How JustAnswer Works:

  • Ask an Expert
    Experts are full of valuable knowledge and are ready to help with any question. Credentials confirmed by a Fortune 500 verification firm.
  • Get a Professional Answer
    Via email, text message, or notification as you wait on our site.
    Ask follow up questions if you need to.
  • 100% Satisfaction Guarantee
    Rate the answer you receive.

Ask Ryan Your Own Question

Ryan
Ryan, Engineer
Category: Calculus and Above
Satisfied Customers: 7506
Experience:  B.S. in Civil Engineering
40260889
Type Your Calculus and Above Question Here...
Ryan is online now
A new question is answered every 9 seconds

find the margin of error for c s and n c+.90 s=3.1, n=100

Resolved Question:

find the margin of error for c s and n
c+.90 s=3.1, n=100
Submitted: 2 years ago.
Category: Calculus and Above
Expert:  Ryan replied 2 years ago.
Hi,

Welcome! Thank you for using JustAnswer.

I assume that the notation "c + .90" was supposed to be "c = 0.90" and is meant to indicate a 90% confidence interval.

If so, the margin of error is:

E = z * s / √n

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

Substituting the value for z, s, and n gives:

E = (1.645)(3.1) / √100

E = 0.5100
(rounded to four decimal places)


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

Thanks,

Ryan
Customer: replied 2 years ago.


i have to pay $45 for answering one question?

Expert:  Ryan replied 2 years ago.
Hi,

The pricing is set by the customer, so I don't have any control over it.

If you have additional math questions I'll be happy to answer those.

Ryan
Customer: replied 2 years ago.


so I can pay at the end of all my questions? Ill gladly pay $45. But I have 19 more to go. If not worth your time let me know. If it is here is the next question.


 


Construct the confidence interval for the population


c=0.90, x=8.1 s=0.2 and n=53

Expert:  Ryan replied 2 years ago.
Hi,

It's hard to say without seeing all of the questions. If they are all more or less about the same amount of work as the first one, then that sounds fine.

Here is the solution for this one:

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

The margin of error is then:

E = z * s / √n = (1.645)(0.2)/√53 = 0.0452

The lower limit of the confidence interval is then:

x-bar - E = 8.1 - 0.0452 = 8.0548


The upper limit of the confidence interval is then:

x-bar + E = 8.1 + 0.0452 = 8.1452


The confidence interval is then (8.0548, 8.1452)

(If all you want is the final answer, please let me know. It will speed things up if I don't have to type the solution out. Of course, if you're interested in the steps, I'm fine with including it.)


Thanks,

Ryan
Customer: replied 2 years ago.

Final answer works...


 


construct the confidence interval for population. Round to one decimal place as needed.


 


c=0.95, x (x has line over it)=15.9 , s=4.0, and n=70

Expert:  Ryan replied 2 years ago.
Hi,

The confidence interval is (14.9629, 16.8371).

Thanks,

Ryan
Customer: replied 2 years ago.


please adjust for a 95% confidence interval for u is? round to one decimal place

Expert:  Ryan replied 2 years ago.
Hi,

I'm sorry, I overlooked the part about rounding off.

The interval is (15.0, 16.8).

Thanks,

Ryan
Customer: replied 2 years ago.


question 6


 


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


 


a biologist reports a confidence interval of (4.7,5.1) when estimating the mean height (in centimeters) of a sample of seedlings


 


The estimated margin of error is?

Expert:  Ryan replied 2 years ago.
Hi,

The estimated margin of error is 0.2

The sample mean is 4.9

Thanks,

Ryan
Customer: replied 2 years ago.


Q 7 of 20


 


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


 


C=0.98,s=7.4,E=1


 


Assume that a preliminary sample of has at least 30 members


n=? (round up to the nearest whole number)

Expert:  Ryan replied 2 years ago.
Hi,

The minimum sample size would be 297.

