16. You randomly select 16 coffee shops and measure the temperature of the coffee sold at each. The sample mean temperature is $162.0 \mathrm{~F}$ with a sample standard deviation of $10.0 \mathrm{~F}$. Construct a $95 \%$ confidence interval for the population mean temperature of coffee sold. Assume the temperatures are
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approximately normally distributed. (4 points)

Step 1 ：Given that the sample mean temperature is $162.0 \mathrm{~F}$, the sample standard deviation is $10.0 \mathrm{~F}$, and the sample size is 16 coffee shops.

Step 2 ：We are asked to construct a 95% confidence interval for the population mean temperature of coffee sold. The formula for a confidence interval is given by: \[ \bar{x} \pm z \frac{s}{\sqrt{n}} \]

Step 3 ：In this formula, \(\bar{x}\) is the sample mean, \(z\) is the z-score corresponding to the desired confidence level, \(s\) is the sample standard deviation, and \(n\) is the sample size.

Step 4 ：For a 95% confidence level, the z-score is 1.96.

Step 5 ：Substituting the given values into the formula, we get: \[162.0 \pm 1.96 \frac{10.0}{\sqrt{16}} \]

Step 6 ：Solving the above expression, we get the margin of error as 4.9.

Step 7 ：Subtracting and adding the margin of error from the sample mean, we get the lower and upper bounds of the confidence interval as 157.1 and 166.9 respectively.

Step 8 ：Final Answer: The 95% confidence interval for the population mean temperature of coffee sold is \(\boxed{[157.1, 166.9]}\) degrees Fahrenheit.