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QUESTION 6

Find the $99\mathrm{\%}$ 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 $\overline{x}$ using the formula $\overline{x}=\frac{\sum x_i}{n}$ where $x_i$ are the data points and $n$ is the sample size.

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

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 $t^{\ast }$ 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 $E=t^{\ast }\times SEM$.

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

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