Question

the state test scores for 12 randomly selected high school seniors are shown on the right. complete parts (a) through (c) below.
assume the population is normally distributed.
(a) find the sample mean.
\\( \overline { x } = 909.3 \\) (round to one decimal place as needed.)
(b) find the sample standard deviation.
\\( s = 304.7 \\) (round to one decimal place as needed.)
(c) construct a 99\\% confidence interval for the population mean \\( \mu \\).
a 99\\% confidence interval for the population mean is \\( ( \square, \square ) \\)
(round to one decimal place as needed.)

Explanation:

Step1: Determine the critical value

Since the population standard deviation is unknown and the sample size \(n = 12\), we use the \(t -\)distribution. The degrees of freedom \(df=n - 1=12-1 = 11\). For a \(99\%\) confidence interval, the significance level \(\alpha=1 - 0.99=0.01\), and \(\frac{\alpha}{2}=0.005\).
From the \(t -\)distribution table or a calculator, \(t_{\frac{\alpha}{2},df}=t_{0.005,11}=3.106\)

Step2: Calculate the margin of error \(E\)

The formula for the margin of error for a confidence interval for the population mean when \(\sigma\) is unknown is \(E = t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\)
We know that \(s = 304.7\), \(n = 12\), and \(t_{\frac{\alpha}{2}}=3.106\)

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Step3: Calculate the confidence interval

The formula for the confidence interval is \(\bar{x}-E<\mu<\bar{x} + E\)
We know that \(\bar{x}=909.3\) and \(E = 246.1\)

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Answer:

$(663.2, 1155.4)$