Calculus and Above

Calculus Questions? Ask a Mathematician for Answers ASAP

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Calculus and Above

Calculus Questions? Ask a Mathematician for Answers ASAP

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I dont understand how...

I dont understand how to do this question The claim is that the white blood cell counts of adult females are normally distributed, with a standard deviation equal to 2.022.02. A random sample of 3737 adult females has white blood cell counts with a mean of 7.97.9 and a standard deviation of 2.732.73. Find the value of the test statistic.

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Sorry, all the numbers repeated for some reason. stan dev is supposed to be 2.02, random sample is 37, mean is 7.9 and standard dev for hat sample is 2.73

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Customer reply replied 3 years ago

It has to be rounded to three decimal places

Answered in 14 minutes by:

5/1/2017

Math-John, PhD in Statistics

Category: Calculus and Above

Satisfied Customers: 1,322

Experience: MS and BS in Math

Verified

Hi, welcome to Just Answer. I would be happy to assist you in this question. By when do you need this to be completed?

Customer reply replied 3 years ago

As soon as possible. Its a practice questions, so im more concerned with learning the formula. I cant move on in my until i learn this formula.

OK I will provide the answer in about 20 minutes.

Customer reply replied 3 years ago

Okay, thank you!

You are welcome.

Customer reply replied 3 years ago

How much would it cost to have a total of 40 questions answered, with formulas written out in the answer. IE write the formula, then plug in the numbers, and then solve? Im trying to create a study guide for my finals on Wednesday, and our professor has given us 2 take homes to work on as preparations for the finals. We will be allowed to use those two works sheets as references during the finals, so i want to make sure I have all the appropriate formulas.

Do you only need the answer to this question or a total of 40?

Customer reply replied 3 years ago

If you can do the 40, I wont need the answer to this specific questions. My only hang up has been identify which formula to use so solve questions. If im able to get all 40, ill be able to use that as a reference to answer the rest of the questions.

Can you post all questions? I will have a look and let you know.

Customer reply replied 3 years ago

will a document upload work?

I think so because other customers have done this but I don't know how. You may ask Just Answer support team. I have finished the above question but I will wait for your document.

You may also create a free account at www.mediafire.com, upload your document and post the link here.

Customer reply replied 3 years ago

Its actually 50 questions, sorry.

I have got both files. I am reviewing now and will get back to you shortly.

Customer reply replied 3 years ago

Thank you!

I have no problem in doing these questions. How about the current price for this post plus $20 bonus for all 40 questions? You will have an opportunity to review my answer file before making the payment.

Customer reply replied 3 years ago

that will include the formulas/showing work?

Yes. I will provide detailed work.

Customer reply replied 3 years ago

Im going to use these as my study guide, so the formulas and the process of how you got the answer is really what i'm looking for, not the answers themselves per say.awesome!I'll gladly pay that! when do you think I could expect to have it back by?

After paying the fee and $20 bonus you can still ask questions about these 40 problems and I will answer for free.

Customer reply replied 3 years ago

to fair, I just looked. its 40 questions in one and 10 in the other, if you would like to adjust the price accordingly.awesome! how do i pay and get this started?

It's okay, 50 questions for the price I stated above. I need 2-3 hours to finish and post the answer. You will then be able to review my answer. If you are satisfied you can rate my answer and pay the bonus.

Customer reply replied 3 years ago

Okay. This is my first time using this, what would the total come out too? including the bonus

I only know I will receive $23 for this post and $15 for the $20 bonus. So I will get $38. You will pay the price of this post + $20 bonus. You may ask Just Anser support team. Please let me know if the price is a concern to you.

Customer reply replied 3 years ago

Price is great, just wanted to make sure i understood the the billing correctly. Thank you so much!

You are welcome. I will start working on it and post the answer ASAP.

Customer reply replied 3 years ago

I look forward to it! Ill check back in 2 hours, and periodically after that.

Customer reply replied 3 years ago

Just checking in to see how it's going.

I have finished most of them and will post the answer shortly.

Customer reply replied 3 years ago

Awesome! Thank you so much!

You are welcome.

For number 34, is it p>0.377?

Customer reply replied 3 years ago

With H1: p (does not equal) .377, the test statistic is z =3.06.its the "=" sign with a "/" through it.

The two answer files can be downloaded at

Sorry for the delay since I underestimated the workload. Please review the answer and let me know if you have any questions. If you are satisfied please accept my answer and pay the bonus. Again, you are welcome to ask questions about these 50 problems before and after you have made the payment and I will answer them for free.

In the future you may request me by writing "For John Only" in the post so that the questions are directed to me.

Customer reply replied 3 years ago

I dont see the link to download them

Sorry I don't know why the link didn't show up. I will type it now.

Customer reply replied 3 years ago

still no link. not sure why... maybe ill try to open in windows instead of chromeJust opened it in explorer, still no link

One second, still trying.

http://www.mediafire.com/file/zebayrfxghd6h53/Practice_Final_Exam_%281%29.doc.docx

http://www.mediafire.com/file/hy3uhcuswhvu9y5/Practice_Final_Exam_%282%29.doc.docx

The above two links are for two files.

Please let me know if they don't work.

Customer reply replied 3 years ago

Both links work, thank you! and how do i do the bonus/pay? Also, I was hoping you could write out the actually formulas for me to the problems, not the formulas with the date already input. I wasnt overly concerned with the answers, more so how you got the answers by writing out the formula and then inputting the information. Like Z = x-u/s/Square root of n. And then input the data and solve the problem. If i pay extra would you be able to do that? Like add the appropriate formula, no data, next to the question? not solving it or anything. Ill gladly pay you for your time!

Sure. I will add. You may ask customer support on how to rate and pay the bonus. I will update these two files shortly.

Math-John, PhD in Statistics

Category: Calculus and Above

Satisfied Customers: 1,322

Experience: MS and BS in Math

Verified

Math-John and 87 other Calculus and Above Specialists are ready to help you

Ask your own question now

Customer reply replied 3 years ago

Like question 8 on the 2nd one is perfect for that. Write the actually formula for how to figure out all those answers, like how to get the values of each individual letter and then how to put them all together to solve. because I will be using this as a study guide and i honestly wont know which information to put where without the formula its self.oh okay, thank you!

Can you let me know for which questions you want to formula?

Customer reply replied 3 years ago

I'm going to add a $50 bonus because it was more than you expected and because you are adding the formulas! I really appreciate it!

Thank you very much.

Customer reply replied 3 years ago

I think it messed up! and only paid $50 total, I contact costumer support to try and resolve it. Its supposed to be $23 plus bonus, correct? I input $50 for the bonus part, but it never paid the $23. I promise I will get it resolved.

You have paid both $23 and $50. When you gave me a positive rating the system automatically paid the price of this post.

If you can let me know the question numbers for which you need formula I can update the files.

Customer reply replied 3 years ago

Its not letting me add a file, one sec

You don't need to add a file. You can lust let me know the question numbers.

Customer reply replied 3 years ago

Here is a list of the questions i need formulas for. if its a repeat, like if questions 21 and 22 are the same, can add the formula to 21 and write (see 21) on question 22 so i know its the same? thank you!

Customer reply replied 3 years ago

Practice test 1 formulas

Questions,

4

5

7

8

10-16

17

18

19

20

21-22 (looks like same formula?)

23-36

Questions on practice exam 2

4

5

6

7

8-10

Questions,

4

5

7

8

10-16

17

18

19

20

21-22 (looks like same formula?)

23-36

Questions on practice exam 2

4

5

6

7

8-10

Got the file. I am working on it and will post the updated answer shortly.

Customer reply replied 3 years ago

can you write out an sub formuls that are required to answer the questions? not just the overall one, but if i need other formulas to find Z or t or p^ and they are not given in the actually question, will you write the formula to find those?

OK I will.

<a href="https://www.mediafire.com/folder/f8bim7t0iia7767,3tfk1doah8vp53q/shared" target="_blank">https://www.mediafire.com/folder/f8bim7t0iia7767,3tfk1doah8vp53q/shared</a>

Please let me know if you need further assistance.

You may use the following two links in case the above link doesn't work.

http://www.mediafire.com/file/f8bim7t0iia7767/Practice_Final_Exam_%281%29.doc%282%29.docx

http://www.mediafire.com/file/3tfk1doah8vp53q/Practice_Final_Exam_%282%29.doc%282%29.docx

Customer reply replied 3 years ago

I'm confused about question 4 and 9 on practice test 2. I don't understand how you got those answers.

For number 4, the number of defected items follows a binomial distribution with n = 23, p = 0.08. Fewer than 3 means X = 0, 1, or 2. The computation of

P(X=0), P(X=1) and P(X=2) are given in the file. Please let me know which part you have difficulty with. Thanks.

Customer reply replied 3 years ago

I don't understand what to do with those numbers to solve the equation. And I think there was a typo maybe. One fraction is 23/122 and the next is 23/221

P(X=0) = (23! / (0! * 23!)) * 0.08^0 * (1-0.08)^23 ,

0! = 1.

23!/(0! * 23!) = 23! / 23! = 1.

0.08^0 = 1.

So P(X =0 ) = (1-0.08)^23

P(X=1) = (23! / (1! * 22!)) * 0.08^1 * (1-0.08)^22

23! / (1! * 22!) = (23*22*21*20 * ... * 2 * 1) / (1 * 22*21*20 * ... * 2*1) = 23.

So P(X = 1) = 23 * 0.08 * 0.92^22

P(X=2) = (23! / (2! * 21!)) * 0.08^2 * (1-0.08)^21

23! / (2! * 21!) = (23*22*21*20 * ... * 2 * 1) / (2*1 * 21*20*19 * ... * 2*1) = 23 * 22 / 2 = 23 * 11 = 253.

So P(X = 2) = 253 * 0.08^2 * 0.92^21.

Customer reply replied 3 years ago

im sorry, i still dont understand how you get the 23 or the 253 in X=1 and x=2Im so lost on this problemBecause even just 253 * 0.08^2 * 0.92^21 = 3.513621, which is much higher than the 0.7218894 that is down for the answer

No problem. I would be happy to explain to you.

n! is defined as n*(n-1)*(n-2) *... * 2 * 1.

For example

4! = 4*3*2*1 = 24,

5! = 5*4*3*2*1 = 120.

For P(X=1). in the expression 23! / (1! * 22!), the numerator is 23*22*21*... * 2*1.

