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Chapter 9: Problem 28

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 is between \(59.7\%\) and \(64.3\%\). The sample has to be a random sample and should be large enough (np and n(1-p) should be greater than or equal to 10) for normal approximation to be valid for this interval to be valid.

Step by step solution

01

Identify the Relevant Data

Firstly, the relevant data must be identified from the problem. The sample size (n) is 1710 children, and the number of successes (X), i.e., the number of children who watch TV before school, is 1060.
02

Calculate the Sample Proportion (p虃)

The sample proportion (p虃) is calculated as the ratio of successes to the sample size, i.e., \( p虃 = X/n \). In this case, \( p虃 = 1060 / 1710 = 0.62 \) or \(62\% \). This is the proportion of children in the sample who watch TV before school.
03

Calculate the Margin of Error

To construct a \(95\%\) confidence interval, we need to calculate the margin of error. For a \(95\%\) confidence interval, the value of z (from standard normal distribution table) is 1.96. The formula to calculate the margin of error (E) is \( E = z * sqrt( (p虃 * (1- p虃) )/ n) \). Substituting in the given values, we find that \( E = 1.96 * sqrt((0.62 * 0.38) / 1710 ) = 0.023 \) or \(2.3\% \).
04

Construct the Confidence Interval

Now that we have all values, we construct the confidence interval as \( p虃 卤 E \). Lower limit of the interval is \(0.62 - 0.023 = 0.597\) or \(59.7\%\) and upper limit is \(0.62 + 0.023 = 0.643\) or \(64.3\%\). Therefore, we are \(95\%\) confident that the true proportion of all Australian children who watch TV before school lies between \(59.7\%\) and \(64.3\%\).
05

State the Assumption

The assumption that must hold true for this method to be valid is that the sample must be a random sample and it should be large enough (np and n(1-p) should be greater than or equal to 10 for normal approximation to be valid). In this case, it seems reasonable to believe that the sample is random and large enough to use this method.

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Most popular questions from this chapter

The article "National Geographic, the Doomsday Machine," which appeared in the March 1976 issue of the Journal of Irreproducible Results (yes, there really is a journal by that name -it's a spoof of technical journals!) predicted dire consequences resulting from a nationwide buildup of National Geographic magazines. The author's predictions are based on the observation that the number of subscriptions for National Geographic is on the rise and that no one ever throws away a copy of National Geographic. A key to the analysis presented in the article is the weight of an issue of the magazine. Suppose that you were assigned the task of estimating the average weight of an issue of National Geographic. How many issues should you sample to estimate the average weight to within \(0.1 \mathrm{oz}\) with \(95 \%\) confidence? Assume that \(\sigma\) is known to be 1 oz.

The formula used to compute a confidence interval for the mean of a normal population when \(n\) is small is $$ \bar{x} \pm(t \text { critical value }) \frac{s}{\sqrt{n}} $$ What is the appropriate \(t\) critical value for each of the following confidence levels and sample sizes? a. \(95 \%\) confidence, \(n=17\) b. \(90 \%\) confidence, \(n=12\) c. \(99 \%\) confidence, \(n=24\) d. \(90 \%\) confidence, \(n=25\) e. \(90 \%\) confidence, \(n=13\) f. \(95 \%\) confidence, \(n=10\)

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