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Chapter 9: Problem 58

In a study of 1710 schoolchildren in Australia (Herald Sun, October 27,1994 ), 1060 children indicated that they normally watch TV before school in the morning. (Interestingly, only \(35 \%\) of the parents said their children watched TV before school!) Construct a \(95 \%\) confidence interval for the true proportion of Australian children who say they watch TV before school. What assumption about the sample must be true for the method used to construct the interval to be valid?

Short Answer

Expert verified
The 95% confidence interval for the true proportion of Australian children who say they watch TV before school will be calculated using the formula and substituting the calculated sample proportion, known z-value and sample size. The assumption is that the sample was a simple random sample and the central limit theorem applies.

Step by step solution

01

Calculate the Sample Proportion

The sample proportion \(p\) is calculated by dividing the number of success outcomes (children who watch TV before school) by the total number of outcomes (total number of children). In this case, \(p = \frac{1060}{1710}\), compute the value to get a fractional value of \(p\).
02

Use the Formula for Confidence Interval

The formula for the confidence interval is \(p \pm z * \sqrt{\frac{p*(1-p)}{n}}\), where \(p\) is the sample proportion, \(n\) is the sample size and \(z\) is the z-score associated with the desired confidence level, in this case 95%. Find the z-value in a Standard Normal Distribution Table, or use a mathematical software or calculator that can do this. For a 95% confidence level, the z-value is about 1.96. Substitute the known values into the formula and compute the value for the confidence interval.
03

State the Assumption

The assumption for this method to be valid is that the sample is a simple random sample; that every possible sample has an equally likely chance of occurring. The central limit theorem (which allows the formula to work) applies reasonably well if the number in the sample that are success outcomes (children who watch TV before school) \(np > 5\) and the number that are failures \(n(1-p) > 5\).

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Most popular questions from this chapter

Because of safety considerations, in May 2003 the Federal Aviation Administration (FAA) changed its guidelines for how small commuter airlines must estimate passenger weights. Under the old rule, airlines used 180 pounds as a typical passenger weight (including carry-on luggage) in warm months and 185 pounds as a typical weight in cold months. The Alaska Journal of Commerce (May 25,2003 ) reported that Frontier Airlines conducted a study to estimate average passenger plus carry-on weights. They found an average summer weight of 183 pounds and a winter average of 190 pounds. Suppose that each of these estimates was based on a random sample of 100 passengers and that the sample standard deviations were 20 pounds for the summer weights and 23 pounds for the winter weights. a. Construct and interpret a \(95 \%\) confidence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers. b. Construct and interpret a \(95 \%\) confidence interval for the mean winter weight (including carry-on luggage) of Frontier Airlines passengers. c. The new FAA recommendations are 190 pounds for summer and 195 pounds for winter. Comment on these recommendations in light of the confidence interval estimates from Parts (a) and (b).

The following quote is from the article "Credit Card Debt Rises Faster for Seniors" (USA Today. July \(28,2009)\) : The study, which will be released today by Demos, a liberal public policy group, shows that low- and middle-income consumers 65 and older carried \(\$ 10,235\) in average credit card debt last year. What additional information would you want in order to evaluate the accuracy of this estimate? Explain.

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One thousand randomly selected adult Americans participated in a survey conducted by the Assodated Press (June 2006 ). When asked "Do you think it is sometimes justified to lie or do you think lying is never justified?" \(52 \%\) responded that lying was never justified. When asked about lying to avoid hurting someone's feelings, 650 responded that this was often or sometimes okay. a. Construct a \(90 \%\) confidence interval for the proportion of adult Americans who think lying is never justified. b. Construct a \(90 \%\) confidence interval for the proportion of adult American who think that it is often or sometimes okay to lie to avoid hurting someone's feelings. c. Comment on the apparent inconsistency in the responses given by the individuals in this sample.

In an AP-AOL sports poll (Assodated Press. December 18,2005\(), 394\) of 1000 randomly selected U.S. adults indicated that they considered themselves to be baseball fans. Of the 394 baseball fans, 272 stated that they thought the designated hitter rule should either be expanded to both baseball leagues or eliminated. a. Construct a \(95 \%\) confidence interval for the proportion of U.S. adults who consider themselves to be baseball fans. b. Construct a \(95 \%\) confidence interval for the proportion of those who consider themselves to be baseball fans who think the designated hitter rule should be expanded to both leagues or eliminated. c. Explain why the confidence intervals of Parts (a) and (b) are not the same width even though they both have a confidence level of \(95 \%\).
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