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Chapter 7: Problem 82

Pew Research reported that \(46 \%\) of Americans surveyed in 2016 got their news from local television. A similar survey conducted in 2017 found that \(37 \%\) of Americans got their news from local television. Assume the sample size for each poll was 1200 . a. Construct the \(95 \%\) confidence interval for the difference in the proportions of Americans who get their news from local television in 2016 and 2017 . b. Based on your interval, do you think there has been a change in the proportion of Americans who get their news from local television? Explain.

Short Answer

Expert verified
The solution to part 'a' will be the calculated confidence interval, while the answer to 'b' will be a statement indicating whether there was a significant change in the proportions, based on the interpretation of the confidence interval results.

Step by step solution

01

Calculate the sample proportions

The sample proportion for 2016 is given as \(46\%\) or \(0.46\) and for 2017 it is \(37\%\) or \(0.37\).
02

Calculate the standard error

The standard error (SE) for the difference of two proportions is calculated using the formula: \[SE = \sqrt{ \frac{ p_{1}(1 - p_{1}) }{ n_{1} } + \frac{ p_{2}(1 - p_{2}) }{ n_{2} } }\]where \(p_{1}\) and \(p_{2}\) are the sample proportions, and \(n_{1}\) and \(n_{2}\) are the sample sizes for the two groups. In this case, the given sample size for both years (2016 and 2017) is 1200. Substitute these values into the formula to get the SE.
03

Construct the confidence interval

The formula for a \(95\%\) confidence interval is \((p_{1} - p_{2}) 卤 (Z*SE)\), where Z is the z-score corresponding to the desired level of confidence. For a \(95\%\) confidence interval, the z-score is approximately 1.96. This z-score states that about \(95\%\) of the area under the normal curve lies within plus or minus 1.96 standard deviations from the mean. Thus substitute the calculated values of standard error (SE), \(p_{1}\), \(p_{2}\), and \(Z=1.96\) into the formula to get the confidence interval.
04

Interpret the Confidence Interval

If the confidence interval includes 0, it would imply that there is no significant difference between the two proportions. If it does not include 0, it would suggest that there is a significant difference. This interpretation will determine the answer to part 'b' of the problem.

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Most popular questions from this chapter

According to a 2017 Pew Research Center report on voting issues, \(59 \%\) of Americans feel that the everything should be done to make it easy for every citizen to vote. Suppose a random sample of 200 Americans is selected. We are interested in finding the probability that the proportion of the sample who feel with way is greater than \(55 \%\). a. Without doing any calculations, determine whether this probability will be greater than \(50 \%\) or less than \(50 \%\). Explain your reasoning. b. Calculate the probability that the sample proportion is \(55 \%\) or more.

Has trust in the executive branch of government declined? A Gallup poll asked U.S. adults if they trusted the executive branch of government in 2008 and again in 2017 . The results are shown in the table. $$\begin{array}{|l|r|}\hline & \mathbf{2 0 0 8} & \mathbf{2 0 1 7} \\ \hline \text { Yes } & 623 & 460 \\\\\hline \text { No } & 399 & 562 \\\\\hline \text { Total } & 1022 & 1022 \\ \hline\end{array}$$ a. Find and compare the sample proportion for those who trusted the executive branch in 2008 and in 2017 . b. Find the \(95 \%\) confidence interval for the difference in the population proportions. Assume the conditions for using the confidence interval are met. Based on the interval, has there been a change in the proportion of U.S. adults who trust the executive branch? Explain.

Suppose it is known that \(20 \%\) of students at a certain college participate in a textbook recycling program each semester. a. If a random sample of 50 students is selected, do we expect that exactly \(20 \%\) of the sample participates in the textbook recycling program? Why or why not? b. Suppose we take a sample of 500 students and find the sample proportion participating in the recycling program. Which sample proportion do you think is more likely to be closer to \(20 \%\) : the proportion from a sample size of 50 or the proportion from a sample size of \(500 ?\) Explain your reasoning.

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In the 2018 study Closing the STEM Gap, researchers wanted to estimate the percentage of middle school girls who planned to major in a STEM field. a. If a \(95 \%\) confidence level is used, how many people should be included in the survey if the researchers wanted to have a margin of error of \(3 \%\) ? b. How could the researchers adjust their margin of error if they want to decrease the number of study participants?
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