A simple random sample of size $n$ 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: 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: 106.2 ; Upper bound: 121.8 (Use ascending order. Round to one decimal place as needed.) How does decreasing the sample size affect the margin of error, E? A. As the sample size decreases, the margin of error decreases. B. As the sample size decreases, the margin of error stays the same. c. As the sample size decreases, the margin of error increases.
Solution
Step 1 ：We are given a sample size of 28, a sample mean of 114, and a sample standard deviation of 10. We are asked to construct a 98% confidence interval for the population mean.
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 corresponding to the desired level of confidence, \(s\) is the sample standard deviation, and \(n\) is the sample size.
Step 3 ：We can use a t-distribution table or a statistical software to find the t-score. For a 98% confidence level and degrees of freedom equal to 27 (which is the sample size minus 1), the t-score is approximately 2.4727.
Step 4 ：Substituting the given values into the formula, we get \(114 \pm 2.4727 \cdot \frac{10}{\sqrt{28}}\).
Step 5 ：Calculating the margin of error, we get approximately 4.673.
Step 6 ：Subtracting and adding the margin of error from the sample mean, we get the lower and upper bounds of the confidence interval, which are approximately 109.3 and 118.7, respectively.
Step 7 ：Final Answer: The 98% confidence interval for the population mean, given a sample size of 28, a sample mean of 114, and a sample standard deviation of 10, is approximately \(\boxed{[109.3, 118.7]}\).
