Questions: The mean and standard deviation of a random sample of n measurements are equal to 33.2 and 3.3, respectively. a. Find a 99% confidence interval for μ if n=81. b. Find a 99% confidence interval for μ if n=324. c. Find the widths of the confidence intervals found in parts a and b. What is the effect on the width of a confidence interval of quadrupling the sample size while holding the confidence coefficient fixed?

Solution Steps

To find the \( 99\% \) confidence interval for the mean \( \mu \) when \( n = 81 \), we use the formula:

\[ \bar{x} \pm z \frac{s}{\sqrt{n}} \]

Where:

• \( \bar{x} = 33.2 \)
• \( z = 2.576 \) (Z-score for \( 99\% \) confidence level)
• \( s = 3.3 \)
• \( n = 81 \)

Calculating the margin of error:

\[ \text{Margin of Error} = z \frac{s}{\sqrt{n}} = 2.576 \cdot \frac{3.3}{\sqrt{81}} = 2.576 \cdot \frac{3.3}{9} = 2.576 \cdot 0.3667 \approx 0.944 \]

Thus, the confidence interval is:

\[ (33.2 - 0.944, 33.2 + 0.944) = (32.256, 34.144) \]

For \( n = 324 \), we apply the same formula:

\[ \bar{x} \pm z \frac{s}{\sqrt{n}} \]

Where:

• \( n = 324 \)

Calculating the margin of error:

\[ \text{Margin of Error} = z \frac{s}{\sqrt{n}} = 2.576 \cdot \frac{3.3}{\sqrt{324}} = 2.576 \cdot \frac{3.3}{18} = 2.576 \cdot 0.1833 \approx 0.472 \]

Thus, the confidence interval is:

\[ (33.2 - 0.472, 33.2 + 0.472) = (32.728, 33.672) \]

The width of the confidence interval for \( n = 81 \) is calculated as:

\[ \text{Width}_{n=81} = 34.144 - 32.256 = 1.888 \]

The width of the confidence interval for \( n = 324 \) is:

\[ \text{Width}_{n=324} = 33.672 - 32.728 = 0.944 \]

Quadrupling the sample size from \( n = 81 \) to \( n = 324 \) results in a decrease in the width of the confidence interval. Specifically, the widths are:

• Width for \( n = 81 \): \( 1.888 \)
• Width for \( n = 324 \): \( 0.944 \)

This demonstrates that increasing the sample size reduces the width of the confidence interval while holding the confidence coefficient fixed.

Final Answer