nothing comma nothing right parenthesis .

,

.

(Round to the nearest tenth as needed.)

cequals=0.9999, x overbarxequals=113113, sequals=1010, nequals=2222

The confidence interval is left parenthesis nothing comma nothing right parenthesis .

,

.

(Round to the nearest tenth as needed.)

(107.0, 119.0)

Customer reply replied 3 years ago

Use the given confidence interval to find the margin of error and the sample mean.

(13.313.3,22.922.9)

The sample mean is

nothing. (Type an integer or a decimal.

(13.313.3,22.922.9)

The sample mean is

nothing. (Type an integer or a decimal.

The sample mean is the midpoint of the interval:

mean = (13.3 + 22.9) / 2 = **18.1**

The margin of error is half of the width of the interval:

margin of error = (22.9 - 13.3) / 2 = **4.8**

Customer reply replied 3 years ago

do you round any number, say incorrectIn a random sample of sixsix microwave ovens, the mean repair cost was $65.0065.00 and the standard deviation was $13.5013.50. Assume the variable is normally distributed and use a t-distribution to construct a 9595% confidence interval for the population mean muμ. What is the margin of error of muμ?

The 9595% confidence interval for the population mean muμ is left parenthesis nothing comma nothing right parenthesis .

,

.

(Round to two decimal places as needed.)your answer was right

The 9595% confidence interval for the population mean muμ is left parenthesis nothing comma nothing right parenthesis .

,

.

(Round to two decimal places as needed.)your answer was right

For the "incorrect" one, which order did you enter the answers in? I think I gave them to you in the reverse order.

Customer reply replied 3 years ago

You did rightare you still with me

Yes. The site seems to be having a bit of an issue (or maybe it's just on my end). I attempted to post an answer, but it seems to have gotten lost somewhere. I'll post it again.

Customer reply replied 3 years ago

ok

Confidence interval: **(50.83, 79.17)**

The margin of error is **14.17**

Customer reply replied 3 years ago

n a random sample of 5454 bolts, the mean length was 1.491.49 inches and the standard deviation was 0.080.08 inch. Use a normal distribution or a t-distribution to construct a 8080% confidence interval for the mean.

Which distribution should be used to construct the 8080% confidence interval?

A.

Use a t-distribution because the lengths are normally distributed and sigmaσ is known.

B.

Use a t-distribution because ngreater than or equals≥30.

C.

Use a normal distribution because the lengths are normally distributed and sigmaσ is known.

D.

Use a normal distribution because ngreater than or equals≥30.

Which distribution should be used to construct the 8080% confidence interval?

A.

Use a t-distribution because the lengths are normally distributed and sigmaσ is known.

B.

Use a t-distribution because ngreater than or equals≥30.

C.

Use a normal distribution because the lengths are normally distributed and sigmaσ is known.

D.

Use a normal distribution because ngreater than or equals≥30.

**D. Use a normal distribution because n ≥ 30. **

Customer reply replied 3 years ago

he 8080% confidence interval is left parenthesis nothing comma nothing right parenthesis .

,

. (Round to two decimal places as needed.

,

. (Round to two decimal places as needed.

**(1.48, 1.50)**

Customer reply replied 3 years ago

6.2.29 Question Help

You take a random survey of 25 sports cars and record the miles per gallon for each. The data are listed to the right. Assume the miles per gallon are normally distributed. Use a normal distribution or a t-distribution to construct a 9090% confidence interval for the mean.

2222

2424

1313

1616

2323

2424

2020

1818

1818

2424

1616

2424

2323

1616

2323

2323

2626

2121

2727

1717

1919

2121

1313

2424

2727

Which distribution should be used to construct the 9090% confidence interval?

A.

Use a t-distribution because nless than<30, the miles per gallon are normally distributed and sigmaσ is unknown.

B.

Use a normal distribution because nless than<30, the miles per gallon are normally distributed and sigmaσ is unknown.

C.

Use a t-distribution because nless than<30 and sigmaσ is known.

D.

Use a normal distribution because nless than<30 and sigmaσ is known.

Click to select your answer and then click Check Answer.

1

part remaining

Clear All Check AnswerA doctor wants to estimate the HDL cholesterol of all 20- to 29-year-old females. How many subjects are needed to estimate the HDL cholesterol within 44 points with 99 %99% confidence assuming sigma equals 11.7 question mark σ=11.7? Suppose the doctor would be content with 90 %90% confidence. How does the decrease in confidence affect the sample size required?

A 99% confidence level requires

46924692 subjects.

(Round up to the nearest whole number as needed

You take a random survey of 25 sports cars and record the miles per gallon for each. The data are listed to the right. Assume the miles per gallon are normally distributed. Use a normal distribution or a t-distribution to construct a 9090% confidence interval for the mean.

2222

2424

1313

1616

2323

2424

2020

1818

1818

2424

1616

2424

2323

1616

2323

2323

2626

2121

2727

1717

1919

2121

1313

2424

2727

Which distribution should be used to construct the 9090% confidence interval?

A.

Use a t-distribution because nless than<30, the miles per gallon are normally distributed and sigmaσ is unknown.

B.

Use a normal distribution because nless than<30, the miles per gallon are normally distributed and sigmaσ is unknown.

C.

Use a t-distribution because nless than<30 and sigmaσ is known.

D.

Use a normal distribution because nless than<30 and sigmaσ is known.

Click to select your answer and then click Check Answer.

1

part remaining

Clear All Check AnswerA doctor wants to estimate the HDL cholesterol of all 20- to 29-year-old females. How many subjects are needed to estimate the HDL cholesterol within 44 points with 99 %99% confidence assuming sigma equals 11.7 question mark σ=11.7? Suppose the doctor would be content with 90 %90% confidence. How does the decrease in confidence affect the sample size required?

A 99% confidence level requires

46924692 subjects.

(Round up to the nearest whole number as needed

A. Use a t-distribution because n < 30, the miles per gallon are normally distributed and σ is unknown.

Customer reply replied 3 years ago

The 9090% confidence interval is left parenthesis nothing comma nothing right parenthesis .

,

. (Round to one decimal place as needed.)

,

. (Round to one decimal place as needed.)

**(19.5, 22.3)**

Customer reply replied 3 years ago

Let p be the population proportion for the following condition. Find the point estimates for p and q.

In a survey of 12811281 adults from country A, 299299 said that they were not confident that the food they eat in country A is safe.

The point estimate for p, ModifyingAbove p with caretp, is

nothing.

(Round to three decimal places as needed.)

In a survey of 12811281 adults from country A, 299299 said that they were not confident that the food they eat in country A is safe.

The point estimate for p, ModifyingAbove p with caretp, is

nothing.

(Round to three decimal places as needed.)

*A doctor wants to estimate the HDL cholesterol of all 20- to 29-year-old females. How many subjects are needed to estimate the HDL cholesterol within 44 points with 99 %99% confidence assuming sigma equals 11.7 question mark σ=11.7? Suppose the doctor would be content with 90 %90% confidence. How does the decrease in confidence affect the sample size required?A 99% confidence level requires46924692 subjects.(Round up to the nearest whole number as needed*

For the 90% confidence interval, the required sample size is:

n = (1.645 * 11.7 / 0.44)^2 = **1914**

The decrease in confidence leads to a decrease in the required sample size.

p = 299 / 1281 = **0.233**

q = 1 - p = 1 - 0.233 =** 0.767**

Customer reply replied 3 years ago

Let p be the population proportion for the following condition. Find the point estimates for p and q.

Of 808808 children surveyed, 118118 plan to join the armed forces in the future.

The point estimate for p, ModifyingAbove p with caretp, is

nothing.

(Round to three decimal places as needed

Of 808808 children surveyed, 118118 plan to join the armed forces in the future.

The point estimate for p, ModifyingAbove p with caretp, is

nothing.

(Round to three decimal places as needed

p = 118 / 808 = **0.146**

q = 1 - p = 1 - 0.146 =** 0.854**

Customer reply replied 3 years ago

Use the given confidence interval to find the margin of error and the sample proportion.

(0.7530.753,0.7790.779)

Eequals=

nothing (Type an integer or a decim

(0.7530.753,0.7790.779)

Eequals=

nothing (Type an integer or a decim

The margin of error is (0.779 - 0.753)/2 = **0.013**

The sample proportion is (0.753 + 0.779)/2 = **0.776**

Customer reply replied 3 years ago

0.776 incorrect

I'm sorry. That was either a typo or a brain hiccup. It should be 0.7__ 6__6 instead.

Customer reply replied 3 years ago

you go itIn a survey of 90009000 women, 64316431 say they change their nail polish once a week. Construct a 99 %99% confidence interval for the population proportion of women who change their nail polish once a week.

A 9999% confidence interval for the population proportion is (

nothing,

nothing).

(Round to three decimal places as needed.)

A 9999% confidence interval for the population proportion is (

nothing,

nothing).

(Round to three decimal places as needed.)

