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 $90 \%$ confidence interval about $\mu$ if the sample size, $n$, is 29 .
(b) Construct a $90 \%$ confidence interval about $\mu$ if the sample size, $n$, is 11 .
(c) Construct an $80 \%$ confidence interval about $\mu$ if the sample size, $n$, is 29 .
(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.
(Use ascending order. Round to one decimal place as needed)
Compare the results to those obtained in part (a). How does decreasing the level of confidence affect the size of the margin of error, E?
A. As the level of confidence decreases, the size of the interval decreases.
B. As the level of confidence decreases, the size of the interval stays the same.
C. As the level of confidence decreases, the size of the interval increases.
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Step 1 ：To construct a confidence interval, we use the formula: \(CI = \bar{x} ± t(\alpha/2, n-1) * (s/\sqrt{n})\), where \(\bar{x}\) is the sample mean, \(t(\alpha/2, n-1)\) is the t-score for a given level of confidence and degrees of freedom (n-1), s is the sample standard deviation, and n is the sample size.

Step 2 ：We can find the t-score from a t-distribution table. For a 90% confidence level, the t-score is 1.699 for n=29 (28 degrees of freedom) and 1.812 for n=11 (10 degrees of freedom). For an 80% confidence level, the t-score is 1.311 for n=29 (28 degrees of freedom).

Step 3 ：For n=29 and a 90% confidence level, we calculate the confidence interval as follows: \(CI = 114 ± 1.699 * (10/\sqrt{29}) = 114 ± 3.14\). So, the 90% confidence interval is \(\boxed{(110.86, 117.14)}\).

Step 4 ：For n=11 and a 90% confidence level, we calculate the confidence interval as follows: \(CI = 114 ± 1.812 * (10/\sqrt{11}) = 114 ± 5.46\). So, the 90% confidence interval is \(\boxed{(108.54, 119.46)}\).

Step 5 ：For n=29 and an 80% confidence level, we calculate the confidence interval as follows: \(CI = 114 ± 1.311 * (10/\sqrt{29}) = 114 ± 2.43\). So, the 80% confidence interval is \(\boxed{(111.57, 116.43)}\).

Step 6 ：If the population had not been normally distributed, we could not have computed the confidence intervals in parts (a)-(c) using this method. The method assumes that the population is normally distributed.

Step 7 ：As for the effect of decreasing the level of confidence on the size of the interval, the answer is A. As the level of confidence decreases, the size of the interval decreases. This is because a lower level of confidence means we are less certain about our estimate, so the range of possible values is narrower.