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Chapter 9: Problem 16

The article "Students Increasingly Turn to Credit Cards" (San Luis Obispo Tribune, July 21,2006 ) reported that \(37 \%\) of college freshmen and \(48 \%\) of college seniors carry a credit card balance from month to month. Suppose that the reported percentages were based on random samples of 1000 college freshmen and 1000 college seniors. a. Construct a \(90 \%\) confidence interval for the proportion of college freshmen who carry a credit card balance from month to month. b. Construct a \(90 \%\) confidence interval for the proportion of college seniors who carry a credit card balance from month to month. c. Explain why the two \(90 \%\) confidence intervals from Parts (a) and (b) are not the same width.

Short Answer

Expert verified
The 90% confidence intervals for the proportion of college freshmen and seniors who carry a credit card balance from month to month are calculated separately due to differing proportions and equal sample sizes. The varying widths of the intervals are due to these factors.

Step by step solution

01

Calculate standard error for both populations

The standard error for proportion is calculated as \(SE = sqrt[p(1-p)/n]\), where p is the proportion and n is the sample size. For college freshmen, \(SE_1 = sqrt[0.37(1-0.37)/1000]\). For college seniors, \(SE_2 = sqrt[0.48(1-0.48)/1000]\). Respectively, calculate these values.
02

Find the z-value corresponding to a 90% confidence level

Using a z-table or an online calculator, find the z-value corresponding to a confidence level of 90%. This z-value will be the same for both groups as they both have a 90% confidence level. The z-value for a 90% confidence level is approximately 1.645.
03

Construct confidence intervals

A confidence interval for a proportion is calculated as \(p 卤 z*SE\). For freshmen, the confidence interval would be \(0.37 卤 1.645*SE_1\). For seniors, it would be \(0.48 卤 1.645*SE_2\). Calculate these ranges.
04

Explain the difference in widths

The width of a confidence interval is determined by the product of the z-value and the standard error. Since the z-value is constant for both groups, the difference in width must be due to differences in the standard errors, which in turn are influenced by the different proportions and equal sample sizes for the two groups.

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Most popular questions from this chapter

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.

In the article "Fluoridation Brushed Off by Utah" (Associated Press, August 24,1998 ), it was reported that a small but vocal minority in Utah has been successful in keeping fluoride out of Utah water supplies despite evidence that fluoridation reduces tooth decay and despite the fact that a clear majority of Utah residents favor fluoridation. To support this statement, the article included the result of a survey of Utah residents that found \(65 \%\) to be in favor of fluoridation. Suppose that this result was based on a random sample of 150 Utah residents. Construct and interpret a \(90 \%\) confidence interval for \(\pi\), the true proportion of Utah residents who favor fluoridation. Is this interval consistent with the statement that fluoridation is favored by a clear majority of residents?

The interval from \(-2.33\) to \(1.75\) captures an area of \(.95\) under the \(z\) curve. This implies that another largesample \(95 \%\) confidence interval for \(\mu\) has lower limit \(\bar{x}-2.33 \frac{\sigma}{\sqrt{n}}\) and upper limit \(\bar{x}+1.75 \frac{\sigma}{\sqrt{n}} .\) Would you recommend using this \(95 \%\) interval over the \(95 \%\) interval \(\bar{x} \pm 1.96 \frac{\sigma}{\sqrt{n}}\) discussed in the text? Explain. (Hint: Look at the width of each interval.)

Five students visiting the student health center for a free dental examination during National Dental Hygiene Month were asked how many months had passed since their last visit to a dentist. Their responses were as follows: \(\begin{array}{lllll}6 & 17 & 11 & 22 & 29\end{array}\) Assuming that these five students can be considered a random sample of all students participating in the free checkup program, construct a \(95 \%\) confidence interval for the mean number of months elapsed since the last visit to a dentist for the population of students participating in the program.

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.
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