**(0.702, 0.727)**

Customer reply replied 3 years ago

A researcher wishes to estimate, with 9090% confidence, the proportion of adults who have high-speed Internet access. Her estimate must be accurate within 11% of the true proportion.

a) Find the minimum sample size needed, using a prior study that found that 3434% of the respondents said they have high-speed Internet access.

b) No preliminary estimate is available. Find the minimum sample size needed.

a) What is the minimum sample size needed using a prior study that found that 3434% of the respondents said they have high-speed Internet access?

nequals=

nothing (Round up to the nearest whole number as needed.)

a) Find the minimum sample size needed, using a prior study that found that 3434% of the respondents said they have high-speed Internet access.

b) No preliminary estimate is available. Find the minimum sample size needed.

a) What is the minimum sample size needed using a prior study that found that 3434% of the respondents said they have high-speed Internet access?

nequals=

nothing (Round up to the nearest whole number as needed.)

a) 51

b) 56

Customer reply replied 3 years ago

a in correctA researcher wishes to estimate, with 9090% confidence, the proportion of adults who have high-speed Internet access. Her estimate must be accurate within 11% of the true proportion.

a) Find the minimum sample size needed, using a prior study that found that 3434% of the respondents said they have high-speed Internet access.

b) No preliminary estimate is available. Find the minimum sample size needed.

a) What is the minimum sample size needed using a prior study that found that 3434% of the respondents said they have high-speed Internet access?

nequals=

5151 (Round up to the nearest whole number as needed.)

a) Find the minimum sample size needed, using a prior study that found that 3434% of the respondents said they have high-speed Internet access.

b) No preliminary estimate is available. Find the minimum sample size needed.

a) What is the minimum sample size needed using a prior study that found that 3434% of the respondents said they have high-speed Internet access?

nequals=

5151 (Round up to the nearest whole number as needed.)

Oops, I missed correcting the "11%" in your post to "1%". Most of the numbers are showing up doubled on my end, as in "3434%" instead of "34%".

The corrected answers are:

a) 6073

b) 6766

Customer reply replied 3 years ago

A beverage company uses a machine to fill one-liter bottles with water (see figure). Assume that the population of volumes is normally distributed.

(a) The company wants to estimate the mean volume of water the machine is putting in the bottles within 1 milliliter. Determine the minimum sample size required to construct a 9090% confidence interval for the population mean. Assume the population standard deviation is 55 milliliters.

(b) Repeat part (a) using an error tolerance of 22 milliliters. Which error tolerance requires a larger sample size? Explain.

(a) The minimum sample size required for an error tolerance of 1 milliliter is

81868186 bottles.

(Round up to the nearest integer.)Find the margin of error for the given values of c, s, and n.

cequals=0.950.95, sequals=2.32.3, nequals=3636

Eequals=

nothing

(Round to three decimal places as needed

(a) The company wants to estimate the mean volume of water the machine is putting in the bottles within 1 milliliter. Determine the minimum sample size required to construct a 9090% confidence interval for the population mean. Assume the population standard deviation is 55 milliliters.

(b) Repeat part (a) using an error tolerance of 22 milliliters. Which error tolerance requires a larger sample size? Explain.

(a) The minimum sample size required for an error tolerance of 1 milliliter is

81868186 bottles.

(Round up to the nearest integer.)Find the margin of error for the given values of c, s, and n.

cequals=0.950.95, sequals=2.32.3, nequals=3636

Eequals=

nothing

(Round to three decimal places as needed

b) required sample size for an error tolerance of 22 milliliters =** 17 bottles**

E = **0.751**

Customer reply replied 3 years ago

0.751 wrong answer

Margin of error = z * s / √n

For a c = 0.95, the value of z is 1.96.

Then E = 1.96 * 2.3 / √36 = 0.751

Customer reply replied 3 years ago

6.1.17 Question Help

Match the level of confidence, cequals=0.950.95, with its representation on the number line, given x overbar equals 56.7x=56.7, sequals=8.88.8, and nequals=9090.

Choose the correct number line below.

A.

53

54

55

56

57

58

59

60

61

54.2

59.2

x overbar equals 56.7x=56.7

B.

53

54

55

56

57

58

59

60

61

54.9

58.5

x overbar equals 56.7x=56.7

C.

53

54

55

56

57

58

59

60

61

54.5

58.9

x overbar equals 56.7x=56.7

D.

53

54

55

56

57

58

59

60

61

55.2

58.2

x overbar equals 56.7x=56.7

Click to select your answer and then click Check Answer.

All parts showing

Clear All Check Answerforget about this question

Match the level of confidence, cequals=0.950.95, with its representation on the number line, given x overbar equals 56.7x=56.7, sequals=8.88.8, and nequals=9090.

Choose the correct number line below.

A.

53

54

55

56

57

58

59

60

61

54.2

59.2

x overbar equals 56.7x=56.7

B.

53

54

55

56

57

58

59

60

61

54.9

58.5

x overbar equals 56.7x=56.7

C.

53

54

55

56

57

58

59

60

61

54.5

58.9

x overbar equals 56.7x=56.7

D.

53

54

55

56

57

58

59

60

61

55.2

58.2

x overbar equals 56.7x=56.7

Click to select your answer and then click Check Answer.

All parts showing

Clear All Check Answerforget about this question

ok

Customer reply replied 3 years ago

A doctor wants to estimate the HDL cholesterol of all 20- to 29-year-old females. How many subjects are needed to estimate the HDL cholesterol within 44 points with 99 %99% confidence assuming sigma equals 11.7 question mark σ=11.7? Suppose the doctor would be content with 90 %90% confidence. How does the decrease in confidence affect the sample size required?

A 99% confidence level requires

5757 subjects.

(Round up to the nearest whole number as needed.)

A 90 %90% confidence level requires

nothing subjects.

(Round up to the nearest whole number as neede

A 99% confidence level requires

5757 subjects.

(Round up to the nearest whole number as needed.)

A 90 %90% confidence level requires

nothing subjects.

(Round up to the nearest whole number as neede

Required sample size = **24**

Customer reply replied 3 years ago

for which questin 24 will be an answer

*A doctor wants to estimate the HDL cholesterol of all 20- to 29-year-old females. How many subjects are needed to estimate the HDL cholesterol within 4 points with 99% confidence assuming σ=11.7? Suppose the doctor would be content with 90% confidence. How does the decrease in confidence affect the sample size required?A 99% confidence level requires 57 subjects.(Round up to the nearest whole number as needed.)A 90% confidence level requiresnothing subjects.(Round up to the nearest whole number as needed)*

Required sample size = **24**

Customer reply replied 3 years ago

Use the confidence interval to find the estimated margin of error. Then find the sample mean.A biologist reports a confidence interval of

left parenthesis 2.1 comma 2.9 right parenthesis(2.1,2.9)

when estimating the mean height (in centimeters) of a sample of seedlings.The estimated margin of error is

nothingm.Use the confidence interval to find the margin of error and the sample mean.left parenthesis 0.426 comma 0.550 right parenthesis(0.426,0.550)The margin of error is

nothingFind the minimum sample size n needed to estimate

muμ

for the given values of c, s, and E.cequals=0.980.98,sequals=8.18.1,

and

Eequals=22Assume that a preliminary sample has at least 30 members.nequals=nothingm (Round up to theA beverage company uses a machine to fill one-liter bottles with water (see figure). Assume that the population of volumes is normally distributed.(a) The company wants to estimate the mean volume of water the machine is putting in the bottles within 1 milliliter. Determine the minimum sample size required to construct a

9090%

confidence interval for the population mean. Assume the population standard deviation is

55

milliliters.

(b) Repeat part (a) using an error tolerance of

22

milliliters. Which error tolerance requires a larger sample size? Explain.(a) The minimum sample size required for an error tolerance of 1 milliliter is

6868 bottles.

(Round up to the nearest integer.)(b) The minimum sample size required for an error tolerance of

22

milliliters is

nothingm bottles.

(Round up to the nearest integer.)Construct the confidence interval for the population mean

muμ.cequals=0.950.95,x overbar equals 16.6x=16.6,sequals=7.07.0,

and

nequals=5050A

9595%

confidence interval for

muμ

is

left parenthesis nothing comma nothing right parenthesis .m,m. (Round to one decimal place as needed.)

left parenthesis 2.1 comma 2.9 right parenthesis(2.1,2.9)

when estimating the mean height (in centimeters) of a sample of seedlings.The estimated margin of error is

nothingm.Use the confidence interval to find the margin of error and the sample mean.left parenthesis 0.426 comma 0.550 right parenthesis(0.426,0.550)The margin of error is

nothingFind the minimum sample size n needed to estimate

muμ

for the given values of c, s, and E.cequals=0.980.98,sequals=8.18.1,

and

Eequals=22Assume that a preliminary sample has at least 30 members.nequals=nothingm (Round up to theA beverage company uses a machine to fill one-liter bottles with water (see figure). Assume that the population of volumes is normally distributed.(a) The company wants to estimate the mean volume of water the machine is putting in the bottles within 1 milliliter. Determine the minimum sample size required to construct a

9090%

confidence interval for the population mean. Assume the population standard deviation is

55

milliliters.

(b) Repeat part (a) using an error tolerance of

22

milliliters. Which error tolerance requires a larger sample size? Explain.(a) The minimum sample size required for an error tolerance of 1 milliliter is

6868 bottles.

(Round up to the nearest integer.)(b) The minimum sample size required for an error tolerance of

22

milliliters is

nothingm bottles.

