Question 1
$10 \mathrm{pts}$
In 2021, a random sample of twenty-five test-takers is chosen from the takers of the Civil Service Exam. Their average score was 79 and their standard deviation was 3.8 points. . Assume the test scores are normally distributed. Create a $98 \%$ confidence interval for the mean. You must know which error formula to use to create the confidence interval. The Standard Normal Distribution and T-distribution are linked below, but you will only use one of them. You must know which to use.
Error Formulae
\[
E=z_{c} \cdot \frac{\sigma}{\sqrt{n}} \text { or } E=t_{c} \cdot \frac{s}{\sqrt{n}}
\]
Table Lists Various Levels of C along with their Z-score Values
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline$c$ & .80 & .85 & .90 & .95 & .97 & .98 & 4.99 \\
\hline$z_{c}$ & 1.28 & 1.44 & 1.645 & 1.96 & 2.17 & 2.33 & 2.575 \\
\hline
\end{tabular}
Answer
The 98% confidence interval for the mean is \(\boxed{(77.23, 80.77)}\).
Step 1 ：Given values are: mean score (\(\mu\)) = 79, standard deviation (\(\sigma\)) = 3.8, sample size (\(n\)) = 25, and Z-score for 98% confidence interval (\(Z_c\)) = 2.33.
Step 2 ：Calculate the margin of error (\(E\)) using the formula: \(E = Z_c \times \frac{\sigma}{\sqrt{n}}\).
Step 3 ：Substitute the given values into the formula: \(E = 2.33 \times \frac{3.8}{\sqrt{25}}\) which simplifies to \(E = 1.77\).
Step 4 ：Calculate the confidence interval using the formula: \(\mu \pm E\).
Step 5 ：Substitute the given values into the formula: lower bound = \(79 - 1.77 = 77.23\) and upper bound = \(79 + 1.77 = 80.77\).
Step 6 ：The 98% confidence interval for the mean is \(\boxed{(77.23, 80.77)}\).
