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QUESTION 6

Find the $99 \%$ confidence interval for the true mean for the grade point average (GPA) of 13 randomly selected college students with data are: $3.3,2.6,1.8,0.2,3.1,4.0,0.7,2.3,2.0,3.1,3.4,1.3,2.6$
$(1.40,3.28)$
$(1.79,2.89)$
$(1.67,3.01)$
$(1.51,3.17)$
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Step 1 ：Calculate the sample mean \( \bar{x} \) using the formula \( \bar{x} = \frac{\sum x_i}{n} \) where \( x_i \) are the data points and \( n \) is the sample size.

Step 2 ：Calculate the sample standard deviation \( s \) using the formula \( s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \).

Step 3 ：Calculate the standard error of the mean (SEM) using the formula \( SEM = \frac{s}{\sqrt{n}} \).

Step 4 ：Determine the degrees of freedom \( df \) which is \( n - 1 \).

Step 5 ：Find the t-critical value \( t^* \) for a 99% confidence level and \( df \) degrees of freedom using the t-distribution.

Step 6 ：Calculate the margin of error \( E \) using the formula \( E = t^* \times SEM \).

Step 7 ：Calculate the confidence interval using the formula \( CI = (\bar{x} - E, \bar{x} + E) \).

Step 8 ：Conclude the final answer: The 99% confidence interval for the true mean GPA of the 13 randomly selected college students is \( \boxed{(1.40, 3.28)} \).