Thanks,

Ryan
Ryan, Engineer
Satisfied Customers: 7506
Experience: B.S. in Civil Engineering
Ryan and other Calculus and Above Specialists are ready to help you
Customer: replied 2 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 result and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals.A random sample of 55 home theather systems has a mean price of $143.00 and a standard deviation is $16.80.Construct 90% confidence interval for the poplulation mean.The 90% confidence interval is? (round to two decimal places as needed)
Expert:  Ryan replied 2 years ago.
Hi,

The 90% confidence interval is (139.27, 146.73)

The interpretation is that we are 90% certain that the true population mean lies within the limits of the 90% confidence interval.


The 95% confidence interval is (138.56, 147.44)


The interpretation is that we are 95% certain that the true population mean lies within the limits of the 95% confidence interval.


The 95% confidence interval is wider than the 90% confidence interval.


Thanks,

Ryan
Customer: replied 2 years ago.


q9 of 20 ( i will give you max tip as well hang in there lol)


 


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 convienient, use technology to construct the confidence intervals.


 


A random sample of 44 gas grills has a mean price of $646.30 and a standard deviation of $57.20.


 


The 90% confidence interval is? (round to two decimal places as needed)

Expert:  Ryan replied 2 years ago.
Hi,

The 90% confidence interval is (632.11, 660.49)

The 95% confidence interval is (624.09, 668.51)

The 95% confidence interval is wider.


Thanks,

Ryan
Customer: replied 2 years ago.


FYI 95% interval was not correct...


 


q10 of 20


 


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


 


A random sample pf 32 eight ounce servings of different juice drinks has a mean of 76.6 calories and a standard deviation of 45.2 calories.


 


The 90% confidence interval is ? (round to one decimal place as needed)

Expert:  Ryan replied 2 years ago.
Hi,

I'm sorry. I looked at the wrong line in the spreadsheet I set up. The one I typed was the 99% interval.

It may be too late, but the 95% interval should have been (629.40, 662.20)


For #10:

The 90% confidence interval is (63.5, 89.7)

The 95% confidence interval is (60.9, 92.3)

The 95% confidence interval is wider.


Thanks,

Ryan
Customer: replied 2 years ago.


q11


 


People were polled on how many books they read the previous year. How many subjects are needed to estimate the number of books read the previous year within one book with 90% confidence? The intial survey results indicate that O (looks kind of like an o symbol) o=13.9 books.


 


A 90% confidence level requires blank subjects

Expert:  Ryan replied 2 years ago.
Hi,

The answer here depends on which value your course uses for a 90% confidence level.

If it uses z = 1.645, then the answer is 523

If, instead, they round it off to z = 1.65, then the answer is 527

We have used 1.645 in previous problems in this set, so hopefully that is the value that they want you to use. This issue didn't come up earlier because it didn't change the value of the answer like it does here.


Thanks,

Ryan
Customer: replied 2 years ago.


q12 of 20


 


A doctor wants to estimate the HDLcholesterol of all 20- to 29 year old females. How many subjects are needed to estimate the HDL cholesterol within 2 points with 99% confidence assuming o=13.1? Suppose the doctor would be content with 90% confidence. How does the degrease in confiendence affect the sample size required?


 


A 99% confidence level requires Blank subjects


 


 

Expert:  Ryan replied 2 years ago.
Hi,

The 99% confidence level requires 285 subjects.

Decreasing the confidence level would reduce the required sample size.

Thanks,

Ryan

Customer: replied 2 years ago.


pls add the 90% confidence level for q12

Expert:  Ryan replied 2 years ago.
Hi,

The required sample size for the 90% confidence level would be 117. (Using the 1.645 value for z.)


Thanks,

Ryan
Customer: replied 2 years ago.


q13 of 20


 


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


 


c=0.95, x (with line over it) x=13.3, s=3.0 n=10


 


(a) the 95% confidence interval using a t distribution is? ( round to one decimal place as needed)

Expert:  Ryan replied 2 years ago.
Hi,

The 95% confidence interval using the t-distribution is (11.2, 15.4).

If you had used the normal distribution the 95% confidence interval would be (11.4, 15.2). The interval using the normal distrubition is narrower than the interval using the t-distribution.