(Round up to the nearest integer.)Construct the confidence interval for the population mean

muμ.cequals=0.950.95,x overbar equals 16.6x=16.6,sequals=7.07.0,

and

nequals=5050A

9595%

confidence interval for

muμ

is

left parenthesis nothing comma nothing right parenthesis .m,m. (Round to one decimal place as needed.)

Hi,

I'm sorry about the delay in replying. I am away from my desk at the moment. I will send the solutions for these as soon as I can.

Ryan

Customer reply replied 3 years ago

A doctor wants to estimate the HDL cholesterol of all 20- to 29-year-old females. How many subjects are needed to estimate the HDL cholesterol within 44 points with 99 %99% confidence assuming sigma equals 11.7 question mark σ=11.7? Suppose the doctor would be content with 90 %90% confidence. How does the decrease in confidence affect the sample size required?

A 99% confidence level requires

5757 subjects.

(Round up to the nearest whole number as needed.)

A 90 %90% confidence level requires

nothing subjects.

(Round up to the nearest whole number as needeI really need assist when you will et back on line

A 99% confidence level requires

5757 subjects.

(Round up to the nearest whole number as needed.)

A 90 %90% confidence level requires

nothing subjects.

(Round up to the nearest whole number as needeI really need assist when you will et back on line

Probably in three or four hours or so.

Customer reply replied 3 years ago

I dont have many question like yesterday, that will not take you more than 1 hr. but the problem is the due is today

I'll see if I can rearrange my schedule a bit.

Customer reply replied 3 years ago

pleaseI don't have may question you will spend less than hour

For this one:

*Use the confidence interval to find the estimated margin of error. Then find the sample mean.*

*A biologist reports a confidence interval of left parenthesis (2.1,2.9) when estimating the mean height (in centimeters) of a sample of seedlings.*

The estimated margin of error is (2.9 - 2.1)/2 = **0.40**

The sample mean is (2.1 + 2.9)/2 = **2.50**

For this one:

*Use the confidence interval to find the margin of error and the sample mean.*

*(0.426,0.550)*

The margin of error is (0.550 - 0.426)/2 = **0.062**

The sample mean is (0.426 + 0.550)/2 = **0.488**

Customer reply replied 3 years ago

nice to see you againI am not able to much your answer with my qustion

Those were the first two that you posted this morning.

Would it be easier for you if we just started fresh and have you just post the ones that you currently need help with?

Customer reply replied 3 years ago

ok that sounds good

Ok, ready when you are.

Customer reply replied 3 years ago

ind the critical value z Subscript czc necessary to form a confidence interval at the level of confidence shown below.

cequals=0.860.86

z Subscript czcequals=

nothing

(Round to two decimal places as needed.)please show me stepare you with Me

cequals=0.860.86

z Subscript czcequals=

nothing

(Round to two decimal places as needed.)please show me stepare you with Me

To determine the critical value, either look up the value closest to 0.86 in a table of the normal (z) distribution, or use a statistical calculator.

If you have Excel, you can enter "=normsinv(x)" into a cell to get the z-value that has an area to the left of z equal to x.

Since the confidence interval must include 86% of the distribution (c = 0.86), each tail will include (1 - 0.86)/2 = 0.07. This must be added to the value 0.86 to get the total area to the left of the desired value of z. 0.86 + 0.07 = 0.093

Thus, enter "=normsinv(0.93)" into Excel, without the quotation marks.

zc =** 1.48**

Oops, typo. 0.86 + 0.07 = 0.93, not 0.093 as written in the previous post.

Customer reply replied 3 years ago

6.1.17 Question Help

Match the level of confidence, cequals=0.950.95, with its representation on the number line, given x overbar equals 56.7x=56.7, sequals=8.88.8, and nequals=9090.

Choose the correct number line below.

A.

53

54

55

56

57

58

59

60

61

54.2

59.2

x overbar equals 56.7x=56.7

B.

53

54

55

56

57

58

59

60

61

54.9

58.5

x overbar equals 56.7x=56.7

C.

53

54

55

56

57

58

59

60

61

54.5

58.9

x overbar equals 56.7x=56.7

D.

53

54

55

56

57

58

59

60

61

55.2

58.2

x overbar equals 56.7x=56.7

Click to select your answer and then click Check Answer.

All parts showing

Clear All Check AnswerThe pattern is changed when I post it you may dont uderstand what the question is. It was graphical

Match the level of confidence, cequals=0.950.95, with its representation on the number line, given x overbar equals 56.7x=56.7, sequals=8.88.8, and nequals=9090.

Choose the correct number line below.

A.

53

54

55

56

57

58

59

60

61

54.2

59.2

x overbar equals 56.7x=56.7

B.

53

54

55

56

57

58

59

60

61

54.9

58.5

x overbar equals 56.7x=56.7

C.

53

54

55

56

57

58

59

60

61

54.5

58.9

x overbar equals 56.7x=56.7

D.

53

54

55

56

57

58

59

60

61

55.2

58.2

x overbar equals 56.7x=56.7

Click to select your answer and then click Check Answer.

All parts showing

Clear All Check AnswerThe pattern is changed when I post it you may dont uderstand what the question is. It was graphical

The confidence interval is (54.88, 58.52), so it appears that choice B is what they are looking for.

Customer reply replied 3 years ago

Homework: Homework 3 (Read 5.1 to 5.4 and 6.1 to 6.3) Save

Score: 0 of 1 pt 49 of 81 (71 complete)

HW Score: 84.25%, 84.25 of 100 ptssorry my mitack ignor itConstruct the confidence interval for the population mean muμ.

cequals=0.980.98, x overbar equals 4.9x=4.9, sequals=0.30.3, and nequals=4848

A 9898% confidence interval for muμ is left parenthesis nothing comma nothing right parenthesis .

,

. (Round to two decimal places as needed.clease show me calculationConstruct the confidence interval for the population mean muμ.

cequals=0.980.98, x overbar equals 4.9x=4.9, sequals=0.30.3, and nequals=4848

A 9898% confidence interval for muμ is left parenthesis nothing comma nothing right parenthesis .

,

. (Round to two decimal places as needed.)

Score: 0 of 1 pt 49 of 81 (71 complete)

HW Score: 84.25%, 84.25 of 100 ptssorry my mitack ignor itConstruct the confidence interval for the population mean muμ.

cequals=0.980.98, x overbar equals 4.9x=4.9, sequals=0.30.3, and nequals=4848

A 9898% confidence interval for muμ is left parenthesis nothing comma nothing right parenthesis .

,

. (Round to two decimal places as needed.clease show me calculationConstruct the confidence interval for the population mean muμ.

cequals=0.980.98, x overbar equals 4.9x=4.9, sequals=0.30.3, and nequals=4848

A 9898% confidence interval for muμ is left parenthesis nothing comma nothing right parenthesis .

,

. (Round to two decimal places as needed.)

The z-value for a 98% confidence interval is z = 2.326.

The margin of error for the confidence interval is then E = (z * s / √n) = (2.326 * 0.3 / √48) = 0.1007

Lower limit = mean - E = 4.9 - 0.1007 = 4.80

Upper limit = mean + E = 4.9 + 0.1007 = 5.00

The 98% confidence interval is then **(4.80, 5.00)**

Customer reply replied 3 years ago

Construct the confidence interval for the population mean muμ.

cequals=0.950.95, x overbar equals 16.6x=16.6, sequals=7.07.0, and nequals=5050

A 9595% confidence interval for muμ is left parenthesis nothing comma nothing right parenthesis .

,

. (Round to one decimal place as needed.)

cequals=0.950.95, x overbar equals 16.6x=16.6, sequals=7.07.0, and nequals=5050

A 9595% confidence interval for muμ is left parenthesis nothing comma nothing right parenthesis .

,

. (Round to one decimal place as needed.)

Same process as the last one. Here, the z-value is 1.96.

The resulting confidence interval is **(14.7, 18.5).**

Customer reply replied 3 years ago

se the confidence interval to find the estimated margin of error. Then find the sample mean.

A biologist reports a confidence interval of left parenthesis 2.7 comma 4.5 right parenthesis(2.7,4.5) when estimating the mean height (in centimeters) of a sample of seedlings.

The estimated margin of error is

nothing.

A biologist reports a confidence interval of left parenthesis 2.7 comma 4.5 right parenthesis(2.7,4.5) when estimating the mean height (in centimeters) of a sample of seedlings.

The estimated margin of error is

nothing.

Margin of error = (4.5 - 2.7)/2 = **0.90**

Sample mean = (2.7 + 4.5)/2 = **3.60**

Customer reply replied 3 years ago

Use the confidence interval to find the margin of error and the sample mean.

left parenthesis 0.426 comma 0.550 right parenthesis(0.426,0.550)

The margin of error isPlease show me calculation

left parenthesis 0.426 comma 0.550 right parenthesis(0.426,0.550)

The margin of error isPlease show me calculation

Margin of error = (0.550 - 0.426)/2 = **0.062**

Sample mean = (0.426 + 0.550)/2 = **0.488**

Customer reply replied 3 years ago

Find the minimum sample size n needed to estimate muμ for the given values of c, s, and E.

cequals=0.900.90, sequals=8.78.7, and Eequals=22

Assume that a preliminary sample has at least 30 members.

nequals=

nothing (Round up to the nearest whole number.)show me calculation pleaseshw me calculation plsshow me clculation pls

cequals=0.900.90, sequals=8.78.7, and Eequals=22

Assume that a preliminary sample has at least 30 members.

nequals=

nothing (Round up to the nearest whole number.)show me calculation pleaseshw me calculation plsshow me clculation pls

For a 90% confidence level, use z = 1.645.