The first factor in the denominator is 1! = 1.

The second factor in the denominator is 22! = 22*21*20 * ... * 2 * 1. So this expression is

(23*22*21*... * 2*1) / (1 * 22*21*20*...*2*1).

After cancelling the common factors we only have 23 left.

For P(X=2), in the expression 23! / (2! * 21!),

the numerator is 23! = 23 * 22 * 21 * .. * 2 * 1.

the denominator is 2! * 21!. The first factor is 2! = 2* 1 = 2. The second factor is 21! = 21 * 20 * 19 * ... * 2 * 1.

So the expression is

(23 * 22 * 21 * .. * 2 * 1) / (2 * 21 * 20 * 19 * ... * 2 * 1).

After cancelling the common factors (21*20*19*... *2*1), we only have

23 * 22/2 = 23 * 11 = 253.

253 * 0.08^2 * 0.92^21 = 253 * 0.08 * 0.08 * 0.92 * 0.92 * 0.92 * 0.92 *0.92 * 0.92 *0.92 * 0.92 *0.92 * 0.92 *0.92 * 0.92 *0.92 * 0.92 *0.92 * 0.92 *0.92 * 0.92 *0.92 * 0.92 *0.92 = 0.2810897

Customer reply replied 3 years ago

I'm sorry, this wasn't the way I was taught so it's very confusing for me. im beyond lost because that doesn't look anything like the formula

Customer reply replied 3 years ago

Can you just write out the equation shorthand for each 0-2 and the answer? I'll just try to mimic the answer.

Which method did you use? To compute binomial probabilities, we can use Excel, calculator, a table.

Customer reply replied 3 years ago

Excel

Open Excel workbook.

Method 1: In any empty cell, type "=BINOM.DIST(2, 23, 0.08, 1)" to get P(X<=2) = P(X =0) + P(X=1) + P(X=2) = 0.721889.

Method 2:

In any empty cell type "=BINOM.DIST(0, 23, 0.08, 0)" to get P(X=0) = 0.146933.

In another empty cell type "=BINOM.DIST(1, 23, 0.08, 0)" to get P(X=1) = 0.293866.

In the third empty cell type "=BINOM.DIST(2, 23, 0.08, 0)" to get P(X=2) = 0.28109.

Then add up these three numbers to get P(X<=2) = 0.721889.

Customer reply replied 3 years ago

the answer, if rounded to 2 decimal points would be 72.19%?If the answer to 4 is that there is a 72.19% of 2 or fewer being defective, can you help me with number 9?

Correct for number 4, 79.19%.

Customer reply replied 3 years ago

79.19 or 72.19?

Sorry 72.19% or 0.7219.

For number 9, can you let me know which part you have difficulty with?

Customer reply replied 3 years ago

on nine, i used p ̂=x/n (40/160) = 0.25 to get P-hatthen i tried to use the (p ̂-p)/√(pq/n) to find the Z statistic. But its saying its 3.54 which is over the limit on the chart.0.25-0.15)/√((0.15*0.85 )/160)= and i get 3.54, which i dont think is right. Because you got 9.9209

The Z statistic you got is correct. Sorry about the error. Anyway the probability is approximately 0 if the number is ***** large and is over the limit of the chart.

Customer reply replied 3 years ago

if the number is ***** what is P-value? and how do i write the answer out.thank you so much for your help by the way! i havnt slept in over 40 hours trying to prep for this test!

You are welcome. It's my pleasure to assist you in preparing the exam.

p-value is 2 * 0 = 0 since it's a two tailed test. Just replace 9.9209 with 3.54 in my answer.

p-value = 2 * P(Z > 3.54) = 2*0 = 0.

Compare p-value with 0.01. Since p-value < 0.01, we reject the null hypothesis and conclude that there is evidence to show the proportion is NOT 15%.

Customer reply replied 3 years ago

p-value = 2*(Z>3.54)=0<0.01.is that correct?

Yes correct.

Customer reply replied 3 years ago

or is it this?p-value = 2 * (Z > 3.54) = 2*0 = 0.

p-value = 2 * P(Z > 3.54) = 2 * 0 = 0 < 0.01.

Don't forget letter "P" which means "probability of".

Customer reply replied 3 years ago

just wondering, how did we get a P-value of 2? Im going to have to repeat a question very similar and will probably need to know

p-value is a probability. It must be greater than 0 and less than 1. For this problem, we first compute Z statistic 3.54, then find the probability Z > 3.54, which is approximately 0.

Since it's a two tailed test, p-value 2 * P(Z > 3.54), which is approximately 2 * 0 = 0. So the p-value is approximately 0, not 2.

Customer reply replied 3 years ago

ohhh! okay. so the 2 is just there because it was a two tail test?

Yes. p-value is equal to TWICE of P(Z > 3.54), which is 2 times probability of Z > 3.54.

Customer reply replied 3 years ago

would draw a bell shape and each tail put in .0050 percent shaded, correct?

Correct for number 9.

Method 1. Use p-values as we did above.

Method 2. Use critical values as following.

The cutoff points (called critical values) are -2.58 and 2.58 for the standard normal distribution. Since the Z statistic is 3.54 > 2.58 (in the shaded area), we reject the null hypothesis.

-2.58 and 2.58 can be found in a table using 0.005 or 0.995.

Customer reply replied 3 years ago

perfect! and one more thing on 90.25- 2.58*√((0.25*(1-0.25))/160)=0.1617. when i do this equation, i come up with -0.0798, but you have 0.1617... what am i doing wrong?my up is .0969, yours is .3383

We can check one step at a time.

0.25 * 0.75/ 160 = 0.001171875.

Square root of 0.001171875 = 0.03423266.

2.58 * 0.03423266 = 0.088

Lower bound: 0.25 - 0.088 = 0.162.

Upper bound: 0.25 + 0.088 = 0.338.

Customer reply replied 3 years ago

ohhhh. okay. I did .25 - 2.58 and then did the numbers in the squar root and then multiplied -2.33 by 0.03423266we always add and remove that P-hat at the very end?what the values represent? are they are % or Z statistics? should i leave them at 4 dec?

For computation by order of operation we need to do the multiplication before subtraction.

We use the p-hat 0.25 to obtain the lower and upper bounds. 0.25 is only a point obtained from the sample. 0.1617 and 0.3383 are proportions like 0.25. They give us an interval (0.1617, 0.3383). Now we can say that with 99% confidence the population proportion is somewhere between 0.1617 and 0.3383.

Using 4 decimals is enough for most homework questions like this. You may not need that many decimal places if the question does not ask you.

Customer reply replied 3 years ago

if I had to plot that on a bell, I could find the corisponding z statistics to plot them? Wouldn't they both be on the left side of the mean; and is that okay?

Obtaining a point estimate is important but usually not enough. We cannot say the population proportion is 0.25 only because a sample of n = 160 gives us this estimate. We need a lower and upper bounds to give us a better idea of where the population proportion is.

I don't think it's necessary to plot a graph for the lower and upper bound.

Customer reply replied 3 years ago

okay, Perfect!annnd ONNNNE last question before i take a LOOOONG break. The p-value = 0.35 in question 10... is that just the rounded up number from the Standev?

The question says "greater than". So it's a one tailed test.

degrees of freedom is 8-1= 7.

p-value = P(T > 0.4082) = 0.35. It's a probability rounded from 0.3476621. I used a computer software to obtain this, you can use Excel.

The correct Excel command is T.DIST.RT(0.4082, 7)

Customer reply replied 3 years ago

Thank you for all the help! I'll be back on later.

You are welcome. If you open a new post for other questions you may write "For John only" at the beginning of your post so that the questions are directed to me.

Customer reply replied 3 years ago

Quick question on 9 again. since the Z score is 3.54, wouldn't the answer be .9999, and not .0000?

.9999 is the probability that Z < 3.54. But we are looking for the tail probability P(Z > 3.54). So it's

1- 0.9999 = 0.00001.

We always find tail probabilities for p-values.

Customer reply replied 3 years ago

Would you be able to help me with a few questions from the 40 question one? They are all similar, so when I get one I should get them all

Sure. which questions do you need help with?

Customer reply replied 3 years ago

Question number 24, is there any need to keep the Margin of error formula, because it appears it is already given to us?

We need to use the formula, there are 4 variables: margin of error, z, p-hat and n. Three of them are given and we need to solve for n.

Customer reply replied 3 years ago

right, but isn't this the formula to solve for n=(z^2*p*(1-p))/E^2we have

Z = 2.58

p ̂ = 0.12

E = .04

Z = 2.58

p ̂ = 0.12

E = .04

Yes, it's correct.

n=(z^2*p*(1-p))/E^2

is obtained from

E = z * sqrt(p(1-p)/n).

These two are equivalent. I like to memorized the original one E = z * sqrt(p(1-p)/n) and solve for the n whenever I have to. You can of course memorize both.

Customer reply replied 3 years ago

Okay. can you explain how you came to the answer on 25? it seems simple enough, I'm just not sure how you did it, hahaalso, if you are able to continue to help me finish all these questions so that I have a decent understand of how you got the answers, ill gladly submit another bonus!

No problem. I will continue assist you for all these questions.

For question 25, the population standard deviation is unknown, so we need to use a T distribution (otherwise, use normal distribution).

Degrees of freedom = n - 1 = 10-1 = 9.

Now use Excel command "=T.INV.2T(0.1, 9)" to find the critical number.

Customer reply replied 3 years ago

got it, thanks!

You are welcome.

Customer reply replied 3 years ago

can you tell me the actually formula that was used in 26 in order to get n=(〖1.96〗^2*〖500〗^2)/〖135〗^2 =52.7.or just, what is the symbol that the 500^2 replaced?Because I have no idea what H stands for

For this question. the population standard deviation is known to be

sigma = $500.

It must be a typo and should be population standard deviation sigma.

So use a normal distribution.

The formula is

E = z * sigma / sqrt(n).

Solve for n to obtain

n = z^2 * sigma^2 / E^2 = 1.96^2 * 500^2 / 135^2.

Just like question 24, we have a formula for margin of error. If all variables are given except the sample size n, we can solve for n.

Customer reply replied 3 years ago

in 28, if the confidence level is 95%, should the critical value be 1.96?

Customer reply replied 3 years ago

Or is it a crit val because its a T value. .05 in two tails with 9 degrees of freedom?

Like other problems, the population standard deviation is not known, so we have to use a T distribution. Degrees of freedom = n-1 = 10-1 = 9. Confidence level = 95%.

