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Chapter 10: Problem 41

The article "Americans Seek Spiritual Guidance on Web" (San Luis Obispo Tribune, October 12,2002 ) reported that \(68 \%\) of the general population belong to a religious community. In a survey on Internet use, \(84 \%\) of "religion surfers" (defined as those who seek spiritual help online or who have used the web to search for prayer and devotional resources) belong to a religious community. Suppose that this result was based on a sample of 512 religion surfers. Is there convincing evidence that the proportion of religion surfers who belong to a religious community is different from \(.68\), the proportion for the general population? Use \(\alpha=.05\).

Short Answer

Expert verified
To provide a short answer, the actual computation of the z score and P-value are necessary. After that, based on the comparison of the P-value to \(\alpha = 0.05\), we would know whether to reject or fail to reject the null hypothesis, and hence could conclude if there is convincing evidence that the proportion of religion surfers who belong to a religious community is different from \(0.68\), as that the proportion for the general population.

Step by step solution

01

State the Hypotheses

We start by stating the null and alternative hypotheses. The null hypothesis assumes no difference between the two proportions, while the alternative hypothesis states otherwise. Let \( p \) denote the proportion of 'religion surfers' who belong to a religious community.\n\nNull hypothesis \( H_0 \): \( p = 0.68 \)\nAlternative hypothesis \( H_a \): \( p 鈮� 0.68 \)
02

Calculate the Test Statistic

We'll use the formula for the two-proportion z-test to find the z score:\n\n\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \]\n\nwhere:\n\(\hat{p}\) is the sample proportion (0.84),\n\(p_0\) is the stated population proportion (0.68), and\n\(n\) is the sample size (512).\n\nCarrying out the calculation yields: \n\n\[ z = \frac{0.84 - 0.68}{\sqrt{\frac{0.68(1 - 0.68)}{512}}} \]
03

Find the P-value

Once we find the z score, we can determine the P-value, which is the probability under the null hypothesis of obtaining a z score as extreme as the one calculated. The P-value is found using a z-table or statistical software.
04

Draw Conclusion

We compare the P-value to the significance level, \(\alpha = 0.05\). \n\nIf the P-value is less than \(\alpha\), we reject the null hypothesis and conclude that the proportion of religion surfers belonging to a religious community is significantly different from that of the general population.\n\nOtherwise, if the P-value is greater than \(\alpha\), we fail to reject the null hypothesis, implying that the proportion for religion surfers may be equal to 0.68.

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Most popular questions from this chapter

The city council in a large city has become concerned about the trend toward exclusion of renters with children in apartments within the city. The housing coordinator has decided to select a random sample of 125 apartments and determine for each whether children are permitted. Let \(\pi\) be the true proportion of apartments that prohibit children. If \(\pi\) exceeds . 75 , the city council will consider appropriate legislation. a. If 102 of the 125 sampled apartments exclude renters with children, would a level \(.05\) test lead you to the conclusion that more than \(75 \%\) of all apartments exclude children? b. What is the power of the test when \(\pi=.8\) and \(\alpha=.05\) ?

The article "Fewer Parolees Land Back Behind Bars" (Associated Press, April 11,2006 ) includes the following statement: "Just over 38 percent of all felons who were released from prison in 2003 landed back behind bars by the end of the following year, the lowest rate since 1979." Explain why it would not be necessary to carry out a hypothesis test to determine if the proportion of felons released in 2003 was less than \(.40\).

Use the definition of the \(P\) -value to explain the following: a. Why \(H_{0}\) would certainly be rejected if \(P\) -value \(=.0003\) b. Why \(H_{0}\) would definitely not be rejected if \(P\) -value \(=\) \(.350\)

A survey of teenagers and parents in Canada conducted by the polling organization Ipsos ("Untangling the Web: The Facts About Kids and the Internet," January \(25 .\) 2006) included questions about Internet use. It was reported that for a sample of 534 randomly selected teens, the mean number of hours per week spent online was \(14.6\) and the standard deviation was \(11.6\). a. What does the large standard deviation, \(11.6\) hours, tell you about the distribution of online times for this sample of teens? b. Do the sample data provide convincing evidence that the mean number of hours that teens spend online is greater than 10 hours per week?

According to the article "Workaholism in Organizations: Gender Differences" (Sex Roles [1999]: \(333-346\) ), the following data were reported on 1996 income for random samples of male and female MBA graduates from a certain Canadian business school: \begin{tabular}{lccc} & \(\boldsymbol{N}\) & \(\overline{\boldsymbol{x}}\) & \(\boldsymbol{s}\) \\ \hline Males & 258 & \(\$ 133,442\) & \(\$ 131,090\) \\ Females & 233 & \(\$ 105,156\) & \(\$ 98,525\) \\ \hline \end{tabular} Note: These salary figures are in Canadian dollars. a. Test the hypothesis that the mean salary of male MBA graduates from this school was in excess of \(\$ 100,000\) in \(1996 .\) b. Is there convincing evidence that the mean salary for all female MBA graduates is above \(\$ 100,000 ?\) Test using \(\alpha=.10\) c. If a significance level of \(.05\) or \(.01\) were used instead of \(.10\) in the test of Part (b), would you still reach the same conclusion? Explain.
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