Eat your cereal: Boxes of cereal are labeled as containing 14 ounces. Following are the weights, in ounces, of a sample of 14 boxes. It is reasonable to assume that the population is approximately normal.
\begin{tabular}{lllllll}
\hline 13.99 & 13.96 & 14.08 & 14.09 & 14.07 & 13.99 & 14.12 \\
13.96 & 14.02 & 14.01 & 14.08 & 14.09 & 14.02 & 14.01 \\
\hline
\end{tabular}

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Part: $0 / 2$

Part 1 of 2
(a) Construct a $90 \%$ confidence interval for the mean weight. Round the answers to at least three decimal places.

A $90 \%$ confidence interval for the mean weight is $\square< \mu< \square$.
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Step 1 ：Given the weights of the 14 boxes of cereal, we first calculate the sample mean and the standard deviation. The sample mean is the sum of all the weights divided by the number of weights, and the standard deviation is the square root of the variance, which is the average of the squared differences from the mean.

Step 2 ：Next, we calculate the standard error of the mean, which is the standard deviation divided by the square root of the sample size.

Step 3 ：We then use the t-distribution to find the critical value for a 90% confidence interval. The critical value is the t-score that cuts off the upper 5% of the distribution (since we are constructing a two-tailed confidence interval).

Step 4 ：We multiply the critical value by the standard error to find the margin of error.

Step 5 ：Finally, we construct the 90% confidence interval for the mean weight by adding and subtracting the margin of error from the sample mean. The confidence interval is \(\boxed{14.010}<\mu<\boxed{14.060}\).