In Excel, Use "=T.INV.2T(0.05, 9)" to obtain the critical value.

If we knew population standard deviation and could use a normal distribution, we would use 1.96.

Customer reply replied 3 years ago

Again, THANK YOU! I'm dying over here. Could you explain how you got the answer on 34?

It's a two tailed test because of the "not equal to" sign. In Excel type "=2 * NORM.DIST(-3.06,0,1,1)" to obtain the p-value.

We are looking for

p-value = P(X < -3.06) + P(X > 3.06) since it's a two tailed test.

Normal distribution is symmetric and so the above two tailed probabilities are equal to each other. So

p-value = 2 * P(X < -3.06) = "2 * NORM.DIST(-3.06,0,1,1)" in Excel.

Customer reply replied 3 years ago

Thank you!!!!!If I submitted my complete (I re did it so I understood it) 10 question would you proof check it? I'll gladly pay a bonus!

Customer reply replied 3 years ago

Im going to go to sleep, my finals is tomorrow and i have been doing math for over 34 straight hours... im going to attach a file in case you read this later, if you are able to check it and provide any tips (such as better excel formats or something) id gladly pay another bonus!

Customer reply replied 3 years ago

Or if you know how to do the histogram or impericle bell using word, i'll DEFINITELY pay extra for you to have that add in as needed. My professor said he wants to see a bell every time. Either way, thank you for all of your help!OR! a spread sheet with all the formulas and meanings of symbols used in the 2 documents! literally, name your price for those things and I'll gladly pay it. After tomorrow I dont ever want to have to do statistics again, hahaha, so im willing to do anything i can to just be done with it.

How about a bonus of $40?

Customer reply replied 3 years ago

Sounds good to me!But I leave for my finals at noon. So that's in 3 hours and 40 minutes.

https://www.mediafire.com/folder/5umdd9nbve4m8ft,iasig425f1jdzg9/shared

The above link is for the WORD file and EXCEL file. I will post a formula sheet shortly.

Customer reply replied 3 years ago

On question 5, when i use the excel the answer is .7202, but when i use the Z chart, its .7190. Which one is correct?in question 8, the problem is to test the claim that "weights is less than 200 lb" so shouldnt the claim be "p < 200 lb"

For question 5, both are correct.

1. In excel, it does not round (33-32.3)/1.2 to two decimal places and gives the exact answer P(Z < (33-32.3)/1.2)

2. When you use the chart, you have to round 0.7/1.2 to 0.58. The chart gives P(Z < 0.58)

I think both are acceptable.

For question 8, the alternative hypothesis is p < 200lb. I crossed out "(1) p>200" and kept "(2) p<200".

Customer reply replied 3 years ago

okay, im going to leave both, never hurts to have extra to study.for step 1, we have to write out the claim. just the claim. step 2 he wants us to do the Ho and H1

Then the claim is p<200.

Customer reply replied 3 years ago

here are the two charts he has us using to complete 8-10

Customer reply replied 3 years ago

He said he wants us to complete ALL the steps, not just get the answer correctly. So should i delete the stuff you put in a line in or just keep it?for the T.dist in excel, he should us how to do a right tail, 2 tail and left tail (normal). So should i change the excels to fit that? the numbers change

In the first step, write the alternative hypothesis.

I have include all Excel command in my Excel file above. Again using Excel or a chart will give you different numbers but they should be very close. I don't know what your instructor wants but just follow his/her instructions.

Customer reply replied 3 years ago

1. Identify the Claim and write it in symbolic form

2. Give the symbolic form for the Null and Alternative Hypotheses.

2. Give the symbolic form for the Null and Alternative Hypotheses.

1. p<200

2. H0: p=200

Customer reply replied 3 years ago

Yea, for some reason I can mimic your commands, when i inpute them they are invalid, it wont allow me to us a negative number in the formula on excel

1. p<200

2. H0: p=200

Ha: p<200.

For which problem in Excel?

Customer reply replied 3 years ago

any of the ones that have a negative in the x value for the formula, mine will just say invalid. but when i just put the number in as a positive, i get the same results as you

Which version of Excel are you using? I am using 2013, maybe you are using an old version?

Customer reply replied 3 years ago

Im not sure. but thats probably it. As long as the answer is coming out the same, im happy. hahain questions 10, you got a 1.41 for crit value using excel, when i looked up on the Z chart i get a .84

For question 9 and 10, write the alternative hypothesis Ha in (1).

9. (1) p (is not equal to) 0.15

10. (1). mu > 14.

For 10, we have a T distribution (not normal). So use a T chart, not Z chart.

Customer reply replied 3 years ago

ohhhh. Gotcha! why are we using t? I know there was a rule, like when the stand dev is known or something, correct? i cant remember off the top of my head

When the population (not sample) standard deviation is known, we use normal Z chart (this is better when we have more information).

Otherwise use a T distribution (and T chart) with degrees of freedom n-1. Not as good as normal since we have less information.

The formula sheet can be downloaded using the following link.

http://www.mediafire.com/file/mgi44ynslyc36e4/Formula_Sheet.doc.docx

Please let me know if you have any questions. Thanks.

Customer reply replied 3 years ago

For question 8-10, how do i find the actual Z statistic for the 2nd part of the questions? the low and highs. I have to plot them.for the cheat sheet, could you add a thing at the bottom that just says what each symbol means? I get confused with X and x bar and stuff like that. I'll make it $50 since i keep asking for so much. Thank you so much for all the help!for question 8, part 2. The Z scores for the low and high would be -.21 and .21 correct?Or does the critical value determine where the actually lines are?

For question 8-10, all the computation for finding test statistics are correct in the file.

I think you are asking the critical values (not test statistic) for drawing graphs.

Yes critical values determine where the actual lines are.

Customer reply replied 3 years ago

will you look at question 8 and see if i did it correctly before i move on and chart them all? I will shade in the tails after i print it off.

For question 8, it's a left tailed Z test. The critical value is =NORM.INV(0.05,0,1) = -1.644853627.

Since the test statistic is -1.13, not in the shaded region, fail to reject the null hypothesis.

Customer reply replied 3 years ago

yea, i just wanted to make sure i drew the lines in the right place and labeled them correctly under itThe answers are good, just making sure i depict that information correctly on the bell shape

Step (1), (2) are okay.

Graph looks okay, need to shade the area to the left of -1.645.

In (4), just write "Z critical value = -1.645" and delete everything else. The value should be the same as the value on your graph. This is a normal distribution, not T distribution.

(5) is okay.

(6). Correct but you may want to write since -1.13 > -1.645, we fail to reject the null hypothesis. This is equivalent to using the p-value but it seems your instructor wants this. It doesn't hurt to write both (and get the same conclusion).

All the rest are okay.

Customer reply replied 3 years ago

perfect, fixed it.First half of questions 9, i would plot the Z values all the way to one side?

It's a two sided test. Need to find two critical values and plot two shaded regions.

First, it's a Z test. Use 0.01 significance level. On each side we have 0.01/2 = 0.005 area shaded.

To find the two critical values, use =NORM.INV(0.005,0,1) to find -2.58. So the two critical values are -2.58 and 2.58.

Customer reply replied 3 years ago

yea, the shaded areas are on both size, but where do i plot the answer? which i just fixed because it was a T dis, not a norm.dis

It's a normal distribution, not T distribution.

The Z statistic is computed as 3.54. So it's to the right of 2.58.

Customer reply replied 3 years ago

yea, sorry. What i meant to say is I had it at a T.dis and i had to change it to a Norm.dis

No problem.

Customer reply replied 3 years ago

Z score is plotted all the way to the right, correct? inside the shaded area, yes?

Label 3 points, -2.58, 2.58, 3.54. Shade two areas, one before -2.58 and one after 2.58.

Customer reply replied 3 years ago

here is my answer, p-value = 2*=NORM.DIST(3.54,0,1,TRUE)=2*.0002= .0004< .01 after the edit on the Norm.dist, is that correct?Im lost again. Sorry. I used the T table chart that he gave me, and got a T value of = 2.708, not 2.58

Question 9 is about the proportion. We use Z test (not T test) and Z chart (not T chart).

Customer reply replied 3 years ago

ohhh! got it! So i can delete eveything in 4 and just put Z crit value is =2.575? (he gave us that to use specifically for a 99% value)

Yes. But we have two critical values: -2.575 and 2.575 since it's a two tailed test.

Method 1. p-value = 2 * NORM.DIST(-3.54, 0, 1, 1) = 0.0004 < 0.01. So reject the null.

Method 2. 3.54 > 2.58 and is in the shaded area, so reject the null.

Again always use method 2 since your instructor asked that. You can use both of course.

Customer reply replied 3 years ago

perfect! I think i got it

Actually there should be 4 points on the graph: -2.575, 2.575, -3.54, 3.54.

-2.575 and 2.575 are cutoff points used as standards.

3.54 is computed Z test statistics. But we need to have -3.54 since it's a two tailed test.

You may want to label 3.54 ONLY as "Z test statistic".

-3.54 is not computed from data but is used to obtain the p-value. In the command 2 * NORM.DIST(-3.54, 0, 1, 1), we are computing

The area to the left of -3.54 + The area to the right of 3.54.

The are same and so we do 2 times one of them.

Customer reply replied 3 years ago

Im not sure how to plot the first part of question 1. I have α = .005 on the right side as the rejection are. But do i put the T score also?

On the right 0.005, on the left also 0.005. Label -2.575, 2.575 and 3.54.

Customer reply replied 3 years ago

but isnt question 10 just a right tail?

Sorry I thought you were talking about question 9.

Customer reply replied 3 years ago

I think i am good on 9.Just not sure how to label the Critical value on 10 when plotting it

The significance level is not given in the problem. So use 0.05. The critical value is 1.895.

Customer reply replied 3 years ago

that critical is a Z value, correct?

Use "=T.INV(0.95,7)" in Excel since the area to the left is 0.95 and the degrees of freedom is 8-1 = 7.

It's a t distribution (not Z) since the population standard deviation is NOT known.

Shade the area to the right of 1.895. The area is 0.05.

Customer reply replied 3 years ago

The critical value is known, "Test the claim at the 0.10 significance level" so will crit be 1.645?

Also label the test statistic 0.41 on the graph.

In (6). Write

Method 1: Since 0.41 < 1.895 and is not in the shaded region, we fail to reject the null hypothesis.

