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Chapter 8: Problem 26

A random sample of size 300 is to be selected from a population. Determine the mean and standard deviation of the sampling distribution of \(\hat{p}\) for each of the following population proportions. a. \(p=0.20\) b. \(p=0.45\) c. \(p=0.70\) d. \(p=0.90\)

Short Answer

Expert verified
The mean and standard deviation for each case are: a. \(\mu_{\hat{p}} = 0.20, \sigma_{\hat{p}} = \sqrt{\frac{0.20 * (1-0.20)}{300}}\)b. \(\mu_{\hat{p}} = 0.45, \sigma_{\hat{p}} = \sqrt{\frac{0.45 * (1-0.45)}{300}}\)c. \(\mu_{\hat{p}} = 0.70, \sigma_{\hat{p}} = \sqrt{\frac{0.70 * (1-0.70)}{300}}\)d. \(\mu_{\hat{p}} = 0.90, \sigma_{\hat{p}} = \sqrt{\frac{0.90 * (1-0.90)}{300}}\)

Step by step solution

01

Calculate mean for each population proportion

The mean \(\mu_{\hat{p}}\) of the sampling distribution of the sample proportion is equal to the population proportion. So, for each case: a. \(\mu_{\hat{p}} = p = 0.20\)b. \(\mu_{\hat{p}} = p = 0.45\)c. \(\mu_{\hat{p}} = p = 0.70\)d. \(\mu_{\hat{p}} = p = 0.90\)
02

Calculate standard deviation for each population proportion

The standard deviation \(\sigma_{\hat{p}}\) of the sampling distribution of the sample proportion is calculated using the formula \(\sqrt{\frac{p(1-p)}{n}}\), where \(n\) is the sample size and \(p\) is the population proportion. For each case:a. \(\sigma_{\hat{p}} = \sqrt{\frac{0.20 * (1-0.20)}{300}}\)b. \(\sigma_{\hat{p}} = \sqrt{\frac{0.45 * (1-0.45)}{300}}\)c. \(\sigma_{\hat{p}} = \sqrt{\frac{0.70 * (1-0.70)}{300}}\)d. \(\sigma_{\hat{p}} = \sqrt{\frac{0.90 * (1-0.90)}{300}}\)

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Most popular questions from this chapter

For which of the following combinations of sample size and population proportion would the standard deviation of \(\hat{p}\) be smallest? $$ \begin{array}{ll} n=40 & p=0.3 \\ n=60 & p=0.4 \\ n=100 & p=0.5 \end{array} $$

Consider the following statement: The proportion of all calls made to a county \(9-1-1\) emergency number during the year 2011 that were nonemergency calls was \(0.14 .\) a. Is the number that appears in boldface in this statement a sample proportion or a population proportion? b. Which of the following use of notation is correct, \(p=0.14\) or \(\hat{p}=0.14 ?\)

The article referenced in the previous exercise also reported that \(38 \%\) of the 1,200 social network users surveyed said it was OK to ignore a coworker's friend request. If \(p=0.38\) is used as an estimate of the proportion of all social network users who believe this, is it likely that this estimate is within 0.05 of the actual population proportion? Use what you know about the sampling distribution of \(\hat{p}\) to support your answer.

The article "Unmarried Couples More Likely to Be Interracial" (San Luis Obispo Tribune, March 13,2002 ) reported that \(7 \%\) of married couples in the United States are mixed racially or ethnically. Consider the population consisting of all married couples in the United States. a. A random sample of \(n=100\) couples will be selected from this population and \(\hat{p},\) the proportion of couples that are mixed racially or ethnically, will be calculated. What are the mean and standard deviation of the sampling distribution of \(\hat{p} ?\) b. Is the sampling distribution of \(\hat{p}\) approximately normal for random samples of size \(n=100 ?\) Explain. c. Suppose that the sample size is \(n=200\) rather than \(n=\) \(100 .\) Does the change in sample size affect the mean and standard deviation of the sampling distribution of \(\hat{p} ?\) If so, what are the new values for the mean and standard deviation? If not, explain why not. d. Is the sampling distribution of \(\hat{p}\) approximately normal for random samples of size \(n=200 ?\) Explain.

The report "California's Education Skills Gap: Modest Improvements Could Yield Big Gains" (Public Policy Institute of California, April \(16,2008,\) www.ppic.org) states that nationwide, \(61 \%\) of high school graduates go on to attend a two-year or four-year college the year after graduation. The proportion of high school graduates in California who go on to college was estimated to be \(0.55 .\) Suppose that this estimate was based on a random sample of 1,500 California high school graduates. Is it reasonable to conclude that the proportion of California high school graduates who attend college the year after graduation is different from the national figure? (Hint: Use what you know about the sampling distribution of \(\hat{p}\). You might also refer to Example \(8.5 .)\)
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