Then, n = (z * s / E)^2 = (1.645 * 8.7 / 2)^2 =** 52 **(rounded up to next whole number)

Customer reply replied 3 years ago

ind the minimum sample size n needed to estimate muμ for the given values of c, s, and E.

cequals=0.980.98, sequals=8.18.1, and Eequals=22

Assume that a preliminary sample has at least 30 members.

nequals=

nothing (Round up to the nearest whole number.)show me calculation pleae

cequals=0.980.98, sequals=8.18.1, and Eequals=22

Assume that a preliminary sample has at least 30 members.

nequals=

nothing (Round up to the nearest whole number.)show me calculation pleae

For c = 0.98, use z = 2.326.

n = (z * s / E)^2 = (2.326 * 8.1 / 2)^2 = 88.7 -> **89**

Customer reply replied 3 years ago

Use the confidence interval to find the estimated margin of error. Then find the sample mean.

A biologist reports a confidence interval of left parenthesis 2.1 comma 2.9 right parenthesis(2.1,2.9) when estimating the mean height (in centimeters) of a sample of seedlings.

The estimated margin of error is

nothing.calc as well please

A biologist reports a confidence interval of left parenthesis 2.1 comma 2.9 right parenthesis(2.1,2.9) when estimating the mean height (in centimeters) of a sample of seedlings.

The estimated margin of error is

nothing.calc as well please

Estimate margin of error = (2.9 - 2.1)/2 = **0.40**

Sample mean = (2.1 + 2.9)/2 = **2.50**

Customer reply replied 3 years ago

You are given the sample mean and the sample standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals.

A random sample of 6060 home theater systems has a mean price of $119.00119.00 and a standard deviation is $15.3015.30.

Construct a 90% confidence interval for the population mean.

The 90% confidence interval is left parenthesis nothing comma nothing right parenthesis .

,

. You are given the sample mean and the sample standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals.

A random sample of 6060 home theater systems has a mean price of $119.00119.00 and a standard deviation is $15.3015.30.

Construct a 90% confidence interval for the population mean.

The 90% confidence interval is left parenthesis nothing comma nothing right parenthesis .

,

.

A random sample of 6060 home theater systems has a mean price of $119.00119.00 and a standard deviation is $15.3015.30.

Construct a 90% confidence interval for the population mean.

The 90% confidence interval is left parenthesis nothing comma nothing right parenthesis .

,

. You are given the sample mean and the sample standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals.

A random sample of 6060 home theater systems has a mean price of $119.00119.00 and a standard deviation is $15.3015.30.

Construct a 90% confidence interval for the population mean.

The 90% confidence interval is left parenthesis nothing comma nothing right parenthesis .

,

.

n = 60, x-bar = 119.00, s = 15.30, z = 1.645 for a 90% confidence interval

Margin of error = E = (z * s / √n) = (1.645 * 15.30 / √60) = 3.2492

Lower limit = x-bar - E = 119.00 - 3.2492 = **115.75**

Upper limit = x-bar + E = 119.00 + 3.2492 = **122.25**

The 90% confidence interval is **(115.75, 122.25).**

Customer reply replied 3 years ago

You are given the sample mean and the sample standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals.

A random sample of 6060 home theater systems has a mean price of $119.00119.00 and a standard deviation is $15.3015.30.

Construct a 90% confidence interval for the population mean.

The 90% confidence interval is left parenthesis nothing comma nothing right parenthesis .

115.75115.75,

122.25122.25.

(Round to two decimal places as needed.)

Construct a 95% confidence interval for the population mean.

The 95% confidence interval is left parenthesis nothing comma nothing right parenthesis .

,

.

(Round to two decimal places as needed

A random sample of 6060 home theater systems has a mean price of $119.00119.00 and a standard deviation is $15.3015.30.

Construct a 90% confidence interval for the population mean.

The 90% confidence interval is left parenthesis nothing comma nothing right parenthesis .

115.75115.75,

122.25122.25.

(Round to two decimal places as needed.)

Construct a 95% confidence interval for the population mean.

The 95% confidence interval is left parenthesis nothing comma nothing right parenthesis .

,

.

(Round to two decimal places as needed

For the 95% confidence interval, do the same thing, except with z = 1.96.

Margin of error = E = (z * s / √n) = (1.96 * 15.30 / √60) = 3.8714

Lower limit = x-bar - E = 119.00 - 3.8714 = **115.13**

Upper limit = x-bar + E = 119.00 + 3.8714 = **122.87**

The 90% confidence interval is **(115.13, 122.87).**

Customer reply replied 3 years ago

nterpret the results. Choose the correct answer below.

A.

With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the 90%.

B.

With 90% confidence, it can be said that the sample mean price lies in the first interval. With 95% confidence, it can be said that the sample mean price lies in the second interval. The 95% confidence interval is wider than the 90%.

C.

With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is narrower than the 90%.

A.

With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the 90%.

B.

With 90% confidence, it can be said that the sample mean price lies in the first interval. With 95% confidence, it can be said that the sample mean price lies in the second interval. The 95% confidence interval is wider than the 90%.

C.

With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is narrower than the 90%.

A. With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the 90%.

Customer reply replied 3 years ago

In a survey of 10201020 adults, 803803 disapprove of the job the legislature is doing.

Use technology to construct 90%, 95%, and 99% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals.

A 90% confidence interval for the population proportion is left parenthesis nothing comma nothing right parenthesis

,

.In a survey of 10201020 adults, 803803 disapprove of the job the legislature is doing.

Use technology to construct 90%, 95%, and 99% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals.

A 90% confidence interval for the population proportion is left parenthesis nothing comma nothing right parenthesis

,

.

Use technology to construct 90%, 95%, and 99% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals.

A 90% confidence interval for the population proportion is left parenthesis nothing comma nothing right parenthesis

,

.In a survey of 10201020 adults, 803803 disapprove of the job the legislature is doing.

Use technology to construct 90%, 95%, and 99% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals.

A 90% confidence interval for the population proportion is left parenthesis nothing comma nothing right parenthesis

,

.

90% interval: **(0.7662, 0.8083)**

95% interval:** (0.7621, 0.8124)**

99% interval: **(0.7542, 0.8203)**

Round off values as needed.

Customer reply replied 3 years ago

ca you try for 90% again it is nicorrect

Does it say how many digits to round off to?

Customer reply replied 3 years ago

sorry it was rounding IssueI will have other optionIn a survey of 10271027 adults, 805805 disapprove of the job the legislature is doing.

Use technology to construct 90%, 95%, and 99% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals.

A 90% confidence interval for the population proportion is left parenthesis nothing comma nothing right parenthesis

,

.In a survey of 10271027 adults, 805805 disapprove of the job the legislature is doing.

Use technology to construct 90%, 95%, and 99% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals.

A 90% confidence interval for the population proportion is left parenthesis nothing comma nothing right parenthesis

,

.In a survey of 10271027 adults, 805805 disapprove of the job the legislature is doing.

Use technology to construct 90%, 95%, and 99% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals.

A 90% confidence interval for the population proportion is left parenthesis nothing comma nothing right parenthesis

,

.

Use technology to construct 90%, 95%, and 99% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals.

A 90% confidence interval for the population proportion is left parenthesis nothing comma nothing right parenthesis

,

.In a survey of 10271027 adults, 805805 disapprove of the job the legislature is doing.

Use technology to construct 90%, 95%, and 99% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals.

A 90% confidence interval for the population proportion is left parenthesis nothing comma nothing right parenthesis

,

.In a survey of 10271027 adults, 805805 disapprove of the job the legislature is doing.

Use technology to construct 90%, 95%, and 99% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals.

A 90% confidence interval for the population proportion is left parenthesis nothing comma nothing right parenthesis

,

.

How do they want you to round this?

Customer reply replied 3 years ago

that was my fault . it should be 3 digit

Ok...

90% interval = **(0.763, 0.805)**

95% interval = **(0.759, 0.809)**

99% interval = **(0.751, 0.817)**

Customer reply replied 3 years ago

With 90% confidence it can be said that the population proportion of adults who disapprove of the job legislature is doing is between

nothing and

nothing.

B.

With 90% confidence it can be said that the population proportion of adults who disapprove of the job legislature is doing is not between

nothing and

nothing

nothing and

nothing.

B.

With 90% confidence it can be said that the population proportion of adults who disapprove of the job legislature is doing is not between

nothing and

nothing

between **0.763 and 0.805**

Customer reply replied 3 years ago

that is for A right

yes

Customer reply replied 3 years ago

A 95% confidence interval for the population proportion is left parenthesis nothing comma nothing right parenthesis

,

.

(Round to three decimal places as needed.)

,

.

(Round to three decimal places as needed.)

95% interval = **(0.759, 0.809)**

Customer reply replied 3 years ago

Round to three decimal places as needed.)

A.

With 95% confidence it can be said that the population proportion of adults who disapprove of the job legislature is doing is between

nothing and

nothing.

