Question 7 (1 point)
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Provide an appropriate response.

Construct a $90 \%$ confidence interval for the population mean, $\mu$. Assume the population has a normal distribution. A sample of 15 randomly selected math majors has a grade point average of 2.86 with a standard deviation of 0.78 . Round to the nearest hundredth.
A) $(2.51,3.21)$
B) $(2.28,3.66)$
C) $(2.37,3.56)$
D) $(2.41,3.42)$

Step 1 ：Given values are sample mean (\(x_{bar}\)) = 2.86, standard deviation (\(\sigma\)) = 0.78, sample size (n) = 15, and Z-score for 90% confidence interval (Z) = 1.645.

Step 2 ：Calculate the margin of error using the formula: Margin of Error = Z * \(\sigma/\sqrt{n}\).

Step 3 ：Substitute the given values into the formula: Margin of Error = 1.645 * (0.78/\(\sqrt{15}\)) = 0.3312949954345824.

Step 4 ：Calculate the confidence interval using the formula: Confidence Interval = \(x_{bar}\) ± Margin of Error.

Step 5 ：Substitute the values into the formula: Lower Limit = 2.86 - 0.3312949954345824 = 2.53 (rounded to the nearest hundredth), Upper Limit = 2.86 + 0.3312949954345824 = 3.19 (rounded to the nearest hundredth).

Step 6 ：Final Answer: The 90% confidence interval for the population mean is \(\boxed{(2.53, 3.19)}\).