Question

the grade point averages (gpa) for 12 randomly selected college students are shown on the right. complete parts (a) through (c) below. assume the population is normally distributed. 2.3 3.3 2.9 1.5 0.7 4.0 2.4 1.3 3.8 0.2 2.4 3.3 (a) find the sample mean. \\( \overline { x } = 2.34 \\) (round to two decimal places as needed.) (b) find the sample standard deviation. \\( s = 1.21 \\) (round to two decimal places as needed.) (c) construct a 90% confidence interval for the population mean \\( \mu \\). a 90% confidence interval for the population mean is ( ) (round to two decimal places as needed.)

Explanation:

Step1: Determine the critical value

Since the population is normally distributed and the sample size \(n = 12\) (small - sample, \(n<30\)), we use the \(t\) - distribution. The degree of freedom \(df=n - 1=12-1 = 11\). For a \(90\%\) confidence interval, the significance level \(\alpha=1 - 0.90=0.10\), and \(\frac{\alpha}{2}=0.05\).
Using the \(t\) - table or a calculator, the critical value \(t_{\frac{\alpha}{2},df}=t_{0.05,11}=1.796\)

Step2: Calculate the margin of error

The formula for the margin of error \(E\) is \(E = t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\)
We know that \(s = 1.21\), \(n = 12\), and \(t_{\frac{\alpha}{2}}=1.796\)

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Step3: Construct the confidence interval

The formula for the confidence interval for the population mean \(\mu\) (when \(\sigma\) is unknown) is \(\bar{x}-E<\mu<\bar{x} + E\)
We know that \(\bar{x}=2.34\) and \(E = 0.63\)

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Answer:

\((1.71,2.97)\)