B.

With 95% confidence it can be said that the population proportion of adults who disapprove of the job legislature is doing is not between

nothing and

nothing.

A.

With 95% confidence it can be said that the population proportion of adults who disapprove of the job legislature is doing is between

nothing and

nothing.

B.

With 95% confidence it can be said that the population proportion of adults who disapprove of the job legislature is doing is not between

nothing and

nothing.

A. between 0.759 and 0.809

Customer reply replied 3 years ago

With 99% confidence it can be said that the population proportion of adults who disapprove of the job legislature is doing is not between

0.7630.763 and

0.8050.805.

B.

With 99% confidence it can be said that the population proportion of adults who disapprove of the job legislature is doing is between

nothing and

nothing

0.7630.763 and

0.8050.805.

B.

With 99% confidence it can be said that the population proportion of adults who disapprove of the job legislature is doing is between

nothing and

nothing

**B.** between **0.751 and 0.817**

Customer reply replied 3 years ago

Which interval is widest?

A.

The 95% confidence interval

B.

The 99% confidence interval

C.

The 90% confidence interval

A.

The 95% confidence interval

B.

The 99% confidence interval

C.

The 90% confidence interval

B. The 99% confidence interval

Customer reply replied 3 years ago

A beverage company uses a machine to fill one-liter bottles with water (see figure). Assume that the population of volumes is normally distributed.

(a) The company wants to estimate the mean volume of water the machine is putting in the bottles within 1 milliliter. Determine the minimum sample size required to construct a 9090% confidence interval for the population mean. Assume the population standard deviation is 55 milliliters.

(b) Repeat part (a) using an error tolerance of 22 milliliters. Which error tolerance requires a larger sample size? Explain.

(a) The company wants to estimate the mean volume of water the machine is putting in the bottles within 1 milliliter. Determine the minimum sample size required to construct a 9090% confidence interval for the population mean. Assume the population standard deviation is 55 milliliters.

(b) Repeat part (a) using an error tolerance of 22 milliliters. Which error tolerance requires a larger sample size? Explain.

Is the standard deviation 5, or 55?

Also, in part b, is the tolerance 2 milliliters, or 22 milliliters?

Customer reply replied 3 years ago

5

2

2

a) n = (1.645 * 5 / 1)^2 =** 68**

b) n = (1.645 * 5 / 2)^2 = **17**

Customer reply replied 3 years ago

I Think 68 is incorrecr

I don't know what else to say there. This problem is a repeat, and the issue with the previous answers was due to whether the values were 5 and 2 or 55 and 22. This doubling of digits is a bit of a pain to deal with.

The calculation itself is pretty simple. Are you getting some kind of an error message?

Customer reply replied 3 years ago

I think let me give you other similar question8203;(b) The minimum sample size required for an error tolerance of 22 milliliters is

1717 bottles.

(Round up to the nearest integer.)

Which error tolerance requires a larger sample size? Explain.

The error tolerance of

▼

2 milliliters2 milliliters

1 milliliter1 milliliter

requires a larger sample size. As the error size decreases, a

▼

larger

smaller

sample must be taken to obtain sufficient information from the population to ensure the desired accuracy.

1717 bottles.

(Round up to the nearest integer.)

Which error tolerance requires a larger sample size? Explain.

The error tolerance of

▼

2 milliliters2 milliliters

1 milliliter1 milliliter

requires a larger sample size. As the error size decreases, a

▼

larger

smaller

sample must be taken to obtain sufficient information from the population to ensure the desired accuracy.

The error tolerance of __ 1 milliliter __requires a larger sample size. As the error size decreases, a

Customer reply replied 3 years ago

error =2 million is 17 botele

yes, that's correct

Customer reply replied 3 years ago

Use the normal distribution of SAT critical reading scores for which the mean is 507507 and the standard deviation is 120120. Assume the variable x is normally distributed.

left parenthesis a right parenthesis(a)

What percent of the SAT verbal scores are less than 550550?

left parenthesis b right parenthesis(b)

If 1000 SAT verbal scores are randomly selected, about how many would you expect to be greater than 525525?

left parenthesis a right parenthesis(a) Approximately

nothing% of the SAT verbal scores are less than 550550.

(Round to two decimal places as needed.)mean =507

standard divination=120

SAT verbal score are less than 550show me step leasewhen I copy and past numbers are repeated ;lease ignore repeat number

left parenthesis a right parenthesis(a)

What percent of the SAT verbal scores are less than 550550?

left parenthesis b right parenthesis(b)

If 1000 SAT verbal scores are randomly selected, about how many would you expect to be greater than 525525?

left parenthesis a right parenthesis(a) Approximately

nothing% of the SAT verbal scores are less than 550550.

(Round to two decimal places as needed.)mean =507

standard divination=120

SAT verbal score are less than 550show me step leasewhen I copy and past numbers are repeated ;lease ignore repeat number

a) z = (x - mu) / sigma = (550 - 507) / 120 = 0.3583

P(x < 550) = P(z < 0.3583) = 0.6399 = 0.6399 * 100% = **63.99%**

b) z = (x - mu) / sigma = (525 - 507) / 120 = 0.15

P(x > 525) = P(z > 0.15) = 0.4404

Expected number greater than 525 is 0.4404 * 1000 = ≈ **440.4 **

How many more do you have? We are well beyond an hour at this point.

Customer reply replied 3 years ago

Can I make it 440.40 to two decimal place because the question asked to answer two decimal place

Yes

Customer reply replied 3 years ago

If a random variable x is normally distributed, you can find the probability that x will fall in a given interval by calculating the area under the normal curve for the given interval. To find the area under the normal curve, first convert the upper bound of the interval to a z-score. Use either technology or the standard normal table to find the area corresponding to the z-score. To convert the area to a percent multiply by 100.the answer was incorrect

Try 440.38 instead. That's the only thing I can think of.

Customer reply replied 3 years ago

still not rightOk, I am just about to finish let Us do other example

Ok, I don't know what else we can do with that last one that would be different from what I have already done.

Customer reply replied 3 years ago

The population mean and standard deviation are given below. Find the required probability and determine whether the given sample mean would be considered unusual.

For a sample of nequals 62, find the probability of a sample mean being less than 23.323.3 if muμequals=2323 and sigmaσequals=1.31.3.

LOADING... Click the icon to view page 1 of the standard normal table.

LOADING... Click the icon to view page 2 of the standard normal table.

For a sample of nequals=62, the probability of a sample mean being less than 23.3 if mu equals 23μ=23 and sigmaσequals1.3 is

nothing.

(Round to four decimal places as needed.)Don't worry about it

For a sample of nequals 62, find the probability of a sample mean being less than 23.323.3 if muμequals=2323 and sigmaσequals=1.31.3.

LOADING... Click the icon to view page 1 of the standard normal table.

LOADING... Click the icon to view page 2 of the standard normal table.

For a sample of nequals=62, the probability of a sample mean being less than 23.3 if mu equals 23μ=23 and sigmaσequals1.3 is

nothing.

(Round to four decimal places as needed.)Don't worry about it

z = (x-bar - mu) / (sigma / √n) = (23.3 - 23) / (1.3 / √62) = 1.8171

P(x-bar < 23.3) = P(z < 1.8171) = **0.9654**

This would **not** be considered unusual.

Customer reply replied 3 years ago

For a sample of nequals=6262, the probability of a sample mean being less than 23.323.3 if mu equals 23μ=23 and sigmaσequals=1.31.3 is

0.96540.9654.

(Round to four decimal places as needed.)

Would the given sample mean be considered unusual?

The sample mean

▼

would

would not

be considered unusual because it has a probability that is

▼

greater

less

than 5%.

0.96540.9654.

(Round to four decimal places as needed.)

Would the given sample mean be considered unusual?

The sample mean

▼

would

would not

be considered unusual because it has a probability that is

▼

greater

less

than 5%.

The sample mean **would not **be considered unusual because it has a probability that is **greater **than 5%.

Customer reply replied 3 years ago

The population mean and standard deviation are given below. Find the required probability and determine whether the given sample mean would be considered unusual.

For a sample of nequals=6565, find the probability of a sample mean being greater than 212212 if muμequals=211211 and sigmaσequals=3.43.4.

For a sample of nequals=6565, the probability of a sample mean being greater than 212212 if mu equals 211μ=211 and sigmaσequals=3.43.4 is

nothing.

(Round to four decimal places as needed.)sample mean bigger than =212

For a sample of nequals=6565, find the probability of a sample mean being greater than 212212 if muμequals=211211 and sigmaσequals=3.43.4.

For a sample of nequals=6565, the probability of a sample mean being greater than 212212 if mu equals 211μ=211 and sigmaσequals=3.43.4 is

nothing.

(Round to four decimal places as needed.)sample mean bigger than =212

z = (212 - 211) / (3.4 / √65) = 2.3713

P(x-bar > 212) = P(z > 2.3713) =** 0.0089**

Customer reply replied 3 years ago

The sample mean

▼

would

would not

be considered unusual because it

▼

lies

does not lie

within the range of a usual event, namely within

▼

1 standard deviation

2 standard deviations

3 standard deviations

of the

▼

would

would not

be considered unusual because it

▼

lies

does not lie

within the range of a usual event, namely within

▼

1 standard deviation

2 standard deviations

3 standard deviations

of the

The sample mean **would** be considered unusual because it** does not lie**

within the range of a usual event, namely within **2 standard deviations**

of the mean.