Thanks,

Ryan

Customer: replied 2 years ago.


q14 of 20


 


In the following situation, assume the random variable is normally distributed and use a normal distribution or a t-distribution to construct a 90% confidence interval for the population mean. If convenient, use technology to construct the confidence interval.


 


(a) in a random sample of 10 adults from a near by county, the mean waste generated per person per days was 5.26 punds and the standard deviation was 1.79 pounds.


(b) Repeat part 9a), assuming the same statistics came from a sample size of 600. Compare the results.


 


(a) for the sample of 10 adults, the 90% confidence interval is ? 9round to two decimal places as needed)

Expert:  Ryan replied 2 years ago.
Hi,

a) The 90% confidence interval is (4.22, 6.30).

b) The 90% confidence interval is (4.89, 5.63).

The confidence interval in part b is narrower due to the larger sample size.

Thanks,

Ryan
Customer: replied 2 years ago.


b was wrong not sure if you based if of 600.


 


q15 of 20


 


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


 


(0.786,0.810)


 


E=blank (type an interger or a decimal)

Expert:  Ryan replied 2 years ago.
Hi,

Yes, I did. I don't know what happened there. It should have been (5.14, 5.38). I had the numbers in the right places, but something happened when pushing it all through the calculator. Embarassed

For #15:

E = 0.012

The sample proportion is 0.798


Ryan
Customer: replied 2 years ago.


q16 of 20


 


In a survey of 626 males ages 18-64, 391 say they have gone to the dentist in the past year.


 


Construct 90% and 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals.


 


The 90% confidence interval for the population proportion p is? (round the final answers to the nearest thousandth as needed. Round all intermediate values to the nearest thousandth as needed)


 

Expert:  Ryan replied 2 years ago.
Hi,

The 90% confidence interval for the population proportion is (0.593, 0.657).

The interpretation of this interval is that we can be 90% certain that the true population proportion is between the limits of the interval.


The 95% confidence interval for the population proportion is (0.587, 0.663).

The interpretation of this interval is that we can be 95% certain that the true population proportion is between the limits of the interval.


The 95% confidence interaval is wider.


Thanks,

Ryan
Customer: replied 2 years ago.


q17 of 20


 


In a survey of 6000 women, 3431 say they change their nail polish once a week. Construct a 99% confidence interval for the population proportion of women who change their nail polish once a week.


 


A 99% confidence interval for the population proportion is? Round to the nearsest thousandth

Expert:  Ryan replied 2 years ago.
Hi,

The 99% confidence interval for the population proportion is (0.555, 0.588).

Thanks,

Ryan
Customer: replied 2 years ago.


q18 of 20


 


A researcher wishes to estimate, with 90% confidence the proportion of adults who have high speed internet access. her estimate must be accurate within 1% of the true proportion.


 


(a) find the minimum sample size needed, using a prior student that found that 52% 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 %52 of the respondents said they have highspeed internet access?


 


n=? round up to the nearest whole number as needed

Expert:  Ryan replied 2 years ago.
Hi,

For part (a), the required minimum sample size is 6,755.

For part (b), the required minimum sample size is 6,766.


Thanks,

Ryan
Ryan, Engineer
Satisfied Customers: 7506
Experience: B.S. in Civil Engineering
Ryan and other Calculus and Above Specialists are ready to help you

JustAnswer in the News:

 
 
 
Ask-a-doc Web sites: If you've got a quick question, you can try to get an answer from sites that say they have various specialists on hand to give quick answers... Justanswer.com.
JustAnswer.com...has seen a spike since October in legal questions from readers about layoffs, unemployment and severance.
Web sites like justanswer.com/legal
...leave nothing to chance.
Traffic on JustAnswer rose 14 percent...and had nearly 400,000 page views in 30 days...inquiries related to stress, high blood pressure, drinking and heart pain jumped 33 percent.
Tory Johnson, GMA Workplace Contributor, discusses work-from-home jobs, such as JustAnswer in which verified Experts answer people’s questions.
I will tell you that...the things you have to go through to be an Expert are quite rigorous.
 
