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? ت 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: 109.3; Upper bound: 118.7 (Use ascending order. Round to one decimal place as needed.) (b) Construct a $98 \%$ confidence interval about $\mu$ if the sample size, $n$, is 12 . Lower bound: $\square$; Upper bound: (Use ascending order. Round to one decimal place as needed.)

Step 1 ：Given that the sample mean (\(\bar{x}\)) is 114, the sample standard deviation (s) is 10, and the sample size (n) is 12, we are asked to construct a 98% confidence interval for the population mean (\(\mu\)).

Step 2 ：The formula for a confidence interval is given by: \[\bar{x} \pm t_{\alpha/2, n-1} \times \frac{s}{\sqrt{n}}\] where \(t_{\alpha/2, n-1}\) is the t-score for a given level of confidence and degrees of freedom.

Step 3 ：In this case, we need to find the t-score for a 98% confidence level and 11 degrees of freedom (since the sample size is 12, the degrees of freedom is n-1 = 12-1 = 11).

Step 4 ：Using a t-distribution table or a statistical software, we find that the t-score for a 98% confidence level and 11 degrees of freedom is approximately 2.718.

Step 5 ：Substituting the given values into the formula, we get: \[114 \pm 2.718 \times \frac{10}{\sqrt{12}}\]

Step 6 ：Solving the above expression, we get the lower and upper bounds of the confidence interval as approximately 106.2 and 121.8 respectively.

Step 7 ：Thus, the 98% confidence interval for the population mean when the sample size is 12 is approximately \(\boxed{[106.2, 121.8]}\).