Customer reply replied 3 years ago

Use the Central Limit Theorem to find the mean and standard error of the mean of the sampling distribution. Then sketch a graph of the sampling distribution.

The mean price of photo printers on a website is $231231 with a standard deviation of $6363. Random samples of size 3535 are drawn from this population and the mean of each sample is determined.

The mean of the distribution of sample means is

231231.

The standard deviation of the distribution of sample means is

10.64910.649.

(Type an integer or decimal rounded to three decimal places as needed.)

Sketch a graph of the sampling distribution. Choose the correct answer below.

A.

-2

0

2

Mean price (in dollars)

A bell shaped curve is over a horizontal axis labeled Mean price (in dollars) from negative 3 to 3 in increments of 1 and is centered on 0.

Your answer is not correct.B.

229

231

233

Mean price (in dollars)

A bell shaped curve is over a horizontal axis labeled Mean price (in dollars) from 228 to 234 in increments of 1 and is centered on 231.

C.

105

231

357

Mean price (in dollars)

A bell shaped curve is over a horizontal axis labeled Mean price (in dollars) from 42 to 420 in increments of 63 and is centered on 231.

D.

231

252.3

209.7

Mean price (in dollars)

A bell shaped curve is over a horizontal axis labeled Mean price (in dollars) from 199.1 to 263 in increments of 10.65 and is centered on 231. All values are approximate.

This is the correct answer.

Question is complete. Tap on the red indicators to see incorrect

The mean price of photo printers on a website is $231231 with a standard deviation of $6363. Random samples of size 3535 are drawn from this population and the mean of each sample is determined.

The mean of the distribution of sample means is

231231.

The standard deviation of the distribution of sample means is

10.64910.649.

(Type an integer or decimal rounded to three decimal places as needed.)

Sketch a graph of the sampling distribution. Choose the correct answer below.

A.

-2

0

2

Mean price (in dollars)

A bell shaped curve is over a horizontal axis labeled Mean price (in dollars) from negative 3 to 3 in increments of 1 and is centered on 0.

Your answer is not correct.B.

229

231

233

Mean price (in dollars)

A bell shaped curve is over a horizontal axis labeled Mean price (in dollars) from 228 to 234 in increments of 1 and is centered on 231.

C.

105

231

357

Mean price (in dollars)

A bell shaped curve is over a horizontal axis labeled Mean price (in dollars) from 42 to 420 in increments of 63 and is centered on 231.

D.

231

252.3

209.7

Mean price (in dollars)

A bell shaped curve is over a horizontal axis labeled Mean price (in dollars) from 199.1 to 263 in increments of 10.65 and is centered on 231. All values are approximate.

This is the correct answer.

Question is complete. Tap on the red indicators to see incorrect

It seems that this question was already answered, since the values for the mean and standard deviation of the distribution of means are already computed, and one of the choices about the graph is marked as "correct".

What do you need me to do here?

Customer reply replied 3 years ago

Sorry, yes it isUse the Central Limit Theorem to find the mean and standard error of the mean of the sampling distribution. Then sketch a graph of the sampling distribution.

The mean price of photo printers on a website is $220220 with a standard deviation of $5757. Random samples of size 2929 are drawn from this population and the mean of each sample is determined.

The mean of the distribution of sample means is

nothing.please use 220 not 220220

The mean price of photo printers on a website is $220220 with a standard deviation of $5757. Random samples of size 2929 are drawn from this population and the mean of each sample is determined.

The mean of the distribution of sample means is

nothing.please use 220 not 220220

The mean of the distribution of sample means is **$220** (the same as the sample mean)

The standard deviation of the distribution of sample means is 57 / √29 = **10.58**

Customer reply replied 3 years ago

he standard deviation of the distribution of sample means is

10.58510.585.

(Type an integer or decimal rounded to three decimal places as needed.)

Sketch a graph of the sampling distribution. Choose the correct answer below.

A.

-2

0

2

Mean price (in dollars)

A bell shaped curve is over a horizontal axis labeled Mean price (in dollars) from negative 3 to 3 in increments of 1 and is centered on 0.

B.

106

220

334

Mean price (in dollars)

A bell shaped curve is over a horizontal axis labeled Mean price (in dollars) from 49 to 391 in increments of 57 and is centered on 220.

C.

220

241.2

198.8

Mean price (in dollars)

A bell shaped curve is over a horizontal axis labeled Mean price (in dollars) from 188.2 to 251.8 in increments of 10.6 and is centered on 220. All values are approximate.

D.

218

220

222

Mean price (in dollars)

10.58510.585.

(Type an integer or decimal rounded to three decimal places as needed.)

Sketch a graph of the sampling distribution. Choose the correct answer below.

A.

-2

0

2

Mean price (in dollars)

A bell shaped curve is over a horizontal axis labeled Mean price (in dollars) from negative 3 to 3 in increments of 1 and is centered on 0.

B.

106

220

334

Mean price (in dollars)

A bell shaped curve is over a horizontal axis labeled Mean price (in dollars) from 49 to 391 in increments of 57 and is centered on 220.

C.

220

241.2

198.8

Mean price (in dollars)

A bell shaped curve is over a horizontal axis labeled Mean price (in dollars) from 188.2 to 251.8 in increments of 10.6 and is centered on 220. All values are approximate.

D.

218

220

222

Mean price (in dollars)

There are four choices (A, B, C, and D), but only three descriptions.

I would select whichever choice goes with the following:

*A bell shaped curve is over a horizontal axis labeled Mean price (in dollars) from 188.2 to 251.8 in increments of 10.6 and is centered on 220. All values are approximate.*

which is either Choice C or Choice D, depending on whether the descriptions come at the beginning of each choice, or at the end.

Customer reply replied 3 years ago

It is bell shaped graph but when I copy and past it the form is changed

Select the one that has values from 188.2 to 251.8 running along the bottom edge, with the value 220 in the middle along the bottom.

Customer reply replied 3 years ago

A, -2 0 2 B, 106 220 334 C 198 220 241.1 D 218 220 222

think these number are on the X axis that are the bell shape of the graph the 3 point

think these number are on the X axis that are the bell shape of the graph the 3 point

Can you take a screenshot of the graphs and post that?

Customer reply replied 3 years ago

i am not able to

I'd lean toward Choice C then.

Customer reply replied 3 years ago

Imagine -2 and 2 are the two end of a bell shaped graph and 0 is the center of the graph ( the left and the right number are the end point of the bell shaped graph the middle number is ***** center ( symmetry) of the graph

The values -2, 0, and 2 would most likely be z-values though. Unless the axis of the graph is marked as something like "z-value", then this choice is not likely to be correct.

A typical layout though would label the mean in the center, and the values that are 2 standard deviations above and below the mean. With a mean of 220, and a standard deviation of the mean of 10.58, the values that are 2 standard deviations away from the mean are:

220 - 2(10.58) = 198.84

220 + 2(10.58) = 241.16

These values match the values given in Choice C.

Customer reply replied 3 years ago

congra C was correct answerhow do i make screen shutUse the normal distribution of SAT critical reading scores for which the mean is 510510 and the standard deviation is 114114. Assume the variable x is normally distributed.

left parenthesis a right parenthesis(a)

What percent of the SAT verbal scores are less than 550550?

left parenthesis b right parenthesis(b)

If 1000 SAT verbal scores are randomly selected, about how many would you expect to be greater than 525525?

left parenthesis a right parenthesis(a) Approximately

nothing% of the SAT verbal scores are less than 550550.

(Round to two decimal places as needed.)

left parenthesis a right parenthesis(a)

What percent of the SAT verbal scores are less than 550550?

left parenthesis b right parenthesis(b)

If 1000 SAT verbal scores are randomly selected, about how many would you expect to be greater than 525525?

left parenthesis a right parenthesis(a) Approximately

nothing% of the SAT verbal scores are less than 550550.

(Round to two decimal places as needed.)

It depends on what kind of computer you have.

If it is a Macintosh, press the "Command" key, the "Shift" key, and the number 3, all at the same time.

If it is a Windows machine, there is likely a "Print Screen" button on the keyboard. Depending on the make and model, you might need to press a function key, or the "Alt" key at the same time to take the screenshot. Doing so will save the screen image to the clipboard, and you can then paste it into a document.

a) z = (x - mu) / sigma = (550 - 510) / 114 = 0.3509

P(x < 550) = P(z < 0.3509) = **0.6372**

b) z = (525 - 510) / 114 = 0.1316

P(x > 525) = P(z > 0.1316) = 0.44765

n = 1000 * 0.44765 = **447.65**

Customer reply replied 3 years ago

Use the normal distribution of SAT critical reading scores for which the mean is 505505 and the standard deviation is 112112. Assume the variable x is normally distributed.

left parenthesis a right parenthesis(a)

What percent of the SAT verbal scores are less than 675?

left parenthesis b right parenthesis(b)

If 1000 SAT verbal scores are randomly selected, about how many would you expect to be greater than 550?

left parenthesis a right parenthesis(a) Approximately

nothing% of the SAT verbal scores are less than 675.

