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Chapter 8: Q 36. (page 522)

Reporting cheating What proportion of students are willing to report cheating by other students? A student project put this question to an SRS of 172172undergraduates at a large university: “You witness two students cheating on a quiz. Do you go to the professor?” Only 19answered “Yes.” Assume the conditions for inference are met.

a. Determine the critical valueZ*for a 96%confidence interval for a proportion.

b. Construct a 96%confidence interval for the proportion of all undergraduates at this university who would go to the professor.

c. Interpret the interval from part (b).

Short Answer

Expert verified

a. The critical value iszc=za/2critical value=2.04

b. 96%confidence interval for proportion is 0.0617<p<0.1593

c. True population proportion is between0.0617and0.1593

Step by step solution

01

Given Information

It is given that x=19

n=172

Confidence Level =0.98

Level of significancea=0.04

02

Critical Value

Using Excel Formula

Using EXCEL FORMULA=ABS(NORMSINV(0.04/2))ZC=Za/2critical value=2.04

03

Constructing Confidence Interval

Sample Proportion p^=xn

p^=19172=0.1105

Margin of Error E=Zα/2×p^(1-p^)n

E=2.04×0.1105(1-0.1105)172=0.0488

The confidence interval is (p^-E,p^+E)

(0.1105-0.0488,0.1105+0.0488)=(0.0617,0.1593)

The 96%confidence interval for population proportion is50.0617<p<0.1593

04

Interpreting Confidence Interval

There is 0.0617<p<0.1593confidence that correct population of all UG at this university going to professor lie between 0.0617and0.1593

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Most popular questions from this chapter

How confident? The figure shows the result of taking 25SRSs from a Normal population and constructing a confidence interval for the population mean using each sample. Which confidence level—80%,90%,95%,or99%—do you think was used? Explain your reasoning.

Explaining confidence The admissions director for a university found that (107.8,116.2)is a 95%confidence interval for the mean IQ score of all freshmen. Discuss whether each of the following explanations is correct, based on that information.

a. There is a 95%probability that the interval from role="math" localid="1654200953396" 107.8to116.2contains μ.

b. There is a 95%chance that the interval(107.8,116.2)contains x¯.

c. This interval was constructed using a method that produces intervals that capture the true mean in 95%of all possible samples.

d. If we take many samples, about 95%of them will contain the interval (107.8,116.2).

e. The probability that the interval (107.8,116.2)captures μ is either 0 or 1, but we don’t know which.

Prayer in school A New York Times/CBS News Poll asked a random sample of U.S. adults the question “Do you favor an amendment to the Constitution that would permit organized prayer in public schools?” Based on this poll, the 95%confidence interval for the population proportion who favor such an amendment is (0.63,0.69).

a. Interpret the confidence interval.

b. What is the point estimate that was used to create the interval? What is the margin of error?

c. Based on this poll, a reporter claims that more than two-thirds of U.S. adults favor such an amendment. Use the confidence interval to evaluate this claim.

Judy is interested in the reading level of a medical journal. She records the length of a random sample of100words. The histogram displays the distribution of word length for her sample. Determine if the Normal/Large Sample condition is met in this context.

Catching goldfish for schoolCarly and Maysem plan to be preschool teachers after they graduate from college. To prepare for snack time, they want to know the mean number of goldfish crackers in a bag of original-flavored goldfish. To estimate this value, they randomly selected12 bags of original-flavored goldfish and counted the number of crackers in each bag. Here are their data:

Construct and interpret a 95% confidence interval for the true mean number of crackers in a bag of original- flavored goldfish.
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