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Chapter 8: Q.2.3 (page 490)

Alcohol abuse has been described by college presidents as the number one problem on campus, and it is an important cause of death in young adults. How common is it? A survey of 10,904 randomly selected U.S. college students collected information on drinking behavior and alcohol-related problems.9 The researchers defined 鈥渇requent binge drinking鈥� as having five or more drinks in row three or more times in the past two weeks. According to this definition, 2486 students were classified as frequent binge drinkers.

Find the critical value for a 99% confidence interval. Show your method. Then calculate the interval.

Short Answer

Expert verified

The confidence interval is(0.21764,0.23834).

Step by step solution

01

Given Information

Given that

Number of students (n)=10904

Number of binger drivers(x)=2486

Confidence level=99%

02

Explanation

The formula to calculate the confidence interval for a population proportion is:

CI=p^卤za/2脳p^(1-p^)n

Where,p^=xn

p^=xn

03

Calculation

From the standard normal table, the Z-score at99% confidence level is calculated as:

za/2=z0.10/2

=2.576

The confidence interval using Ti-83 plus calculator is calculated as:

Therefore, the99%confidence interval is(0.21764,0.23834).

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Most popular questions from this chapter

Equality for women? Have efforts to promote equality for women gone far enough in the United States? A poll on this issue by the cable network MSNBC contacted 1019adults. A newspaper article about the poll said, 鈥淩esults have a margin of sampling error of plus or minus 3 percentage points.鈥�15

(a) The news article said that 65% of men, but only 43% of women, think that efforts to promote equality have gone far enough. Explain why we do not have enough information to give con铿乨ence intervals for men and women separately.

(b) Would a 95% con铿乨ence interval for women alone have a margin of error less than 0.03, about equal to 0.03, or greater than 0.03? Why? (You see that the news article鈥檚 statement about the margin of error for poll results is a bit misleading.)

It's about ME Explain how each of the following would affect the margin of error of a confidence interval, if all other things remained the same.

(a) Increasing the confidence level

(b) Quadrupling the sample size

How confident? The figure at top right shows the result of taking 25SRSs from a Normal population and constructing a confidence interval for each sample. Which confidence level鈥�80%,90%,95%, or 99%鈥攄o you think was used? Explain.

Running red lights A random digit dialing telephone survey of880drivers asked, "Recalling the last ten traffic lights you drove through, how many of them were red when you entered the intersections?" Of the 880respondents, 171admitted that at least one light had been red.15

(a) Construct and interpret a95%confidence interval for the population proportion.

(b) Nonresponse is a practical problem for this survey-only 21.6%of calls that reached a live person were completed. Another practical problem is that people may not give truthful answers. What is the likely direction of the bias: do you think more or fewer than 171of the 880respondents really ran a red light? Why? Are these sources of bias included in the margin of error?

You have an SRS of 23 observations from a Normally distributed population. What critical value would you use to obtain a 98% confidence interval for the mean M of the population if S is unknown?

(a) 2.508

(b) 2.500

(c) 2.326

(d) 2.183

(e) 2.177
See all solutions

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