(Round to two decimal places as needed.)

left parenthesis a right parenthesis(a)

What percent of the SAT verbal scores are less than 675?

left parenthesis b right parenthesis(b)

If 1000 SAT verbal scores are randomly selected, about how many would you expect to be greater than 550?

left parenthesis a right parenthesis(a) Approximately

nothing% of the SAT verbal scores are less than 675.

(Round to two decimal places as needed.)

a) z = (675 - 505) / 112 = 1.5179

P(x < 675) = P(z < 1.5179) = 0.9355 = **93.55%**

b) z = (550 - 505) / 112 = 0.4018

P(x > 550) = P(z > 0.4018) = 0.343915

n = 1000 * 0.343915 ≈ **343.92**

Customer reply replied 3 years ago

A doctor wants to estimate the HDL cholesterol of all 20- to 29-year-old females. How many subjects are needed to estimate the HDL cholesterol within 33 points with 99 %99% confidence assuming sigma equals 11.7 question mark σ=11.7? Suppose the doctor would be content with 95 %95% confidence. How does the decrease in confidence affect the sample size required?

A 99% confidence level requires

nothing subjects.

(Round up to the nearest whole number as needed.)

A 99% confidence level requires

nothing subjects.

(Round up to the nearest whole number as needed.)

n = (2.576 * 11.7 / 3)^2 = **101**

Customer reply replied 3 years ago

A doctor wants to estimate the HDL cholesterol of all 20- to 29-year-old females. How many subjects are needed to estimate the HDL cholesterol within 44 points with 99 %99% confidence assuming sigma equals 13.7 question mark σ=13.7? Suppose the doctor would be content with 90 %90% confidence. How does the decrease in confidence affect the sample size required?

A 99% confidence level requires

nothing subjects.

(Round up to the nearest whole number as needed.)

A 99% confidence level requires

nothing subjects.

(Round up to the nearest whole number as needed.)

n = (2.576 * 13.7 / 4)^2 = **78**

Customer reply replied 3 years ago

thank you for your hard, I am really proud of you

You're welcome.

Customer reply replied 3 years ago

I will see you on Monday if you are available

Ok. You will need to start a new question for that. If you wish to request me specifically, you can do so by simply adding "For Ryan" at the beginning of your new question.

Customer reply replied 3 years ago

ok I will do that

Thanks.

Customer reply replied 3 years ago

Hi Ryan I have been checking you ,if you have time to be online . Can you please reply when you have time I have a few question

thank you

Tiezaz

thank you

Tiezaz

Customer reply replied 3 years ago

An automotive manufacturer claims the mean price of a small SUV

is at mostis at most$26 comma 00626,006.

If a hypothesis test is performed, how should you interpret a decision that (a) rejects the null hypothesis and (b) fails to reject the null hypothesis?(a) Choose the correct answer below.A.There is not enough evidence to support the claim that the mean price of a small SUV

is at mostis at most$26 comma 00626,006.B.There is not enough evidence to reject the claim that the mean price of a small SUV

is at mostis at most$26 comma 00626,006.C.There is enough evidence to reject the claim that the mean price of a small SUV

is at mostis at most$26 comma 00626,006.D.There is enough evidence to support the claim that the mean price of a small SUV

is at mostis at most$26 comma 00626,006.A local chess club claims that the length of time to play a game has a

standard deviationstandard deviation

of

more than 11 minutesmore than 11 minutes.

Write sentences describing type I and type II errors for a hypothesis test of this claim.A type I error will occur if the actual

standard deviationstandard deviation

of the length of time to play a game is

greater than or equal tonot equal togreater thanequal toless than or equal toless thangreater than or equal to

1111

minutes, but you

rejectrejectfail to reject

the null hypothesis,▼Upper H 0 : sigma greater than 11H0: σ>11Upper H 0 : sigma greater than or equals 11H0: σ≥11Upper H 0 : sigma not equals 11H0: σ≠11Upper H 0 : sigma less than 11H0: σ<11Upper H 0 : sigma equals 11H0: σ=11Upper H 0 : sigma less than or equals 11H0: σ≤11

.A type II error will occur if the actual

standard deviationstandard deviation

of the length of time to play a game is▼less thangreater than or equal toless than or equal toequal togreater thannot equal to

1111

minutes, but you▼rejectfail to reject

the null hypothesis,▼Upper H 0 : sigma greater than 11H0: σ>11Upper H 0 : sigma less than 11H0: σ<11Upper H 0 : sigma greater than or equals 11H0: σ≥11Upper H 0 : sigma equals 11H0: σ=11Upper H 0 : sigma not equals 11H0: σ≠11Upper H 0 : sigma less than or equals 11H0: σ≤11

.

is at mostis at most$26 comma 00626,006.

If a hypothesis test is performed, how should you interpret a decision that (a) rejects the null hypothesis and (b) fails to reject the null hypothesis?(a) Choose the correct answer below.A.There is not enough evidence to support the claim that the mean price of a small SUV

is at mostis at most$26 comma 00626,006.B.There is not enough evidence to reject the claim that the mean price of a small SUV

is at mostis at most$26 comma 00626,006.C.There is enough evidence to reject the claim that the mean price of a small SUV

is at mostis at most$26 comma 00626,006.D.There is enough evidence to support the claim that the mean price of a small SUV

is at mostis at most$26 comma 00626,006.A local chess club claims that the length of time to play a game has a

standard deviationstandard deviation

of

more than 11 minutesmore than 11 minutes.

Write sentences describing type I and type II errors for a hypothesis test of this claim.A type I error will occur if the actual

standard deviationstandard deviation

of the length of time to play a game is

greater than or equal tonot equal togreater thanequal toless than or equal toless thangreater than or equal to

1111

minutes, but you

rejectrejectfail to reject

the null hypothesis,▼Upper H 0 : sigma greater than 11H0: σ>11Upper H 0 : sigma greater than or equals 11H0: σ≥11Upper H 0 : sigma not equals 11H0: σ≠11Upper H 0 : sigma less than 11H0: σ<11Upper H 0 : sigma equals 11H0: σ=11Upper H 0 : sigma less than or equals 11H0: σ≤11

.A type II error will occur if the actual

standard deviationstandard deviation

of the length of time to play a game is▼less thangreater than or equal toless than or equal toequal togreater thannot equal to

1111

minutes, but you▼rejectfail to reject

the null hypothesis,▼Upper H 0 : sigma greater than 11H0: σ>11Upper H 0 : sigma less than 11H0: σ<11Upper H 0 : sigma greater than or equals 11H0: σ≥11Upper H 0 : sigma equals 11H0: σ=11Upper H 0 : sigma not equals 11H0: σ≠11Upper H 0 : sigma less than or equals 11H0: σ≤11

.

Hi,

I'm sorry that I couldn't reply sooner. I have been having some computer issues today.

You will need to open a new question page for these problems. Please be sure to add "For Ryan" at the beginning of the new question page so that you are sure to have your question directed to me.

Also, please put ALL of the problems that you need help with into a single document and post that as an attachment. I need to know what I am agreeing to before I make a commitment.

Thanks,

Ryan

Customer reply replied 3 years ago

How do I open a new question page ?

How do I add for Ryan?

How do I add for Ryan?

There should be a link to a list of your questions, and there should be a link or button there that you can click on to start a new question.

Once you are able to start a new question, just type in "For Ryan" at the beginning of the text.

If you need further assistance with getting a new question page opened, you should call Customer Service at(###) ###-#### I am limited in what I can offer because I have only a vague idea of how the site looks on your end.

Customer reply replied 3 years ago

Hi Ryan i am not able to open a new page

I'm sorry, I don't know what else to tell you as I don't have any suggestions other than contacting Customer Service. They will be able to walk you through the process. Not only will they have your current account information (in case there is some issue with billing or something like that), they will also know what the screen looks like on your and and should be able to direct you on which page to go to and which buttons to click.

Customer reply replied 3 years ago

I already contact customer service I should bay a new payment

I'm sorry, I guess I don't know what you mean by that.

Customer reply replied 3 years ago

hear is question for you -I can pay now but I am not able to to ask you many question today and tomororwI am not sure if I am still able to ask a question on Thursday on the same chapter with out an other paymentif I pay now it will take me to someone else, I am not able to see your nameI just want to make sure before i make transaction

I don't mind combining questions today with questions on Thursday, as long as we can agree on a reasonable price for a certain number of questions. What happened last time is that there was no definite end to the number of questions, and the end result is that I was committed for several hours with no additional compensation. Particularly on the second day when "an hour" got stretched out to more like four hours. Even though there are gaps between the questions, I'm still tied to my computer, and my ability to get anything else done is very limited. I'm not here to get rich, but I'm also not here to work for free.

I'm ok with how that played out last time, but I don't want to do that again.

If you just type "For Ryan" at the beginning, that should take precedence over any assignment that the system doles out. If someone else replies to you, you are free to tell them that you want to work with me.

Customer reply replied 3 years ago

are you free now

I need to run a quick errand for about 15 minutes, and then I will be.

Customer reply replied 3 years ago

ok

I'm sorry, I don't know if you were waiting to hear back from me, or if something else came up. I'm back, and ready when you are.

Customer reply replied 3 years ago

My connection is too law new can u give 20 minute

No problem.