Method 2 (optional): Since p-value = 0.35 > 0.05, we fail to reject the null hypothesis.

I only found the following

**Question (10)**

**A cereal company claims that the mean weight of the cereal in its packets is greater than 14 oz. The weights (in ounces) of the cereal in a random sample of 8 of its cereal packets are listed below. 14.6 13.8 14.1 13.7 14.0 14.4 13.6 14.2**

**If the question says "**Test the claim at the 0.10 significance level" then the critical value is =T.INV(0.9,7) = 1.415.

1.645 comes from the standard normal distribution, but we are doing a T test and should use T chart.

On the graph, shade the region to the right of 1.419 and label the test statistic 0.41

In (6). Write

Method 1: Since 0.41 < 1.415 and is not in the shaded region, we fail to reject the null hypothesis.

Method 2 (optional): Since p-value = 0.35 > 0.10, we fail to reject the null hypothesis.

Customer reply replied 3 years ago

Here is what I have. Im confused on part 1 of 10. i THINK the other ones are good. (ill shade in the rejections areas after i print off)

Customer reply replied 3 years ago

As soon as this is complete, im going to send the bonus and head to school.

(1)(2) are okay.

For the graph, the critical value is 1.415, not 1.645. This value 1.415 should be same as the one in (4)

(4) Delete the last "=1.415".

(5) okay

(6)

Write

Method 1: Since 0.41 < 1.415 and is not in the shaded region, we fail to reject the null hypothesis.

Method 2 (optional): Since p-value = 0.35 > 0.10, we fail to reject the null hypothesis.

In (6), always include Method 1 since your instructor asked it.

This is what your instructor says

6. If the test statistic (value from step 4) lies inside the critical region (shaded) the Reject the null hypothesis. Otherwise, Fail to Reject the null hypothesis.

This means compare the test statistic with critical value, in question 10, compare 0.41 with 1.415.

Customer reply replied 3 years ago

i dont really need P value in there, do i?

If the question doesn't ask you and you want to do what your instructor wants, then yes. Comparing the test statistic with critical value is enough.

Customer reply replied 3 years ago

just to be sure, that critical value is label as a T statistic on the actual bell chart? because we never did that. we just did value

Obtaining p-value and compare p-value with significance level is another way to do it.

It's actually the preferred way in statistical analysis but in this exam you don't have to do it UNLESS the question ask you.

The critical value is labelled at "T critical value", it's 1.415.

"T statistic" is 0.41, computed from the data.

For other problems like question 9, we may have "Z critical value" and "Z statistic".

Customer reply replied 3 years ago

just to be sure, that critical value is label as a T statistic on the actual bell chart? because we never did that. we just did value. Either way I have to head to school to take my finals! Can i attach my final draft? will you just confirm i didnt make any mistakes when making those changes? thanks so much! im going to send your bonus now.

Everything in problem 10 looks fine.

Customer reply replied 3 years ago

Last time all i had to do was rate you and i was able to pay the bonus. How do i do another bonus? Thats not an option this time.

I don't know. You may ask the customer service after your exam.

Customer reply replied 3 years ago

Okay, i will! Please feel free to make any suggested changes in red on the actual word document, i feel like every time i go to edit i i mess it up, hahah. THANK YOU SO MUCH FOR YOUR HELP!

You are welcome.

Customer reply replied 3 years ago

On questions 8-10, the 2nd part when they ask to construct a confidence values, is it the middle or ends that are shaded???

Middle for confidence interval and tail (or tails) for hypothesis test.

Customer reply replied 3 years ago

Thank you.

You are welcome.

Customer reply replied 3 years ago

i think the best way to pay you is just start a new question? How do i go about doing that and making sure i get you?

It's hard. How about you call the customer services for paying another bonus for this post?

Customer reply replied 3 years ago

okay.Before i do that, I have extra credit that I have to have due by midnight tonight. Its on line and im not sure how many questions its going to be. If i call and pay $100 right now, will you stay on line and answer the questions with me? I dont need proof of working or anything, but I can only ask the Questions one at a time.

Is it still this course? Can you let me know the start and end time?

Customer reply replied 3 years ago

same course. I can start when ever works for you,but i have to be done in 6 1/2 hours from now.the first part will be typing in actually answers, after that its multiple choiceHe said the more questions you get correct, it auto adjusts and the less questions you will be asked. But if you struggle in an area, it will keep prompting you to repeat similar questionsIf you agree, I can call and have the money sent now and we can start when you are ready.He said depending on how many we get correct, it will be between 50-75 questions, most of which will be multiple choice. So ill let you decide what a fair price is for that to be complete, plus the $50 i owe you from this morning.

We can start right now. 50+50 is okay for me but please figure out how to pay another bonus on your side.

Customer reply replied 3 years ago

An online site presented this question, "Would the recent norovirus outbreak deter you from taking a cruise?" Among the 34 comma 40434,404 people who responded, 6666% answered "yes." Use the sample data to construct a 9595% confidence interval estimate for the proportion of the population of all people who would respond "yes" to that question. Does the confidence interval provide a good estimate of the population proportion?

___ <p<____(Round to three decimal places as needed.)

Does the confidence interval provide a good estimate of the population proportion?

A.

Yes, the sample is large enough to provide a good estimate of the population proportion.

B.

No, the responses are not independent.

C.

Yes, all the assumptions for a confidence interval are satisfied.

D.

No, the sample is a voluntary sample and might not be representative of the population.Two part. Need the blanks filled and the multiple choice as well.

___ <p<____(Round to three decimal places as needed.)

Does the confidence interval provide a good estimate of the population proportion?

A.

Yes, the sample is large enough to provide a good estimate of the population proportion.

B.

No, the responses are not independent.

C.

Yes, all the assumptions for a confidence interval are satisfied.

D.

No, the sample is a voluntary sample and might not be representative of the population.Two part. Need the blanks filled and the multiple choice as well.

The numbers repeat. Can you check?

Customer reply replied 3 years ago

An online site presented this question, "Would the recent norovirus outbreak deter you from taking a cruise?" Among the 34,404 people who responded, 66% answered "yes." Use the sample data to construct a 95% confidence interval estimate for the proportion of the population of all people who would respond "yes" to that question. Does the confidence interval provide a good estimate of the population proportion?Sorry. I will pay more attention. Im on line with costumer support right now as well, so hopefully it will be submitted soon.They said they had to forward the request to another department because I have already paid a bonus, and that could take 24-48 hours.

No problem. The answer is 0.655, 0.665. Choice D

Customer reply replied 3 years ago

This next one is a three piece, ill give it to you in sections.Use the sample data and confidence level given below to complete parts (a) through (d).

A research institute poll asked respondents if they acted to annoy a bad driver. In the poll, n equals 2491, and x equals 1164, who said that they honked. Use a 90 % confidence level.8203;a) Find the best point estimate of the population proportion p.

(Round to three decimal places as needed.)b) Identify the value of the margin of error E.

E =

(Round to four decimal places as needed.)

c) Construct the confidence interval.__<P<___8203;d) Write a statement that correctly interprets the confidence interval. Choose the correct answer below.A. 90% of sample proportions will fall between the lower bound and the upper bound.

B. One has 90% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.

C. There is a 90% chance that the true value of the population proportion will fall between the lower bound and the upper bound.

D. One has 90% confidence that the sample proportion is equal to the population proportion.

A research institute poll asked respondents if they acted to annoy a bad driver. In the poll, n equals 2491, and x equals 1164, who said that they honked. Use a 90 % confidence level.8203;a) Find the best point estimate of the population proportion p.

(Round to three decimal places as needed.)b) Identify the value of the margin of error E.

E =

(Round to four decimal places as needed.)

c) Construct the confidence interval.__<P<___8203;d) Write a statement that correctly interprets the confidence interval. Choose the correct answer below.A. 90% of sample proportions will fall between the lower bound and the upper bound.

B. One has 90% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.

C. There is a 90% chance that the true value of the population proportion will fall between the lower bound and the upper bound.

D. One has 90% confidence that the sample proportion is equal to the population proportion.

Customer reply replied 3 years ago

Just wanted to make sure you received the next question

Point estimate: 0.467

E = 0.0164.

CI: 0.4508393, 0.4837251 (don't know the number of decimal places)

Choice B

Customer reply replied 3 years ago

Three. sorry again, ill pay more attention when submitting the question. So it would be 0.451, 0.484 correct?A genetic experiment with peas resulted in one sample of offspring that consisted of 448 green peas and 174 yellow peas.a. Construct a 95% confidence interval to estimate of the percentage of yellow peas.

b. It was expected that 25% of the offspring peas would be yellow. Given that the percentage of offspring yellow peas is not 25%, do the results contradict expectations?Question a is rounded to three decimal points

b. It was expected that 25% of the offspring peas would be yellow. Given that the percentage of offspring yellow peas is not 25%, do the results contradict expectations?Question a is rounded to three decimal points

Yes. 0.451 and 0.484

a. 0.244, 0.315

b. NO.

Customer reply replied 3 years ago

2 out of 3 was Good enough to get through that part! haha. one sec while it loads the next sectionFind the sample size, n, needed to estimate the percentage of adults who have consulted fortune tellers. Use a 0.04 margin of error, use a confidence level of 90%, and use results from a prior poll suggesting that 13% of adults have consulted fortune tellers.A) n = ____ (round up to the nearest integer)

192

Customer reply replied 3 years ago

Use the given data to find the minimum sample size required to estimate a population proportion or percentage.Margin of error: 0.07; confidence level 95%; p^(p-hat) and q^ (q-hat) unknownn = ___ (Round up to the nearest integer.)

196

Customer reply replied 3 years ago

Many states are carefully considering steps that would help them collect sales taxes on items purchased through the Internet. How many randomly selected sales transactions must be surveyed to determine the percentage that transpired over the Internet? Assume that we want to be 99% confident that the sample percentage is within five percentage points of the true population percentage for all sales transactions.n =_____ (Round up to the nearest integer.)

664

Customer reply replied 3 years ago

A clinical trial was conducted using a new method designed to increase the probability of conceiving a girl. As of this writing, 934 babies were born to parents using the new method, and 861 of them were girls. Use a 0.05 significance level to test the claim that the new method is effective in increasing the likelihood that a baby will be a girl. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method and the normal distribution as an approximation to the binomial distribution.A) H0: p>0.5

H1: p=0.5B) Ho: p≠0.5

H1: p=0.5C. H0: p=0.5

H1: p>0.5D. H0: p=0.5

H1: p≠0.5E. H0: p=0.5

H1: p<0.5F. H0: p< 0.5

H1: = 0.52) What is the test statistic?

z = ________ (round to 2 decimal places)3) What is the P-value?

