34. A sample of size $n=50$ is drawn from a population whose standard deviation is $\sigma=26$.
a. Find the margin of error for a $90 \%$ confidence interval for $\mu$.
b. If the sample size were $n=40$, would the margin of error be larger or smaller?
Answer
If the sample size were \(n=40\), the margin of error would be larger because the denominator of the fraction in the formula for \(E\) would be smaller, resulting in a larger value for \(E\).
Step 1 ：We are given a sample size of \(n=50\), a population standard deviation of \(\sigma=26\), and a Z-score of \(Z=1.645\) which corresponds to a 90% confidence interval.
Step 2 ：The margin of error for a confidence interval is calculated using the formula: \[E = Z \cdot \frac{\sigma}{\sqrt{n}}\] where \(E\) is the margin of error, \(Z\) is the Z-score, \(\sigma\) is the standard deviation of the population, and \(n\) is the size of the sample.
Step 3 ：Substituting the given values into the formula, we get: \[E = 1.645 \cdot \frac{26}{\sqrt{50}}\]
Step 4 ：Solving the above expression, we find that \(E\) is approximately 6.05.
Step 5 ：Thus, the margin of error for a 90% confidence interval for \(\mu\) is approximately \(\boxed{6.05}\).
Step 6 ：If the sample size were \(n=40\), the margin of error would be larger because the denominator of the fraction in the formula for \(E\) would be smaller, resulting in a larger value for \(E\).
