/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q. 53. Teen drivers Refer to Exercise 5... [FREE SOLUTION] | 91影视

91影视

Teen drivers Refer to Exercise 51.

a. Construct and interpret a 95% confidence interval for the true proportion p of all teens in the state who passed their driving test on the first attempt. Assume that the conditions for inference are met.

b. Explain why the interval in part (a) provides more information than the test in Exercise 51.

Short Answer

Expert verified

Part a) 0.6068<p<0.7692

There are 95%confident that the true proportion of all teens in the state that passed their driving test on the first attempt is between 0.6068and 0.7692

Part b) There is sufficient proof to reject the DMV's claim.

Step by step solution

01

Part a) Step 1: given information

n=125p=60%=0.60x=86c=95%

02

Part a) Step 2: Formula used

p^=xnE=za2p^(1-p^n

03

Part a) Step 3: The objective is to construct and interpret a  confidence interval for the true proportion of all teens in the state that passed their driving test on the first attempt.

The sample proportion is

p^=xn=86125=0.688

For confidence level determined za/2=z0.025using table II (look up 0.025in the table, the score is then they found z-core with opposite sign):

z/2=1.96

The margin of error is:

E=za/2p^(1-p^)n=1.960.688(1-0.688)125=0.0812

The confidence interval is as follows:

p^-E<p<p^+E=0.688-0.0812<p<0.688+0.0812=0.6068<p<0.7692

The true proportion of all teens in the state who passed their driving test on the first try is estimated to be between 0.6068 and 0.7692

04

Part b) Step 1: Given information

The result from exercise part (a):

0.6068<p<0.7692
05

Part b) Step 2: The objective is to explain the interval in part (a) provides more information than the test exercise 51

The confidence interval is not 60% (or 0.60), as claimed by the DMV. As a result, there is sufficient evidence to reject the DMV's claim.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Water! A blogger claims that U.S. adults drink an average of five 8-ounce glasses (that鈥檚 40 ounces) of water per day. Researchers wonder if this claim is true, so they ask a random sample of 24 U.S. adults about their daily water intake. A graph of the data shows a roughly symmetric shape with no outliers.

a. State an appropriate pair of hypotheses for a significance test in this setting. Be sure to define the parameter of interest.

b. Check conditions for performing the test in part (a).

c. The 90% confidence interval for the mean daily water intake is 30.35 to 36.92 ounces. Based on this interval, what conclusion would you make for a test of the hypotheses in part (a) at the 10% significance level?

d. Do we have convincing evidence that the amount of water U.S. children drink per day differs from 40 ounces? Justify your answer.

Walking to school Refer to Exercise 36.

a. Explain why the sample result gives some evidence for the alternative hypothesis.

b. Calculate the standardized test statistic and P-value.

c. What conclusion would you make?

Based on the P-value in Exercise 31, which of the following would be the most

appropriate conclusion?

a. Because the P-value is large, we reject H0. We have convincing evidence that more than 50%of city residents support the tax increase.

b. Because the P-value is large, we fail to reject H0. We have convincing evidence that more than 50%of city residents support the tax increase.

c. Because the P-value is large, we reject H0. We have convincing evidence that at most 50%of city residents support the tax increase.

d. Because the P-value is large, we fail to reject H0. We have convincing evidence that at most 50%of city residents support the tax increase.

e. Because the P-value is large, we fail to reject H0. We do not have convincing

evidence that more than 50%of city residents support the tax increase.

A random sample of 100 likely voters in a small city produced 59 voters in favor of Candidate A. The observed value of the standardized test statistic for performing a test of H0:p=0.5H0:p=0.5versus Ha:p>0.5Ha:p>0.5

is which of the following?

a)z=0.59-0.50.59(0.41)100

b)z=0.59-0.50.5(0.5)100

c)z=0.5-0.590.59(0.41)100

d)z=0.5-0.590.5(0.5)100

e)z=0.59-0.5100

Reality TVTelevision networks rely heavily on ratings of TV shows when deciding

whether to renew a show for another season. Suppose a network has decided that

鈥淢iniature Golf with the Stars鈥 will only be renewed if it can be established that more than 12%of U.S. adults watch the show. A polling company asks a random sample of 2000U.S. adults if they watch 鈥淢iniature Golf with the Stars.鈥 The network uses the data to perform a test of

H0:p=0.12

Ha:p>0.12

where pis the true proportion of all U.S. adults who watch the show. Describe a Type Ierror and a TypeIIerror in this setting, and give a possible consequence of each.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.