P-value = ______ (Round to four decimal places as needed.)

H1: p=0.5B) Ho: p≠0.5

H1: p=0.5C. H0: p=0.5

H1: p>0.5D. H0: p=0.5

H1: p≠0.5E. H0: p=0.5

H1: p<0.5F. H0: p< 0.5

H1: = 0.52) What is the test statistic?

z = ________ (round to 2 decimal places)3) What is the P-value?

P-value = ______ (Round to four decimal places as needed.)

1) C

2) 25.78

3) 0.0000

Customer reply replied 3 years ago

4) What is the conclusion about the null hypothesis?

A. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, α.

B. Reject the null hypothesis because the P-value is greater than the significance level, α.

C. Reject the null hypothesis because the P-value is less than or equal to the significance level, α.

D. Fail to reject the null hypothesis because the P-value is greater than the significance level, α.Is 2 correct? seems really high

A. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, α.

B. Reject the null hypothesis because the P-value is greater than the significance level, α.

C. Reject the null hypothesis because the P-value is less than or equal to the significance level, α.

D. Fail to reject the null hypothesis because the P-value is greater than the significance level, α.Is 2 correct? seems really high

Yes because 861/934 = 0.92, it's very large compared to 0.5. Also 934 is very large.

So intuitively with a very large sample size and a very high sample proportion, the evidence is very strong, meaning that we have a very high Z statistic and extremely small p-value.

Customer reply replied 3 years ago

sounds good to me! hahafar is A, correct?5) What is the final conclusion?

A. There is sufficient evidence to support the claim that the new method is effective in increasing the likelihood that a baby will be a girl.B. There is not sufficient evidence to warrant rejection of the claim that the new method is effective in increasing the likelihood that a baby will be a girl.C. There is sufficient evidence to warrant rejection of the claim that the new method is effective in increasing the likelihood that a baby will be a girl.D. There is not sufficient evidence to support the claim that the new method is effective in increasing the likelihood that a baby will be a girl.

A. There is sufficient evidence to support the claim that the new method is effective in increasing the likelihood that a baby will be a girl.B. There is not sufficient evidence to warrant rejection of the claim that the new method is effective in increasing the likelihood that a baby will be a girl.C. There is sufficient evidence to warrant rejection of the claim that the new method is effective in increasing the likelihood that a baby will be a girl.D. There is not sufficient evidence to support the claim that the new method is effective in increasing the likelihood that a baby will be a girl.

4) Should be C.

5) A

Customer reply replied 3 years ago

n a recent poll of 740 randomly selected adults, 587 said that it is morally wrong to not report all income on tax returns. Use a 0.01 significance level to test the claim that 70% of adults say that it is morally wrong to not report all income on tax returns. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method. Use the normal distribution as an approximation of the binomial distribution.1)A) H0: p=0.7

H1: p<0.7B. H0: p=0.3

H1: p>0.3C. H0: p=0.7

H1: p≠0.7D. H0: p=0.3

H1: p<0.3E. H0: p=0.3

H1: p≠0.3F. H0: p=0.7

H1: p>0.72)

The test statistic is Z = _____ (Round to two decimal places as needed.)3)

The P-value is = _______ Round to four decimal places as needed.)4)

A) Fail to reject (H0)

B) Reject (H0)5)

A) There IS sufficient evidence to warrant rejection of the claim that 7070% of adults say that it is morally wrong not to report all income on tax returns.B) There IS NOT sufficient evidence to warrant rejection of the claim that 7070% of adults say that it is morally wrong not to report all income on tax returns.

H1: p<0.7B. H0: p=0.3

H1: p>0.3C. H0: p=0.7

H1: p≠0.7D. H0: p=0.3

H1: p<0.3E. H0: p=0.3

H1: p≠0.3F. H0: p=0.7

H1: p>0.72)

The test statistic is Z = _____ (Round to two decimal places as needed.)3)

The P-value is = _______ Round to four decimal places as needed.)4)

A) Fail to reject (H0)

B) Reject (H0)5)

A) There IS sufficient evidence to warrant rejection of the claim that 7070% of adults say that it is morally wrong not to report all income on tax returns.B) There IS NOT sufficient evidence to warrant rejection of the claim that 7070% of adults say that it is morally wrong not to report all income on tax returns.

1) C

2) 5.54

3) 0.0000

4) B

5) A

Customer reply replied 3 years ago

A recent sports game set a record for the number of television viewers. The game had a share of 77%, meaning that among the television sets in use at the time of the game, 77% were tuned to the game. The sample size is 25,303 households. Use a 0.01 significance level to test the claim that more than 74% of television sets in use were tuned to the sports game. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method and the normal distribution as an approximation to the binomial distribution.1)

A. H0: p=0.74

H1: p>0.74B. H0: p=0.74

H1: p<0.74C. H0: p=0.74

H1: p≠0.74D. H0: p>0.74

H1: p=0.74E. H0: p<0.74

H1: p=0.74F. H0: p≠0.74

H1: p=0.742)

z = ____ (Round to two decimal places as needed.)3)

P-value = ____

(Round to four decimal places as needed.)4)

What is the conclusion on the null hypothesis?A) Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, α.B) Reject the null hypothesis because the P-value is less than or equal to the significance level, α.C) Fail to reject the null hypothesis because the P-value is greater than the significance level, α.D) Reject the null hypothesis because the P-value is greater than the significance level, α5)

A. There is not sufficient evidence to support the claim that more than 74% of television sets in use were tuned to the sports game.B. There is not sufficient evidence to warrant rejection of the claim that more than 74% of television sets in use were tuned to the sports game.C. There is sufficient evidence to warrant rejection of the claim that more than 74% of television sets in use were tuned to the sports game.D.There is sufficient evidence to support the claim that more than 74% of television sets in use were tuned to the sports game.

A. H0: p=0.74

H1: p>0.74B. H0: p=0.74

H1: p<0.74C. H0: p=0.74

H1: p≠0.74D. H0: p>0.74

H1: p=0.74E. H0: p<0.74

H1: p=0.74F. H0: p≠0.74

H1: p=0.742)

z = ____ (Round to two decimal places as needed.)3)

P-value = ____

(Round to four decimal places as needed.)4)

What is the conclusion on the null hypothesis?A) Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, α.B) Reject the null hypothesis because the P-value is less than or equal to the significance level, α.C) Fail to reject the null hypothesis because the P-value is greater than the significance level, α.D) Reject the null hypothesis because the P-value is greater than the significance level, α5)

A. There is not sufficient evidence to support the claim that more than 74% of television sets in use were tuned to the sports game.B. There is not sufficient evidence to warrant rejection of the claim that more than 74% of television sets in use were tuned to the sports game.C. There is sufficient evidence to warrant rejection of the claim that more than 74% of television sets in use were tuned to the sports game.D.There is sufficient evidence to support the claim that more than 74% of television sets in use were tuned to the sports game.

1)A

2) 10.88

3) 0.0000

4) B

5) D

Customer reply replied 3 years ago

A certain drug is used to treat asthma. In a clinical trial of the drug, 16 of 265 treated subjects experienced headaches (based on data from the manufacturer). The accompanying calculator display shows results from a test of the claim that less than 99% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.05 significance level to complete parts (a) through (e) below.1)Is the test two-tailed, left-tailed, or right-tailed?Right tailed testTwo-tailed testLeft-tailed test2)

Z = _____ (Round to two decimal places as needed.)3)

P-Value = ______ (Round to four decimal places as needed.)4)

Identify the null hypothesis.

A. H0: p≠0.09B. H0: p<0.09C. H0: p=0.09D. H0: p>0.095)

Decide whether to reject the null hypothesis. Choose the correct answer below.A. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, α.B. Fail to reject the null hypothesis because the P-value is greater than the significance level, α.C. Reject the null hypothesis because the P-value is less than or equal to the significance level, α.D. Reject the null hypothesis because the P-value is greater than the significance level, α.6)

A. There is sufficient evidence to warrant rejection of the claim that less than 99% of treated subjects experienced headaches.B. There is not sufficient evidence to warrant rejection of the claim that less than 99% of treated subjects experienced headaches.C. There is not sufficient evidence to support the claim that less than 99% of treated subjects experienced headaches.D. There is sufficient evidence to support the claim that less than 99% of treated subjects experienced headaches.I had no idea each question was going to be 4-6 questions apiece, im sorry! I'll add in some more money when they get back to me!

Z = _____ (Round to two decimal places as needed.)3)

P-Value = ______ (Round to four decimal places as needed.)4)

Identify the null hypothesis.

A. H0: p≠0.09B. H0: p<0.09C. H0: p=0.09D. H0: p>0.095)

Decide whether to reject the null hypothesis. Choose the correct answer below.A. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, α.B. Fail to reject the null hypothesis because the P-value is greater than the significance level, α.C. Reject the null hypothesis because the P-value is less than or equal to the significance level, α.D. Reject the null hypothesis because the P-value is greater than the significance level, α.6)

A. There is sufficient evidence to warrant rejection of the claim that less than 99% of treated subjects experienced headaches.B. There is not sufficient evidence to warrant rejection of the claim that less than 99% of treated subjects experienced headaches.C. There is not sufficient evidence to support the claim that less than 99% of treated subjects experienced headaches.D. There is sufficient evidence to support the claim that less than 99% of treated subjects experienced headaches.I had no idea each question was going to be 4-6 questions apiece, im sorry! I'll add in some more money when they get back to me!

No problem.

1) Left

2)-1.69

3) 0.0455

4) C

5) C

6) D

Customer reply replied 3 years ago

A survey of 1,641 randomly selected adults showed that 527 of them have heard of a new electronic reader. The accompanying technology display results from a test of the claim that 34% of adults have heard of the new electronic reader. Use the normal distribution as an approximation to the binomial distribution, and assume a 0.05 significance level to complete parts (a) through (e).1. Is the test two-tailed, left-tailed, or right-tailed?Right tailed testTwo-tailed testLeft-tailed testB)

Z = _____ (Round to two decimal places as needed.)C)

P-Value = ______ (Round to four decimal places as needed.)4)

Identify the null hypothesis.

