Question

17 multiple choice 4 points
construct the requested confidence interval.
in a recent study of 42 eighth graders, the mean number of hours per week that they watched television was 19.6 with a standard deviation of 5.8 hours. construct a 98% confidence interval for the population mean.
(14.1, 23.2)
(18.3, 20.9)
(17.4, 21.8)
(19.1, 20.4)

Explanation:

Step1: Determine the critical value

Since the sample size \(n = 42\) (\(n>30\)), we can use the standard normal distribution (approximate \(t -\)distribution). For a \(98\%\) confidence interval, the significance level \(\alpha=1 - 0.98=0.02\), and \(\alpha/2=0.01\). The critical value \(z_{\alpha/2}\) is such that \(P(Z>z_{\alpha/2}) = 0.01\). From the standard - normal table, \(z_{\alpha/2}=2.33\).

Step2: Calculate the margin of error

The formula for the margin of error \(E\) is \(E = z_{\alpha/2}\frac{s}{\sqrt{n}}\), where \(s = 5.8\) (sample standard deviation) and \(n = 42\).

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Step3: Calculate the confidence interval

The formula for the confidence interval for the population mean \(\mu\) (when \(\sigma\) is unknown and \(n\) is large) is \(\bar{x}-E<\mu <\bar{x} + E\), where \(\bar{x}=19.6\) (sample mean).

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Rounding to one - decimal place (to match the options), the confidence interval is approximately \((17.4,21.8)\)

Answer:

(17.4, 21.8)