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Chapter 8: Problem 36

A 2018 Gallup poll of 2228 randomly selected U.S. adults found that \(39 \%\) planned to watch at least a "fair amount" of the 2018 Winter Olympics. In \(2014,46 \%\) of U.S. adults reported planning to watch at least a "fair amount." a. Does this sample give evidence that the proportion of U.S. adults who planned to watch the 2018 Winter Olympics was less than the proportion who planned to do so in 2014 ? Use a \(0.05\) significance level. b. After conducting the hypothesis test, a further question one might ask is what proportion of all U.S. adults planned to watch at least a "fair amount" of the 2018 Winter Olympics. Use the sample data to construct a \(90 \%\) confidence interval for the population proportion. How does your confidence interval support your hypothesis test conclusion?

Short Answer

Expert verified
Using the hypothesis test, one can make an evidence-based conclusion about whether the proportion of U.S. adults planning to watch the 2018 Winter Olympics was less than the proportion that planned to do so in 2014. Also, the confidence interval allows for estimation of the proportion of all U.S. adults who planned to watch at least a 'fair amount' of the 2018 Winter Olympics.

Step by step solution

01

State the Hypotheses

The null hypothesis (\(H_0\)) is that the proportion of U.S. adults planning to watch the 2018 Winter Olympics is same as that of 2014 (i.e., 46%). The alternative hypothesis (\(H_A\)) is that the proportion for 2018 is less than that of 2014. Formally, these can be stated as: \(H_0: p = 0.46\) and \(H_A: p < 0.46\).
02

Calculate the Test Statistic

The test statistic for a proportion is given by \(Z = \frac {\hat{p} - p_0}{\sqrt {p_0(1-p_0) / n}} \) where \(p_0 = 0.46\) (proportion from null hypothesis), \(\hat{p} = 0.39\) (sample proportion) and n = 2228 (sample size). Plug these values into the formula to get the test statistic.
03

Find P-value

The P-value can be found using the standard normal distribution table or a statistical software. The p-value represents the probability of achieving the observed test statistic or more extreme, under the null hypothesis.
04

Make Decision

Compare the P-value to the significance level of 0.05. If the P-value is less than or equal to 0.05, then reject the null hypothesis.
05

Construct the Confidence Interval

Using the formula for the confidence interval for a proportion, which is \(\hat{p} \pm Z_{\alpha/2} * \sqrt{(\hat{p}(1-\hat{p})/n)}\), where \(Z_{\alpha/2}\) is the standard normal distribution value at the desired confidence level (90% in this case), \(\hat{p}\) is the sample proportion, and n is the sample size.
06

Conclusion

The confidence interval provides a range of likely values for the proportion of all U.S. adults who planned to watch at least a 'fair amount' of the 2018 Winter Olympics. This can be used to support the conclusion from the hypothesis test.

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Most popular questions from this chapter

Morse determined that the percentage of \(t\) 's in the English language in the 1800 s was \(9 \%\). A random sample of 600 letters from a current newspaper contained \(48 t\) 's. Using the \(0.10\) level of significance, test the hypothesis that the proportion of \(t\) 's in this modern newspaper is \(0.09\).

The mother of a teenager has heard a claim that \(25 \%\) of teenagers who drive and use a cell phone reported texting while driving. She thinks that this rate is too high and wants to test the hypothesis that fewer than \(25 \%\) of these drivers have texted while driving. Her alternative hypothesis is that the percentage of teenagers who have texted when driving is less than \(25 \%\).$$\begin{aligned} &\mathrm{H}_{0}: p=0.25 \\\&\mathrm{H}_{\mathrm{a}}: p<0.25\end{aligned}$$ She polls 40 randomly selected teenagers, and 5 of them report having texted while driving, a proportion of \(0.125 .\) The p-value is \(0.034\). Explain the meaning of the p-value in the context of this question.

The researchers in a Pew study interviewed two random samples, one in 2015 and one in 2018 . Both samples were asked, "Have you read a print book in the last year?" The results are shown in the table below. $$\begin{array}{|lrrl|}\hline \text { Read a print book } & \mathbf{2 0 1 5} & \mathbf{2 0 1 8} & \text { Total } \\ \hline \text { Yes } & 1201 & 1341 & 2542 \\\\\hline \text { No } & 705 & 661 & 1366 \\ \hline \text { Total } & 1906 & 2002 & \\\\\hline\end{array}$$ a. Find and compare the sample proportions that had read a print book for these two groups. b. Find a pooled estimate of the sample proportion. c. Has the proportion who read print books increased? Find the observed value of the test statistic to test the hypotheses \(\mathrm{H}_{0}: p_{2015}=p_{2018}\) and \(\mathrm{H}_{\mathrm{a}}: p_{2015}

Pew Research reported that in the 2016 presidential election, \(53 \%\) of all male voters voted for Trump and \(41 \%\) voted for Clinton. Among all women voters, \(42 \%\) voted for Trump and \(54 \%\) voted for Clinton. Would it be appropriate to do a two-proportion \(z\) -test to determine whether the proportions of men and women who voted for Trump were significantly different (assuming we knew the number of men and women who voted)? Explain.

Refer to Exercise \(8.97 .\) Suppose 14 out of 20 voters in Pennsylvania report having voted for an independent candidate. The null hypothesis is that the population proportion is \(0.50 .\) What value of the test statistic should you report?
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