A. Ho: p>0.34

B. Ho: p<0.34

C. H0: p=0.34

D. Ho: p≠ 0.345)

A. Reject the null hypothesis because the P-value is greater than the significance level, α.

B. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, α.

C. Reject the null hypothesis because the P-value is less than or equal to the significance level, α.

D. Fail to reject the null hypothesis because the P-value is greater than the significance level, α6)

A. There is sufficient evidence to support the claim that 34% of adults have heard of the new electronic reader.

B. There is not sufficient evidence to warrant rejection of the claim that 34% of adults have heard of the new electronic reader.

C. There is sufficient evidence to warrant rejection of the claim that 34% of adults have heard of the new electronic reader.

D. There is not sufficient evidence to support the claim that 34% of adults have heard of the new electronic reader

Z = _____ (Round to two decimal places as needed.)C)

P-Value = ______ (Round to four decimal places as needed.)4)

Identify the null hypothesis.

A. Ho: p>0.34

B. Ho: p<0.34

C. H0: p=0.34

D. Ho: p≠ 0.345)

A. Reject the null hypothesis because the P-value is greater than the significance level, α.

B. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, α.

C. Reject the null hypothesis because the P-value is less than or equal to the significance level, α.

D. Fail to reject the null hypothesis because the P-value is greater than the significance level, α6)

A. There is sufficient evidence to support the claim that 34% of adults have heard of the new electronic reader.

B. There is not sufficient evidence to warrant rejection of the claim that 34% of adults have heard of the new electronic reader.

C. There is sufficient evidence to warrant rejection of the claim that 34% of adults have heard of the new electronic reader.

D. There is not sufficient evidence to support the claim that 34% of adults have heard of the new electronic reader

1. Two

2. -1.61

3. 0.1074

4. C

5. D

6. B

Customer reply replied 3 years ago

An IQ test is designed so that the mean is 100 and the standard deviation is 16 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of statistics students such that it can be said with 90% confidence that the sample mean is within 7 IQ points of the true mean. Assume that σ=16 and determine the required sample size using technology. Then determine if this is a reasonable sample size for a real world calculation.1)

The required sample size is _____. (Round up to the nearest integer.)2) Would it be reasonable to sample this number of students?A) No. This number of IQ test scores is a fairly small number.

B) Yes. This number of IQ test scores is a fairly large number.

C) Yes. This number of IQ test scores is a fairly small number.

D) No. This number of IQ test scores is a fairly large number.

The required sample size is _____. (Round up to the nearest integer.)2) Would it be reasonable to sample this number of students?A) No. This number of IQ test scores is a fairly small number.

B) Yes. This number of IQ test scores is a fairly large number.

C) Yes. This number of IQ test scores is a fairly small number.

D) No. This number of IQ test scores is a fairly large number.

1) 15

2) C

Customer reply replied 3 years ago

A student wants to estimate the mean score of all college students for a particular exam. First use the range rule of thumb to make a rough estimate of the standard deviation of those scores. Possible scores range from 700 to 1600. Use technology and the estimated standard deviation to determine the sample size corresponding to a 99% confidence level and a margin of error of 100 points. What isn't quite right with this exercise?1)

The range rule of thumb estimate for the standard deviation is _____.

(Type an integer or a fraction.)A confidence level of 99% requires a minimum sample size of ______.

(Round up to the nearest integer.)3) What isn't quite right with this exercise?A. These results don't apply to a test that has multiple choice questions.

B. A minimum sample size of 34 is not feasible to use to estimate the mean test scores.

C. The range rule of thumb introduces too much inaccuracy for this procedure.

D. A margin of error of 100 points seems too high to provide a good estimate of the mean score.

The range rule of thumb estimate for the standard deviation is _____.

(Type an integer or a fraction.)A confidence level of 99% requires a minimum sample size of ______.

(Round up to the nearest integer.)3) What isn't quite right with this exercise?A. These results don't apply to a test that has multiple choice questions.

B. A minimum sample size of 34 is not feasible to use to estimate the mean test scores.

C. The range rule of thumb introduces too much inaccuracy for this procedure.

D. A margin of error of 100 points seems too high to provide a good estimate of the mean score.

1) 225

2) 34

3) C

Customer reply replied 3 years ago

In order to estimate the mean amount of time computer users spend on the internet each month, how many computer users must be surveyed in order to be 90% confident that your sample mean is within 15 minutes of the population mean? Assume that the standard deviation of the population of monthly time spent on the internet is 199 min. What is a major obstacle to getting a good estimate of the population mean? Use technology to find the estimated minimum required sample size.1) The minimum sample size required is _____ computer users.

(Round up to the nearest whole number.)2)

A. The data does not provide information on what the computer users did while on the internet.

B. There may not be 477 computer users to survey.

C. It is difficult to precisely measure the amount of time spent on the internet, invalidating some data values.

D. There are no obstacles to getting a good estimate of the population mean

(Round up to the nearest whole number.)2)

A. The data does not provide information on what the computer users did while on the internet.

B. There may not be 477 computer users to survey.

C. It is difficult to precisely measure the amount of time spent on the internet, invalidating some data values.

D. There are no obstacles to getting a good estimate of the population mean

1) 477

2) C

Customer reply replied 3 years ago

Use technology and the given confidence level and sample data to find the confidence interval for the population mean μ. Assume that the population does not exhibit a normal distribution.

Weight lost on a diet: 99% confidence, n = 61, x bar (line over x) = 3.0 kg, S = 4.1 kg1)

___kg <μ< ___kg (Round to one decimal place as needed.)2)

A. Yes, because the sample size is not large enough.

B. No, because the sample size is large enough.

C. Yes, because the population does not exhibit a normal distribution.

D. No, because the population resembles a normal distribution

Weight lost on a diet: 99% confidence, n = 61, x bar (line over x) = 3.0 kg, S = 4.1 kg1)

___kg <μ< ___kg (Round to one decimal place as needed.)2)

A. Yes, because the sample size is not large enough.

B. No, because the sample size is large enough.

C. Yes, because the population does not exhibit a normal distribution.

D. No, because the population resembles a normal distribution

1) 1.6, 4.4

What's the question for 2)?

Customer reply replied 3 years ago

oh yea! DUH. sorry, it doesnt copy well into hereIs the confidence interval affected by the fact that the data appear to be from a population that is not normally distributed?

2) B

Customer reply replied 3 years ago

Make a decision about the given claim. Do not use any formal procedures and exact calculations. Use only the rare event rule.

Claim: A coin favors heads when tossed, and there are 21 heads in 22 tosses.Which of the following is correct?A. There is not sufficient evidence to support the claim because there are substantially more heads than tails.

B. There is not sufficient evidence to support the claim because there are not substantially more heads than tails.

C. There does appear to be sufficient evidence to support the claim because there are not substantially more heads than tails.

D. There does appear to be sufficient evidence to support the claim because there are substantially more heads than tail

Claim: A coin favors heads when tossed, and there are 21 heads in 22 tosses.Which of the following is correct?A. There is not sufficient evidence to support the claim because there are substantially more heads than tails.

B. There is not sufficient evidence to support the claim because there are not substantially more heads than tails.

C. There does appear to be sufficient evidence to support the claim because there are not substantially more heads than tails.

D. There does appear to be sufficient evidence to support the claim because there are substantially more heads than tail

D

Customer reply replied 3 years ago

Find the critical value z Subscript alpha divided by Zα/2 that corresponds to the given confidence level.

95%Zα/2 = ____ (Round to two decimal places as needed.)

95%Zα/2 = ____ (Round to two decimal places as needed.)

1.96

Customer reply replied 3 years ago

Assume a significance level of α=0.05 and use the given information to complete parts (a) and (b) below.

Original claim: The proportion of male golfers is more than 0.2. The hypothesis test results in a P-value of 0.0541) State a conclusion about the null hypothesis. (Reject Upper H0 or fail to reject H0.) Choose the correct answer below.

A. Fail to reject H0 because the P-value is less than α.

B.Reject H0 because the P-value is greater than α.

C. Fail to reject H0 because the P-value is greater than α.

D. Reject H0 because the P-value is less than α.2) Without using technical terms, state a final conclusion that addresses the original claim. Which of the following is the correct conclusion?A.There is sufficient evidence to reject the claim that the proportion of male golfers is more than 0.2

B. There is sufficient evidence to support the claim that the proportion of male golfers is more than 0.2

C There is not sufficient evidence to support the claim that the proportion of male golfers is more than 0.2

D. There is not sufficient evidence to reject the claim that the proportion of male golfers is more than 0.2

Original claim: The proportion of male golfers is more than 0.2. The hypothesis test results in a P-value of 0.0541) State a conclusion about the null hypothesis. (Reject Upper H0 or fail to reject H0.) Choose the correct answer below.

A. Fail to reject H0 because the P-value is less than α.

B.Reject H0 because the P-value is greater than α.

C. Fail to reject H0 because the P-value is greater than α.

D. Reject H0 because the P-value is less than α.2) Without using technical terms, state a final conclusion that addresses the original claim. Which of the following is the correct conclusion?A.There is sufficient evidence to reject the claim that the proportion of male golfers is more than 0.2

B. There is sufficient evidence to support the claim that the proportion of male golfers is more than 0.2

C There is not sufficient evidence to support the claim that the proportion of male golfers is more than 0.2

D. There is not sufficient evidence to reject the claim that the proportion of male golfers is more than 0.2

C

Customer reply replied 3 years ago

then D, right?

only C

Customer reply replied 3 years ago

But its a two part question

I only saw 2)

Customer reply replied 3 years ago

hahah, no worries. Do you need me to resend 1?I worked it as C and C. But im not running the risk of submitting it myself, haha.

Yes. 1)?

Customer reply replied 3 years ago

03 May 2017 08:08

1) State a conclusion about the null hypothesis. (Reject Upper H0 or fail to reject H0.) Choose the correct answer below.

A. Fail to reject H0 because the P-value is less than α.

B.Reject H0 because the P-value is greater than α.

C. Fail to reject H0 because the P-value is greater than α.

D. Reject H0 because the P-value is less than α.

1) State a conclusion about the null hypothesis. (Reject Upper H0 or fail to reject H0.) Choose the correct answer below.

A. Fail to reject H0 because the P-value is less than α.

B.Reject H0 because the P-value is greater than α.

