A simple random sample of size $n$ is drawn. The sample mean, $\bar{x}$, is found to be 17.6 , and the sample standard deviation, $\mathrm{s}$, is found to be 4.7 .
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C. The margin of error decreases.
(c) Construct a $99 \%$ confidence interval about $\mu$ if the sample size $n$. is 35 .

Lower bound: 15.43 ; Upper bound: 19.77

Compare the results to those obtained in part (a). How does increasing the level of confidence affect the size of the margin of error, E?
A. The margin of error decreases.
B. The margin of error increases.
C. The margin of error does not change.

Step 1 ：A simple random sample of size $n$ is drawn. The sample mean, $\bar{x}$, is found to be 17.6 , and the sample standard deviation, $\mathrm{s}$, is found to be 4.7 .

Step 2 ：Construct a $99 \%$ confidence interval about $\mu$ if the sample size $n$. is 35 . Lower bound: 15.43 ; Upper bound: 19.77

Step 3 ：The question is asking to compare the margin of error of a 99% confidence interval to a previously calculated margin of error (presumably at a lower confidence level).

Step 4 ：The margin of error is the range within which we expect the true population mean to lie, with a certain level of confidence. It is calculated using the formula: Margin of Error = t * (s/√n) where t is the t-score (which depends on the desired level of confidence and degrees of freedom), s is the sample standard deviation, and n is the sample size.

Step 5 ：As the level of confidence increases, the t-score also increases. This means that the margin of error will also increase, because we are multiplying by a larger t-score.

Step 6 ：Therefore, increasing the level of confidence should increase the size of the margin of error.

Step 7 ：However, to confirm this, we need to calculate the margin of error for the 99% confidence interval and compare it to the margin of error from part (a). The question does not provide the necessary information to calculate the margin of error from part (a).

Step 8 ：Final Answer: \(\boxed{\text{B. The margin of error increases.}}\)