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Chapter 9: Problem 12

The formula used to compute a large-sample confidence interval for \(\pi\) is $$ p \pm(z \text { critical value }) \sqrt{\frac{p(1-p)}{n}} $$ What is the appropriate \(z\) critical value for each of the following confidence levels? a. \(95 \%\) d. \(80 \%\) b. \(90 \%\) e. \(85 \%\) c. \(99 \%\)

Short Answer

Expert verified
The z critical values for the confidence levels 95%, 90%, 99%, 80%, and 85% are 1.96, 1.645, 2.575, 1.28, and 1.44 respectively.

Step by step solution

01

Understanding the Confidence Interval

The confidence interval is expressed as \( p \pm z \sqrt{\frac{p(1-p)}{n}} \) where \( p \) is the population proportion, \( n \) is the sample size, and \( z \) is the critical value corresponding to the confidence level. The critical value \( z \) is determined by the desired confidence level and can be looked up in a standard normal distribution table.
02

Calculating the z critical value for a 95% confidence level

The z critical value for a 95% confidence level is 1.96.
03

Calculating the z critical value for a 90% confidence level

The z critical value for a 90% confidence level is 1.645.
04

Calculating the z critical value for a 99% confidence level

The z critical value for a 99% confidence level is 2.575.
05

Calculating the z critical value for an 80% confidence level

The z critical value for an 80% confidence level is 1.28.
06

Calculating the z critical value for an 85% confidence level

The z critical value for an 85% confidence level is 1.44.

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Most popular questions from this chapter

Suppose that a random sample of 50 bottles of a par. ticular brand of cough medicine is selected and the alcohol content of each bottle is determined. Let \(\mu\) denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the sample of 50 results in a \(95 \%\) confidence interval for \(\mu\) of \((7.8,9.4)\). a. Would a \(90 \%\) confidence interval have been narrower or wider than the given interval? Explain your answer. b. Consider the following statement: There is a \(95 \%\) chance that \(\mu\) is between \(7.8\) and \(9.4\). Is this statement correct? Why or why not? c. Consider the following statement: If the process of selecting a sample of size 50 and then computing the corresponding \(95 \%\) confidence interval is repeated 100 times, 95 of the resulting intervals will include \(\mu\). Is this statement correct? Why or why not?

The article "Doctors Cite Burnout in Mistakes" (San Luis Obispo Tribune, March 5,2002 ) reported that many doctors who are completing their residency have financial struggles that could interfere with training. In a sample of 115 residents, 38 reported that they worked moonlighting jobs and 22 reported a credit card debt of more than \(\$ 3000\). Suppose that it is reasonable to consider this sample of 115 as a random sample of all medical residents in the United States. a. Construct and interpret a \(95 \%\) confidence interval for the proportion of U.S. medical residents who work moonlighting jobs. b. Construct and interpret a \(90 \%\) confidence interval for the proportion of U.S. medical residents who have a credit card debt of more than \(\$ 3000\). c. Give two reasons why the confidence interval in Part (a) is wider than the confidence interval in Part (b).

An article in the Chicago Tribune (August 29,1999 ) reported that in a poll of residents of the Chicago suburbs, \(43 \%\) felt that their financial situation had improved during the past year. The following statement is from the article: "The findings of this Tribune poll are based on interviews with 930 randomly selected suburban residents. The sample included suburban Cook County plus DuPage, Kane, Lake, McHenry, and Will Counties. In a sample of this size, one can say with \(95 \%\) certainty that results will differ by no more than 3 percent from results obtained if all residents had been included in the poll." Comment on this statement. Give a statistical argument to justify the claim that the estimate of \(43 \%\) is within \(3 \%\) of the true proportion of residents who feel that their financial situation has improved.

The formula used to compute a confidence interval for the mean of a normal population when \(n\) is small is $$ \bar{x} \pm(t \text { critical value }) \frac{s}{\sqrt{n}} $$ What is the appropriate \(t\) critical value for each of the following confidence levels and sample sizes? a. \(95 \%\) confidence, \(n=17\) b. \(90 \%\) confidence, \(n=12\) c. \(99 \%\) confidence, \(n=24\) d. \(90 \%\) confidence, \(n=25\) e. \(90 \%\) confidence, \(n=13\) f. \(95 \%\) confidence, \(n=10\)

Five hundred randomly selected working adults living in Calgary, Canada were asked how long, in minutes, their typical daily commute was (Calgary Herald Traffic Study, Ipsos, September 17,2005 ). The resulting sample mean and standard deviation of commute time were \(28.5\) minutes and \(24.2\) minutes, respectively. Construct and interpret a \(90 \%\) confidence interval for the mean commute time of working adult Calgary residents.
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