C. Fail to reject H0 because the P-value is greater than α.

D. Reject H0 because the P-value is less than α.

C

Customer reply replied 3 years ago

A 0.1 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender selection, the proportion of baby girls is greater than 0.5. Assume that sample data consists of 78 girls in 144 births, so the sample statistic of 13/24 results in a z score that is 1 standard deviation above 0. Complete parts (a) through (h) below.1) Identify the null hypothesis and the alternative hypothesis. Choose the correct answer below.

A. H0: p=0.5

H1: p≠0.5B. H0: p=0.5

H1: <0.5C. H0: p=0.5

H1: p>0.5D. H0: p ≠0.5

H1: p >0.52)

What is the value of α? ____ (Type an integer or a decimal.)3) What is the sampling distribution of the sample statistic?A) Normal distribution

B) χ2

C) Student (t) distribution4)

Is the test two-tailed, left-tailed, or right-tailed?5) What is the value of the test statistic?

The test statistic is ____. (Type an integer or a decimal.)6) What is the P-value?

The P-value is _____ (Round to four decimal places as needed.)g. What are the critical value(s)?

The critical value(s) is/are _______

(Round to two decimal places as needed. Use a comma to separate answers as needed.)8) What is the area of the critical region?

The area is _____. (Round to two decimal places as needed.)

A. H0: p=0.5

H1: p≠0.5B. H0: p=0.5

H1: <0.5C. H0: p=0.5

H1: p>0.5D. H0: p ≠0.5

H1: p >0.52)

What is the value of α? ____ (Type an integer or a decimal.)3) What is the sampling distribution of the sample statistic?A) Normal distribution

B) χ2

C) Student (t) distribution4)

Is the test two-tailed, left-tailed, or right-tailed?5) What is the value of the test statistic?

The test statistic is ____. (Type an integer or a decimal.)6) What is the P-value?

The P-value is _____ (Round to four decimal places as needed.)g. What are the critical value(s)?

The critical value(s) is/are _______

(Round to two decimal places as needed. Use a comma to separate answers as needed.)8) What is the area of the critical region?

The area is _____. (Round to two decimal places as needed.)

1) C

2) 0.1

3) A

4) Right-tailed

5) 1

6) 0.1587

7) 1.28

8) 0.1

Customer reply replied 3 years ago

Express the confidence interval 0.333 n<p<0.555 in the form p^ (p-hat) +/- E.(the +/- are on top of each other but i cant use that symbol for some reason)p^ (p-hat) +/- E = ____ +/-_____

0.444 +/- 0.111

Customer reply replied 3 years ago

Use the given data to find the minimum sample size required to estimate a population proportion or percentage.

Margin of error: 0.01; confidence level 95%; p^ and q^ with are unknownn = ___ (Round up to the nearest integer.)

Margin of error: 0.01; confidence level 95%; p^ and q^ with are unknownn = ___ (Round up to the nearest integer.)

9604

Customer reply replied 3 years ago

I asked a friend who just completed it. They said there are a total of 13 more questions in this section, and then depending on how many right, there is a total of 30 questions (that you only have to go over if we get them wrong in this section, he only had to answer like 7) and then there is a 25 question exam after that. I just wanted to give you a heads up.Do one of the following, as appropriate.

(a) Find the critical value zα/2,

(b) find the critical value tα/2,

(c) state that neither the normal nor the t distribution applies.

Confidence level 98%; n=2020; σ is known; population appears to be very skewed.A. tα/2=2.205

B. zα/2=2.33

C. tα/2=2.539

D. zα/2=2.055

E. Neither normal nor t distribution applies.Neither normal nor t distribution applies.

(a) Find the critical value zα/2,

(b) find the critical value tα/2,

(c) state that neither the normal nor the t distribution applies.

Confidence level 98%; n=2020; σ is known; population appears to be very skewed.A. tα/2=2.205

B. zα/2=2.33

C. tα/2=2.539

D. zα/2=2.055

E. Neither normal nor t distribution applies.Neither normal nor t distribution applies.

OK.

B

Customer reply replied 3 years ago

I had to screen shoot the problem because it wont let me copy

Customer reply replied 3 years ago

let me know if the document doesnt work.

D and D

Customer reply replied 3 years ago

is it easier if i screen shoot it or message it?

Message but if it takes too much trouble screen shot is fine.

Customer reply replied 3 years ago

okay. This one was already done so ill just send it, then I will do message

This screen shot looks better than the previous one.

a) 0.553

b) 0.0401

c) 0.513, 0.593

d) D

Customer reply replied 3 years ago

Using the simple random sample of weights of women from a data set, we obtain these sample statistics: n =35 and x(bar)=152.08 lb. Research from other sources suggests that the population of weights of women has a standard deviation given by σ=32.85 lb.

a. Find the best point estimate of the mean weight of all women.

b. Find a 95% confidence interval estimate of the mean weight of all women.1. The best point estimate is ____ lb.

(Type an integer or a decimal.)2. The 95% confidence interval estimate is ___ <μ< ____ lb.

(Round to two decimal places as needed.)

a. Find the best point estimate of the mean weight of all women.

b. Find a 95% confidence interval estimate of the mean weight of all women.1. The best point estimate is ____ lb.

(Type an integer or a decimal.)2. The 95% confidence interval estimate is ___ <μ< ____ lb.

(Round to two decimal places as needed.)

a. 152.08

b. 141.20, 162.96

Customer reply replied 3 years ago

A genetic experiment involving peas yielded one sample of offspring consisting of 434 green peas and 150 yellow peas. Use a 0.05 significance level to test the claim that under the same circumstances, 23% of offspring peas will be yellow. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method and the normal distribution as an approximation to the binomial distribution.1) What are the null and alternative hypotheses?

A. H0: p=0.23

H1: p≠0.23B. H0: p=0.23

H1: p>0.23C. H0: p≠0.23

H1: p<0.23D. H0: p=0.23

H1: p<0.23E. H0: p≠0.23

H1: p>0.23F. H0: p≠0.23

H1: p=0.232) z = ____ (Round to two decimal places as needed.)3)

P-values= ______ (Round to four decimal places as needed.)

A. H0: p=0.23

H1: p≠0.23B. H0: p=0.23

H1: p>0.23C. H0: p≠0.23

H1: p<0.23D. H0: p=0.23

H1: p<0.23E. H0: p≠0.23

H1: p>0.23F. H0: p≠0.23

H1: p=0.232) z = ____ (Round to two decimal places as needed.)3)

P-values= ______ (Round to four decimal places as needed.)

1) A

2) 1.54

3) 0.12

4)

Customer reply replied 3 years ago

2)What is the conclusion about the null hypothesis?

A. Reject the null hypothesis because the P-value is greater than the significance level, α.

B. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, α.

C. Reject the null hypothesis because the P-value is less than or equal to the significance level, α.

D. Fail to reject the null hypothesis because the P-value is greater than the significance level, α

A. Reject the null hypothesis because the P-value is greater than the significance level, α.

B. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, α.

C. Reject the null hypothesis because the P-value is less than or equal to the significance level, α.

D. Fail to reject the null hypothesis because the P-value is greater than the significance level, α

D.

Customer reply replied 3 years ago

5) What is the final conclusion?

A. There is sufficient evidence to warrant rejection of the claim that 23% of offspring peas will be yellow.

B. There is sufficient evidence to support the claim that less than 23% of offspring peas will be yellow.

C. There is not sufficient evidence to support the claim that less than 23% of offspring peas will be yellow.

D. There is not sufficient evidence to warrant rejection of the claim that 23% of offspring peas will be yellow.

A. There is sufficient evidence to warrant rejection of the claim that 23% of offspring peas will be yellow.

B. There is sufficient evidence to support the claim that less than 23% of offspring peas will be yellow.

C. There is not sufficient evidence to support the claim that less than 23% of offspring peas will be yellow.

D. There is not sufficient evidence to warrant rejection of the claim that 23% of offspring peas will be yellow.

D

Customer reply replied 3 years ago

In a Harris poll, adults were asked if they are in favor of abolishing the penny. Among the responses, 1276 answered "no," 489 answered "yes," and 388 had no opinion. What is the sample proportion of yes responses, and what notation is used to represent it?A. p^ =0.277 The symbol p^ is used to represent a sample proportion.

B. p=0.2270 The symbol p is used to represent a sample proportion.

C. p^=0.227 The symbol p^ is used to represent a sample proportion.

D. p=0.277 The symbol p is used to represent a sample proportion

B. p=0.2270 The symbol p is used to represent a sample proportion.

C. p^=0.227 The symbol p^ is used to represent a sample proportion.

D. p=0.277 The symbol p is used to represent a sample proportion

C

Customer reply replied 3 years ago

An IQ test is designed so that the mean is 100 and the standard deviation is 22 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of statistics students such that it can be said with 95% confidence that the sample mean is within 8 IQ points of the true mean. Assume that σ=22 and determine the required sample size using technology. Then determine if this is a reasonable sample size for a real world calculation.1)The required sample size is ________ (Round up to the nearest integer.)2) Would it be reasonable to sample this number of students?A) No. This number of IQ test scores is a fairly small number.B) Yes. This number of IQ test scores is a fairly large number.C) Yes. This number of IQ test scores is a fairly small number.D) No. This number of IQ test scores is a fairly large number.

1) 30.

2) C

Customer reply replied 3 years ago

The claim is that the proportion of peas with yellow pods is equal to 0.25 (or 25%). The sample statistics from one experiment include 510 peas with 118 of them having yellow pods. Find the value of the test statistic.1)

The value of the test statistic is _____. (Round to two decimal places as needed.)

The value of the test statistic is _____. (Round to two decimal places as needed.)

1) -0.97

Customer reply replied 3 years ago

For the given claim, complete parts (a) and (b) below.

Claim: The mean weight of beauty pageant winners is 119 pounds. A study of 23 randomly selected beauty pageants resulted in a mean winner weight of 117 pounds.1) Express the original claim in symbolic form2) Identify the null and the alternative hypotheses

Claim: The mean weight of beauty pageant winners is 119 pounds. A study of 23 randomly selected beauty pageants resulted in a mean winner weight of 117 pounds.1) Express the original claim in symbolic form2) Identify the null and the alternative hypotheses

1) mu = 119

Customer reply replied 3 years ago

b) Identify the null and the alternative hypotheses

Customer reply replied 3 years ago

Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. In a manual on how to have a number one song, it is stated that a song must be no longer than 210 seconds. A simple random sample of 40 current hit songs results in a mean length of 229.6 sec and a standard deviation of 54.11 sec. Use a 0.05 significance level and the accompanying display to test the claim that the sample is from a population of songs with a mean greater than 210 sec. What do these results suggest about the advice given in the manual?1) What are the hypotheses?

