A simple random sample of size $\mathrm{n}$ is drawn from a population that is normally distributed. The sample mean, $\bar{x}$, is found to be 114 , and the sample standard deviation, $s$, is found to be 10 . (a) Construct a $98 \%$ confidence interval about $\mu$ if the sample size, $n$, is 28 . (b) Construct a $98 \%$ confidence interval about $\mu$ if the sample size, $n$, is 12 (c) Construct a 95\% confidence interval about $\mu$ if the sample size, $n$, is 28 . (d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed? Click the icon to view the table of areas under the t-distribution. (a) Construct a $98 \%$ confidence interval about $\mu$ if the sample size, $n$, is 28 . Lower bound: $\square$; Upper bound: (Use ascending order. Round to one decimal place as needed.)

Step 1 ：We are given that the sample mean, \(\bar{x}\), is 114, the sample standard deviation, \(s\), is 10, and the sample size, \(n\), is 28. We are asked to construct a 98% confidence interval about \(\mu\).

Step 2 ：The formula for a confidence interval is \(\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(t_{\alpha/2}\) is the t-score for a given level of confidence, \(s\) is the sample standard deviation, and \(n\) is the sample size.

Step 3 ：For a 98% confidence interval, the level of significance, \(\alpha\), is 0.02. Since we are constructing a two-tailed test, we need to divide \(\alpha\) by 2 before looking up the t-score. This gives us \(\alpha/2 = 0.01\).

Step 4 ：The degrees of freedom is \(n - 1 = 28 - 1 = 27\).

Step 5 ：We look up the t-score for \(\alpha/2 = 0.01\) and 27 degrees of freedom, which is approximately 2.47.

Step 6 ：Substituting the values into the formula, we get the lower bound of the confidence interval as \(114 - 2.47 \cdot \frac{10}{\sqrt{28}}\) and the upper bound as \(114 + 2.47 \cdot \frac{10}{\sqrt{28}}\).

Step 7 ：Calculating these values, we find the lower bound to be approximately 109.3 and the upper bound to be approximately 118.7.

Step 8 ：Thus, the 98% confidence interval about \(\mu\) when the sample size, \(n\), is 28 is \(\boxed{[109.3, 118.7]}\).