Question

the grade point averages (gpa) for 12 randomly selected college students are shown on the right. complete parts (a) through (c) below. 2.4 3.1 2.9 1.6 0.9 4.0 2.3 1.2 3.5 0.3 2.1 3.4 assume the population is normally distributed. (a) find the sample mean. \\( \overline { x } = 2.31 \\) (round to two decimal places as needed.) (b) find the sample standard deviation. \\( s = 1.14 \\) (round to two decimal places as needed.) (c) construct a 99% confidence interval for the population mean \\( \mu \\). a 99% confidence interval for the population mean is (round to two decimal places as needed.)

Explanation:

Step1: Determine the degrees of freedom and t - value

The sample size \(n = 12\). 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 \(\frac{\alpha}{2}=0.005\).
Using the t - distribution table or a calculator, the t - value \(t_{\frac{\alpha}{2},df}=t_{0.005,11}=3.106\)

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

The formula for the margin of error for a confidence interval when the population standard deviation \(\sigma\) is unknown (using the sample standard deviation \(s\)) is \(E = t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\)
We know that \(s = 1.14\), \(n = 12\), 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 \(\bar{x}=2.31\) and \(E = 1.022\)

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Answer:

\((1.29,3.33)\)