Customer reply replied 3 years ago

What do u think if if meet on line on thursday instead of today my connection is to slow

Right now I am in library and it will close close soon so I should go home, if you ok for thursday I have plenty of time

Right now I am in library and it will close close soon so I should go home, if you ok for thursday I have plenty of time

I don't currently have anything scheduled for Thursday, so that should be fine. What time did you have in mind? (Include the time zone too please.)

Customer reply replied 3 years ago

The time is Which country are u at

I'm on the west coast of the US. Pacific Daylight Time.

Customer reply replied 3 years ago

3:30 will be perfect for me

3:30pm Pacific Time it is. See you here then.

Customer reply replied 3 years ago

If you can be a best helper like last time you did, I can refer many friends

Customer reply replied 3 years ago

Hi are u there

Yes

Customer reply replied 3 years ago

I do have question for u

ok, what is it?

Customer reply replied 3 years ago

sorry, I was text you from my phone its battery is shut down

Customer reply replied 3 years ago

Can we meet tomorrow?Right now I am ta work but I have time tom orrow to be online if you as wellIf u can just tell me what time you have free timeI am just waiting for ur answer

I have some appointments tomorrow. What time did you have in mind?

(Sorry about the delayed response. This page is so long now that it takes forever to load. Yet another reason why we need to get a new page started.)

Customer reply replied 3 years ago

I will open new page. I will have 3 homework do u thing I will pay per homeworck?

Can you be more specific about what "3 homework" means? Do you mean three separate problem sets?

Customer reply replied 3 years ago

Homework1 , homework 2 ...

..

..

When you have the next set ready, you should open a new question page and set the price at an amount that you are comfortable with, taking into consideration both the number of problems and their complexity. Customer Service can help you if you are having difficulty doing that.

Customer reply replied 3 years ago

Ok sounds that sounds good

Customer reply replied 3 years ago

HI

Are you onlineIf you have time i would like to work with u today

Are you onlineIf you have time i would like to work with u today

Hi. I probably won't be available until after 8:30 pm tonight (Pacific time) or so.

Customer reply replied 3 years ago

What time is now at there?

It is 3:30 pm here now.

Customer reply replied 3 years ago

Ok you remember we have appointment for tomorrow?

24 hours from now.

Customer reply replied 3 years ago

Ok, is there any way we can have phone call?

but I wouldn't be able to do it until later.

Customer reply replied 3 years ago

I know. I would like to explain for about 1 home work. It is timed and I don't think I am able to use copy and past . How you think we can do it?

This is where screenshots come in handy. Having you read the problems to me will be very slow.

Customer reply replied 3 years ago

I don't think I can do screen shut. I will not let me to open other file on the same computer

Can you take pictures with your phone?

Would you be able to take a picture of the problems and send that?

Customer reply replied 3 years ago

I will check and let u know

Ok. That would be a better way to go, because I can almost guarantee that having you read the problems to me over the phone will result in us running out of time.

Customer reply replied 3 years ago

NO WE ARE NOT GOING TO YES PHONE FOR QUESTIONS , I just said I will like to explain about the form of the questions over the phone

Customer reply replied 3 years ago

If you are able to read the file below we can use this method

Customer reply replied 3 years ago

HERE is the file

Customer reply replied 3 years ago

Forget about its visibility I will make it clear

I'm sorry if I misunderstood what you were intending.I am able to open the attached image files on my end. However, the images of the computer screen are out of focus, so the text is unreadable.

Customer reply replied 3 years ago

I WILL ADJUST IT

No problem. Our messages are passing each other. I sent that last part before I saw your comment about the visibility of the text.

Customer reply replied 3 years ago

Any way if we are not able to start working we will talked about it tomorrow

Ok.

Customer reply replied 3 years ago

If possible please let meet 2:00 in plastic timePlastic time

Plastic time"? You mean Pacific time?

Customer reply replied 3 years ago

I sorry

Ok, no problem. I'll be here then.

Customer reply replied 3 years ago

Can you see a full text for this file

Customer reply replied 3 years ago

Hi Ryan

Are you thereI am ready to start new page if you have time?

Are you thereI am ready to start new page if you have time?

I am here.Regarding your earlier post asking if I could see "a full text for this file", there was no new file attached.We had made an appoinment for 2:00 pm, and it is currently only 1:00 pm here. I don't mind starting early, but I will need a bit of time to get ready.

Customer reply replied 3 years ago

For today and next homework am able to copy and past but I am worry about the 3rd homework

I'm sorry, I don't know what else to tell you. There is only so much that I can do.

Customer reply replied 3 years ago

I here the issue I am not able to open new page

I don't mind to pay now but the thing is how can I do it.

It was easy when the first Time I signed up

But now it isn'tIf u call me you can tell me how I can do it

I don't mind to pay now but the thing is how can I do it.

It was easy when the first Time I signed up

But now it isn'tIf u call me you can tell me how I can do it

I can't tell you how to open a new page, because I don't know what the site looks like on your end. Customer Service can help you with this. They will know which links you should use, and will be able to direct you to the correct page.

You can call them at(###) ###-####

Customer reply replied 3 years ago

Ryan do you get my question

Which question? Did you open a new one? If so, I haven't seen it. It is possible that it ended up in a different category for some reason, in which case I wouldn't be able to see it.

Customer reply replied 3 years ago

here is the question

A random sample of 8181 eighth grade students' scores on a national mathematics assessment test has a mean score of 268268 with a standard deviation of 3737. This test result prompts a state school administrator to declare that the mean score for the state's eighth graders on this exam is more than 260260. At alphaαequals=0.120.12, is there enough evidence to support the administrator's claim? Complete parts (a) through (e).

(a) Write the claim mathematically and identify Upper H 0H0 and Upper H Subscript aHa. Choose the correct answer below.

A.

Upper H 0H0: muμless than<260260

Upper H Subscript aHa: muμgreater than or equals≥260260 (claim)

B.

Upper H 0H0: muμless than or equals≤260260

Upper H Subscript aHa: muμgreater than>260260 (claim)

C.

Upper H 0H0: muμequals=260260

Upper H Subscript aHa: muμgreater than>260260 (claim)

D.

Upper H 0H0: muμequals=260260 (claim)

Upper H Subscript aHa: muμgreater than>260260

E.

Upper H 0H0: muμless than or equals≤260260 (claim)

Upper H Subscript aHa: muμgreater than>260260

F.

Upper H 0H0: muμgreater than or equals≥260260 (claim)

Upper H Subscript aHa: muμless than<260260Can we Continent with pageremember, when I copy and past question numbers are repeated by error

A random sample of 8181 eighth grade students' scores on a national mathematics assessment test has a mean score of 268268 with a standard deviation of 3737. This test result prompts a state school administrator to declare that the mean score for the state's eighth graders on this exam is more than 260260. At alphaαequals=0.120.12, is there enough evidence to support the administrator's claim? Complete parts (a) through (e).

(a) Write the claim mathematically and identify Upper H 0H0 and Upper H Subscript aHa. Choose the correct answer below.

A.

Upper H 0H0: muμless than<260260

Upper H Subscript aHa: muμgreater than or equals≥260260 (claim)

B.

Upper H 0H0: muμless than or equals≤260260

Upper H Subscript aHa: muμgreater than>260260 (claim)

C.

Upper H 0H0: muμequals=260260

Upper H Subscript aHa: muμgreater than>260260 (claim)

D.

Upper H 0H0: muμequals=260260 (claim)

Upper H Subscript aHa: muμgreater than>260260

E.

Upper H 0H0: muμless than or equals≤260260 (claim)

Upper H Subscript aHa: muμgreater than>260260

F.

Upper H 0H0: muμgreater than or equals≥260260 (claim)

Upper H Subscript aHa: muμless than<260260Can we Continent with pageremember, when I copy and past question numbers are repeated by error

No, we can't just keep on adding to this page. Most importantly, there is no mechanism for me to get paid if we do that, since this "page" has already been paid for. Secondly, the page is so long at this point that it is very slow to load.

Customer reply replied 3 years ago

I just pay 5 minute ago

You may have made a payment to the site, but, as I understand it, that only deposits funds to your account on the site. Unfortunately, the process is a little more complicated than that. I only get paid when you rate the service that you have received. Since this page has already received a rating, there is no provision for you to add another rating, which means that there is no way on this page for you to “pay me again”. This is why a new page needs to be opened.

Please keep in mind that I am not an employee of the site. I am a user of it, just like you. I simply don’t have either the information or the access to assist you with account issues or other site issues. My role on the site is limited to answering math questions. You really need to contact Customer Service for assistance.

Customer reply replied 3 years ago

Hi Ryan I open a new page and, I am still waiting for youRyan somebody was take over it I will start over it

I never received any notification of the new page being opened, and you did not add "For Ryan" at the beginning, which left another expert free to answer your question. Since you have accepted the offer that was presented to you, a rating has been entered on the page, so even if the other expert were to step aside, there would be no way for me to get paid. At least not without you having to contact Customer Service again.

Customer reply replied 3 years ago

let Us work together nowI going to open a new page with your name now

Ok.

I see your new question, but it got posted in the "General" category for some reason and I cannot post a response to it at this time. I have made a request to a moderator to move it to the proper category. I am confident that it will be moved, but I don't know how quickly it will happen.

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