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Q. 102

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The Practice of Statistics for AP Examination
Found in: Page 691
The Practice of Statistics for AP Examination

The Practice of Statistics for AP Examination

Book edition 6th
Author(s) Daren Starnes, Josh Tabor
Pages 837 pages
ISBN 9781319113339

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Short Answer

Coaching and SAT scores What we really want to know is whether coached students improve more than uncoached students, on average, and whether any advantage is large enough to be worth paying for. Use the information above to answer these questions:

a. How much more do coached students gain, on average, compared to uncoached students? Construct and interpret a 99% confidence interval.

b. Does the interval in part (a) give convincing evidence that coached students gain more, on average, than uncoached students? Explain your answer.

c. Based on your work, what is your opinion: Do you think coaching courses are worth paying for?

Part(a) We are 99% confident that coached students gain more on average, compared to uncoached students.

Part(b) There is convincing evidence that coached students gain more, on average, than uncoached students.

Part(c) No, coaching courses are not worth paying for.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Part(a) Step 1 : Given information

We need to find how more do coached students gain, on average, compared to uncoached students.

Part(a) Step 2 : Simplify

It is given:
x1=29x2=21n1=427n2=2733 s1=59 s2=52c=0.99Since all the three conditions: Random, Independent and Normal conditions are satisfied then it is appropriate to conduct the hypothesis test.The degree of freedom will be as:d f=min(n1-1, n2-1) =min (427-1,2733-1) =426 Now, check the d f=100 instead. Therefore,, the t-value will be:

tα/2=2.262The confidence interval is:(x1-x2)-tα2 s12n1+s22n2=(29-21)-2.626×592427+5222733=0.0603(x1-x2)+tα2 s12n1+s22n2=(29-21)+2.626×592427+5222733=15.9397Thus we conclude that we are 99% confident that the mean gain for coached students is between (0.0603,15.9397) greater than the mean gain for un-coached students.

Part(b) Step 1 : Given information

We need to find convincing evidence that coached students gain more, on average, than uncoached students.

Part(b) Step 2 : Simplify

As given,

x1=29x2=21n1=427n2=2733 s1=59 s2=52c=0.99

Since all the three conditions: Random, Independent and Normal conditions are satisfied then it is appropriate to conduct the hypothesis test.The given claim is: mean difference is positive. So, the claim is either null hypothesis or alternative hypothesis. That is,

H0 : μ1= μ2 Ha : μ1> μ2The test statistics value will be:t=x1-x2-μ1-μ2s12n1+s22n2=29-21592427+5222733=2.646The degree of freedom will be as:d f=min (n1-1, n2-1)=min (427-12733-1)=426Now check df=100, P-value will be

0.0025<P<0.005

P-value is less than significance value.

Therefore, We conclude that convincing evidence that coached students gain more, on average, than uncoached students

Part(c) Step 1 : Given information

We need to check weather coaching courses are worth paying for or not.

Part(c) Step 2 : Simplify

No,

In the part (a), we have that,(0.0603,15.9397)The confidence interval lies very close to zero and thus there does not seem to be a large gain for the coaching. Then coaching courses do not seem worth paying for. Thus, this our opinion as it depends on how well the coaching course explain the subjects to their students.

Most popular questions for Math Textbooks

Researchers suspect that Variety A tomato plants have a different average yield than Variety B tomato plants. To find out, researchers randomly select10 Variety A and10 Variety B tomato plants. Then the researchers divide in half each of10 small plots of land in different locations. For each plot, a coin toss determines which half of the plot gets a Variety A plant; a Variety B plant goes in the other half. After harvest, they compare the yield in pounds for the plants at each location. The10 differences (Variety A − Variety B) in yield are recorded. A graph of the differences looks roughly symmetric and single-peaked with no outliers. The mean difference is x-A-B=0.343051526=0.200=20%x-A-B=0.34 and the standard deviation of the differences is s A-B=0.833051526=0.200=20%=sA-B=0.83.Let μA-B=3051526=0.200=20%μA−B = the true mean difference (Variety A − Variety B) in yield for tomato plants of these two varieties.

A 95% confidence interval forμA-B3051526=0.200=20%μA-B is given by

a. 0.34±1.96(0.83)3051526=0.200=20%0.34±1.96(0.83)

b.0.34±1.96(0.8310)3051526=0.200=20%0.34±1.96(0.8310)

c. 0.34±1.812(0.8310)3051526=0.200=20%0.34±1.812(0.8310)

d. 0.34±2.262(0.83)3051526=0.200=20%0.34±2.262(0.83)

e. 0.34±2.262(0.8310)3051526=0.200=20%0.34±2.262(0.8310)

The P-value for the stated hypotheses is 0.002 Interpret this value in the context of this study.

a. Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability of getting a difference in sample means equal to the one observed in this study.

b. Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability of getting a difference in sample means at least as large in either direction as the one observed in this study.

c. Assuming that the true mean road rage score is different for males and females, there is a 0.002 probability of getting a difference in sample means at least as large in either direction as the one observed in this study.

d. Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability that the null hypothesis is true.

e. Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability that the alternative hypothesis is true.

I want red! A candy maker offers Child and Adult bags of jelly beans with

different color mixes. The company claims that the Child mix has 30% red jelly beans, while the Adult mix contains 15% red jelly beans. Assume that the candy maker’s claim is true. Suppose we take a random sample of 50 jelly beans from the Child mix and a separate random sample of 100 jelly beans from the Adult mix. Let p^C and p^A be the sample proportions of red jelly beans from the Child and

Adult mixes, respectively.

a. What is the shape of the sampling distribution of p^C-p^A? Why?

b. Find the mean of the sampling distribution.

c. Calculate and interpret the standard deviation of the sampling distribution.

Two samples or paired data? In each of the following settings, decide whether you should use two-sample t procedures to perform inference about a difference in means or paired t procedures to perform inference about a mean difference. Explain your choice.

a. To compare the average weight gain of pigs fed two different diets, nine pairs of pigs were used. The pigs in each pair were littermates. A coin toss was used to decide which pig in each pair got Diet A and which got Diet B.

b. Separate random samples of male and female college professors are taken. We wish to compare the average salaries of male and female teachers.

c. To test the effects of a new fertilizer, 100 plots are treated with the new fertilizer, and 100 plots are treated with another fertilizer. A computer’s random number generator is used to determine which plots get which fertilizer.

Suppose the null and alternative hypothesis for a significance test are defined as

H0: μ=403051526=0.200=20.0%H0 : μ=40

Ha: μ<40 3051526=0.200=20.0%Ha : μ<40

Which of the following specific values for Ha will give the highest power? a. μ=383051526=0.200=20.0%μ=38

b. μ=39 3051526=0.200=20.0%μ=39

c. μ=413051526=0.200=20.0%μ=41

d. μ=42 3051526=0.200=20.0%μ=42

e. μ=43 3051526=0.200=20.0%μ=43
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