Questions: Question 4, 9.2.19-T A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x̄, is found to be 194, and the sample standard deviation, s, is found to be 58. (a) Construct a 96% confidence interval about μ if the sample size, n, is 37 (b) Construct a 96% confidence interval about μ if the sample size, n, is 55 How does increasing the sample size affect the margin of error, E? (c) Construct a 99% confidence interval about μ if the sample size, n, is 37 How does increasing the level of confidence affect the size of the margin of error, E? (d) if the sample size is 12, what conditions must be satisfied to compute the confidence interval?

Solution Steps

To construct the 96% confidence interval for the population mean \( \mu \) when the sample size \( n = 37 \), we use the formula:

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

Where:

• \( \bar{x} = 194 \) (sample mean)
• \( s = 58 \) (sample standard deviation)
• \( n = 37 \) (sample size)
• \( z \) for 96% confidence level is approximately \( 2.05 \).

Calculating the margin of error:

\[ E = z \frac{s}{\sqrt{n}} = 2.05 \cdot \frac{58}{\sqrt{37}} \approx 19.58 \]

Thus, the confidence interval is:

\[ (194 - 19.58, 194 + 19.58) = (174.42, 213.58) \]

For the sample size \( n = 55 \), we apply the same formula:

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

Where:

• \( n = 55 \)

Calculating the margin of error:

\[ E = z \frac{s}{\sqrt{n}} = 2.05 \cdot \frac{58}{\sqrt{55}} \approx 16.06 \]

Thus, the confidence interval is:

\[ (194 - 16.06, 194 + 16.06) = (177.94, 210.06) \]

For the 99% confidence interval with \( n = 37 \), we again use the formula:

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

Where:

• \( z \) for 99% confidence level is approximately \( 2.58 \).

Calculating the margin of error:

\[ E = z \frac{s}{\sqrt{n}} = 2.58 \cdot \frac{58}{\sqrt{37}} \approx 24.56 \]

Thus, the confidence interval is:

\[ (194 - 24.56, 194 + 24.56) = (169.44, 218.56) \]

Final Answer