A. H0: μ>210 sec

H1: u≤210 secB. H0: u=210 sec

H1: μ>210 secC. H0: μ<210 sec

H1: μ>210 secD. H0: μ=210 sec

H1: μ ≤210 sec2) t = ____ (Round to three decimal places as needed.)sorry, round to two dec not three for number 2)3) The P-value is ____. (Round to three decimal places as needed.)4)State the final conclusion that addresses the original claim. Choose the correct answer below.

A. Fail to reject H0. There is insufficient evidence to support the claim that the sample is from a population of songs with a mean length greater than 210 sec.B. Reject H0. There is insufficient evidence to support the claim that the sample is from a population of songs with a mean length greater than 210 sec.C. Reject H0. There is sufficient evidence to support the claim that the sample is from a population of songs with a mean length greater than 210 sec.D. Fail to reject H0. There is sufficient evidence to support the claim that the sample is from a population of songs with a mean length greater than 210 sec.5) What do the results suggest about the advice given in the manual?

A. The results suggest that the advice of writing a song that must be no longer than 210 seconds is not sound advice.

B. The results are inconclusive because the average length of a hit song is constantly changing.

C. The results suggest that 229.6 seconds is the best song length.

D. The results do not suggest that the advice of writing a song that must be no longer than 210 seconds is not sound advice.

A. H0: μ>210 sec

H1: u≤210 secB. H0: u=210 sec

H1: μ>210 secC. H0: μ<210 sec

H1: μ>210 secD. H0: μ=210 sec

H1: μ ≤210 sec2) t = ____ (Round to three decimal places as needed.)sorry, round to two dec not three for number 2)3) The P-value is ____. (Round to three decimal places as needed.)4)State the final conclusion that addresses the original claim. Choose the correct answer below.

A. Fail to reject H0. There is insufficient evidence to support the claim that the sample is from a population of songs with a mean length greater than 210 sec.B. Reject H0. There is insufficient evidence to support the claim that the sample is from a population of songs with a mean length greater than 210 sec.C. Reject H0. There is sufficient evidence to support the claim that the sample is from a population of songs with a mean length greater than 210 sec.D. Fail to reject H0. There is sufficient evidence to support the claim that the sample is from a population of songs with a mean length greater than 210 sec.5) What do the results suggest about the advice given in the manual?

A. The results suggest that the advice of writing a song that must be no longer than 210 seconds is not sound advice.

B. The results are inconclusive because the average length of a hit song is constantly changing.

C. The results suggest that 229.6 seconds is the best song length.

D. The results do not suggest that the advice of writing a song that must be no longer than 210 seconds is not sound advice.

1) B

2) t= 2.29

3) 0.014

4) C

5) A

Are there more questions?

Customer reply replied 3 years ago

sorry, took a phone callAssume that the significance level is α=0.1. Use the given information to find the P-value and the critical value(s).

The test statistic of z=1.86 is obtained when testing the claim that p 0.1p>0.11)

P-value= ____ (Round to four decimal places as needed.)2)

The critical value(s) is/are z= _____________8203;(for question 2 Round to two decimal places as needed. Use a comma to separate answers as needed.)I only have 2 more questions after this one, and im not even going to try the rest tonight. Could we finish up tomorrow after this part is over? Im just going to email my professor and try to get partial credit tomorrow.

The test statistic of z=1.86 is obtained when testing the claim that p 0.1p>0.11)

P-value= ____ (Round to four decimal places as needed.)2)

The critical value(s) is/are z= _____________8203;(for question 2 Round to two decimal places as needed. Use a comma to separate answers as needed.)I only have 2 more questions after this one, and im not even going to try the rest tonight. Could we finish up tomorrow after this part is over? Im just going to email my professor and try to get partial credit tomorrow.

No problem.

1) 0.0314

2) 1.28

OK we can do the rest tomorrow.

Customer reply replied 3 years ago

I just have two more for this sectionIn statistics, what does df denote? If a simple random sample of 25 speeds of cars on California Highway 405 is to be used to test the claim that the sample values are from a population with a mean greater than the posted speed limit of 65 mi/h, what is the specific value of df?Choose the correct answer below.

A. df denotes the distribution of freedom. For this sample, df=25.

B. df denotes the number of degrees of freedom. For this sample, df=25.

C. df denotes the number of degrees of freedom. For this sample, d=24.

D. df denotes the distribution of freedom. For this sample, df=24.

A. df denotes the distribution of freedom. For this sample, df=25.

B. df denotes the number of degrees of freedom. For this sample, df=25.

C. df denotes the number of degrees of freedom. For this sample, d=24.

D. df denotes the distribution of freedom. For this sample, df=24.

C

Customer reply replied 3 years ago

A certain drug is used to treat asthma. In a clinical trial of the drug, 21 of 257 treated subjects experienced headaches (based on data from the manufacturer). The accompanying calculator display shows results from a test of the claim that less than 99% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.01 significance level to complete parts (a) through (e) below.1)

Right tailed testLeft-tailed testTwo-tailed test2)

z = ____ (Round to two decimal places as needed.)3)

P-value =

(Round to four decimal places as needed.)Identify the null hypothesis.

A. H0: p=0.09

B. H0: p<0.09

C. H0: p>0.09

D.Ho: not equal to 0.095)Decide whether to reject the null hypothesis. Choose the correct answer below.

A. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, α.

B. Reject the null hypothesis because the P-value is greater than the significance level, α.

C. Fail to reject the null hypothesis because the P-value is greater than the significance level, α.

D. Reject the null hypothesis because the P-value is less than or equal to or equal to the significance level, alphaα6) What is the final conclusion?

A.There is sufficient evidence to warrant rejection of the claim that less than 99% of treated subjects experienced headaches.

B. There is not sufficient evidence to support the claim that less than 99% of treated subjects experienced headaches.

C. There is sufficient evidence to support the claim that less than 99% of treated subjects experienced headaches.

D. There is not sufficient evidence to warrant rejection of the claim that less than 99% of treated subjects experienced headaches.

Right tailed testLeft-tailed testTwo-tailed test2)

z = ____ (Round to two decimal places as needed.)3)

P-value =

(Round to four decimal places as needed.)Identify the null hypothesis.

A. H0: p=0.09

B. H0: p<0.09

C. H0: p>0.09

D.Ho: not equal to 0.095)Decide whether to reject the null hypothesis. Choose the correct answer below.

A. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, α.

B. Reject the null hypothesis because the P-value is greater than the significance level, α.

C. Fail to reject the null hypothesis because the P-value is greater than the significance level, α.

D. Reject the null hypothesis because the P-value is less than or equal to or equal to the significance level, alphaα6) What is the final conclusion?

A.There is sufficient evidence to warrant rejection of the claim that less than 99% of treated subjects experienced headaches.

B. There is not sufficient evidence to support the claim that less than 99% of treated subjects experienced headaches.

C. There is sufficient evidence to support the claim that less than 99% of treated subjects experienced headaches.

D. There is not sufficient evidence to warrant rejection of the claim that less than 99% of treated subjects experienced headaches.

1) Left

2) -0.46

3) 0.3228

4) A

5) C

6) B

Customer reply replied 3 years ago

THANK YOU! see you tomorrow! haha

You are welcome.

Customer reply replied 3 years ago

Good Morning. I have two quizzes left. But they are 25 questions a piece and I can view them all at once. I'm working on what we did yesterday, but for a different chapter to unlock that quiz. While Im doing that, would you be able to complete this one?

OK. One second.

Customer reply replied 3 years ago

I was able To send that $100, you should be receiving it shortlyI should have this other portion done in about 3 hours or so, and then can I send you the last 25 questions?

Thanks. The answer can be downloaded at

http://www.mediafire.com/file/ihxasaz9amb5wqu/Answer_to_25_Questions.docx

Please let me know if you have any questions.

Customer reply replied 3 years ago

Thank you! Will you still be able to do the last 25 after I'm done with all the questions that unlock it? I probably won't be done for a few hours

Can you give me an estimate of start and end time of the last 25?

Customer reply replied 3 years ago

I should have it to you in the next 2-3 hours, depending on how quickly I can do these practice questions. And then if you could have them back at some point this evening.

No problem. I can do it.

Customer reply replied 3 years ago

Thank you!

You are welcome.

Customer reply replied 3 years ago

96% on the last one, ill take it! haha. Last one.

Customer reply replied 3 years ago

Just wanted to confirm you recieved the document

Customer reply replied 3 years ago

Hope I didn't lose you

Got it. What's the deadline?

Customer reply replied 3 years ago

technically it was last night ad midnight but the professor just said "get it in ASAP"

I will post the answer shortly.

1) A

2) C

3) C

4) D

5) B

6) D

7) B

8) C

9) D

10) D

11) B

12) C

13) A

14) B

15) B

16) D

17) D

18) C

19) B

20) D

21)

Answer for problem 21, 22, 23, 24 can be downloaded at

http://www.mediafire.com/file/15zh55hyf5xpor6/Answer_from_Question_21.docx

Please let me know if you have any questions. Thanks.

Customer reply replied 3 years ago

Thank you!

Customer reply replied 3 years ago

sorry! i left question 25 out, can you answer this really quick?Provide an appropriate response.If selecting samples of size n > 30 from a population with a known mean and standard deviation, what requirement, if any, must be satisfied in order to assume that the distribution of the sample means is a normal distribution?A) The population must have a normal distribution.

B) The mean must be equal to the standard deviation.

C) The population must have a standard deviation of 0.

D) None; the distribution of sample means will be approximately normal.

B) The mean must be equal to the standard deviation.

C) The population must have a standard deviation of 0.

D) None; the distribution of sample means will be approximately normal.

D

Customer reply replied 3 years ago

thank you so much! did you get the $100 bonus? I sent the money this morning.

You are welcome. I haven't got it yet.

Customer reply replied 3 years ago

If you don't recieve it, let me know, I have confirmation emails and we will be able to track it down.

OK. I will check it tomorrow.

I have got the bonus. Thank you very much. In the future you may request me by writing "For John Only" in your new posts so that they are directed to me. I will respond as soon as possible.

Customer reply replied 3 years ago

Awesome!

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