 
 

What Customers are Saying:

 
 
 
  • Wonderful service, prompt, efficient, and accurate. Couldn't have asked for more. I cannot thank you enough for your help. Mary C. Freshfield, Liverpool, UK
< Last | Next >
  • Wonderful service, prompt, efficient, and accurate. Couldn't have asked for more. I cannot thank you enough for your help. Mary C. Freshfield, Liverpool, UK
  • This expert is wonderful. They truly know what they are talking about, and they actually care about you. They really helped put my nerves at ease. Thank you so much!!!! Alex Los Angeles, CA
  • Thank you for all your help. It is nice to know that this service is here for people like myself, who need answers fast and are not sure who to consult. GP Hesperia, CA
  • I couldn't be more satisfied! This is the site I will always come to when I need a second opinion. Justin Kernersville, NC
  • Just let me say that this encounter has been entirely professional and most helpful. I liked that I could ask additional questions and get answered in a very short turn around. Esther Woodstock, NY
  • Thank you so much for taking your time and knowledge to support my concerns. Not only did you answer my questions, you even took it a step further with replying with more pertinent information I needed to know. Robin Elkton, Maryland
  • He answered my question promptly and gave me accurate, detailed information. If all of your experts are half as good, you have a great thing going here. Diane Dallas, TX
 
 
 

Meet The Experts:

 
 
 
  • SusanAthena

    Master's Degree

    Satisfied Customers:

    98
    Tutor for Algebra, Geometry, Statistics. Explaining math in plain English.
< Last | Next >
  • http://ww2.justanswer.com/uploads/homeworklady/2008-11-24_125231_susannewavatar.jpg SusanAthena's Avatar

    SusanAthena

    Master's Degree

    Satisfied Customers:

    98
    Tutor for Algebra, Geometry, Statistics. Explaining math in plain English.
  • http://ww2.justanswer.com/uploads/ED/educatortech/2012-6-7_1256_williams4.64x64.jpg Mr. Gregory White's Avatar

    Mr. Gregory White

    Master's Degree

    Satisfied Customers:

    93
    M.A., M.S. Education / Educational Administration
  • http://ww2.justanswer.com/uploads/CO/CosumelEyes/2012-5-4_4923_Profile.64x64.jpg Ray Atkinson's Avatar

    Ray Atkinson

    Bachelor's Degree

    Satisfied Customers:

    45
    Degrees in Math, Education and Accounting, years and years of tutoring.
  • http://ww2.justanswer.com/uploads/SO/sonofGauss/2012-6-6_20151_PortraitLHW.64x64.jpg JACUSTOMER-yrynbdjl-'s Avatar

    JACUSTOMER-yrynbdjl-

    Master's Degree

    Satisfied Customers:

    42
    I have a MS in Mathematics and I have taught Mathematics for 10 years at the college level.
  • http://ww2.justanswer.com/uploads/dhouse1940/2010-09-02_133023_DSCN2406.jpg dhouse1940's Avatar

    dhouse1940

    Master's Degree

    Satisfied Customers:

    33
    BS mathematics, MS biostatistics, 35+ yrs designing & analyzing biological experiments.
  • http://ww2.justanswer.com/uploads/AR/arubin9/2012-2-4_42617_ArthurRubinedit.64x64.jpg Dr Arthur Rubin's Avatar

    Dr Arthur Rubin

    Doctoral Degree

    Satisfied Customers:

    32
    Ph.D. in Mathematics, 1978, from the California Institute of Technology, over 20 published papers
  • http://ww2.justanswer.com/uploads/DO/Donbacongamer/2012-3-6_2337_mathmaster.64x64.jpg Don's Avatar

    Don

    Master's Degree

    Satisfied Customers:

    29
    M.S. Astronautical Engineering. Math/Sci/Comp Tutor
 
 
 

Related Calculus and Above Questions

Chat Now With A Tutor
Ryan
Ryan
Tutor
7506 Satisfied Customers
B.S. in Civil Engineering