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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 1,000 college freshmen and 1,000 college seniors. (Hint: See Example 9.5\()\) a. Construct and interpret a \(90 \%\) confidence interval for the proportion of college freshmen who carry a credit card balance from month to month. b. Construct and interpret 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 interval for the population proportion of college freshmen who carry a credit card balance is between 34.55% and 39.45%. The 90% confidence interval for the population proportion of college seniors who carry a credit card balance is between 45.56% and 50.44%.

Step by step solution

01

Calculate the Sample Proportions

First, calculate the sample proportions. In the sample of 1,000 college freshmen, 37% carry a balance, so the sample proportion (\(\hat{p_1}\)) is 0.37. In the sample of 1,000 college seniors, 48% carry a balance, so the sample proportion (\(\hat{p_2}\)) is 0.48.
02

Construct the Confidence Intervals

Now, construct the 90% confidence intervals using the formula: \(\hat{p} \pm z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\), where \(z\) is the z-score for the desired level of confidence. For a 90% confidence interval, \(z = 1.645\). For college freshmen, the CI is: \(0.37 \pm 1.645\sqrt{\frac{0.37(1-0.37)}{1000}}\) and for college seniors, the CI is: \(0.48 \pm 1.645\sqrt{\frac{0.48(1-0.48)}{1000}}\).
03

Calculate the Confidence Intervals

Next, calculate the actual confidence intervals. For college freshmen: \(0.37 \pm 1.645(0.017) = [0.3455, 0.3945]\) For college seniors: \(0.48 \pm 1.645(0.016) = [0.4556, 0.5044]\).
04

Interpret the Confidence Intervals

The 90% confidence interval for the population proportion of college freshmen who carry a credit card balance is between 0.3455 and 0.3945. This means that we are 90% confident that the true proportion of college freshmen carrying a credit card balance is within this interval. Similarly, the 90% confidence interval for the population proportion of college seniors who carry a credit card balance is between 0.4556 and 0.5044. We are 90% confident that the true proportion of college seniors carrying a credit card balance is within this interval.
05

Explain the Differences in Confidence Intervals

The two confidence intervals are not the same width because the sample proportions are different. A bigger sample proportion means a larger variability, resulting in a wider confidence interval. As a result, the confidence interval for college seniors is wider than for college freshmen.

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Most popular questions from this chapter

Appropriate use of the interval $$ \hat{p} \pm(z \text { critial value }) \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} $$ requires a large sample. For each of the following combinations of \(n\) and \(\hat{p}\), indicate whether the sample size is large enough for this interval to be appropriate. $$ \begin{array}{l} \text { a. } n=100 \text { and } \hat{p}=0.70 \\ \text { b. } n=40 \text { and } \hat{p}=0.25 \\ \text { c. } n=60 \text { and } \hat{p}=0.25 \end{array} $$ d. \(n=80\) and \(\hat{p}=0.10\)

Consider taking a random sample from a population with \(p=0.70\). a. What is the standard error of \(\hat{p}\) for random samples of size \(100 ?\) b. Would the standard error of \(\hat{p}\) be smaller for samples of size 100 or samples of size \(400 ?\) c. Does decreasing the sample size by a factor of \(4,\) from 400 to 100 , result in a standard error of \(\hat{p}\) that is four times as large?

A researcher wants to estimate the proportion of students enrolled at a university who are registered to vote. Would the standard error of the sample proportion \(\hat{p}\) be larger if the actual population proportion was \(p=0.4\) or \(p=0.8\) ?

Based on data from a survey of 1,200 randomly selected Facebook users (USA Today, March 24, 2010), a \(90 \%\) confidence interval for the proportion of all Facebook users who say it is not OK to "friend" someone who reports to you at work is (0.60,0.64) . What is the meaning of the \(90 \%\) confidence level associated with this interval?

Use the formula for the standard error of \(\hat{p}\) to explain why a. The standard error is greater when the value of the population proportion \(p\) is near 0.5 than when it is near \(1 .\) b. The standard error of \(\hat{p}\) is the same when the value of the population proportion is \(p=0.2\) as it is when \(p=0.8\)

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