Questions: A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x̄, is found to be 108, and the sample standard deviation, s, is found to be 10 (a) Construct an 80% confidence interval about μ if the sample size, n, is 15. (b) Construct an 80% confidence interval about μ if the sample size, n, is 27. (c) Construct a 99% confidence interval about μ if the sample size, n, is 15. (d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed? (c) Construct a 99% confidence interval about μ if the sample size, n, is 15. Lower bound: 100.3; Upper bound: 115.7 (Use ascending order. Round to one decimal place as needed.) 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. As the level of confidence increases, the size of the interval increases. B. As the level of confidence increases, the size of the interval stays the same. C. As the level of confidence increases, the size of the interval decreases. (d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed? A. Yes, the population does not need to be normally distributed. B. No, the population does not need to be normally distributed. C. No, the population needs to be normally distributed.

Solution Steps

Given:

• Sample mean \( \bar{x} = 108 \)
• Sample standard deviation \( s = 10 \)
• Sample size \( n = 15 \)

To calculate the 80% confidence interval, we use the formula: \[ \bar{x} \pm t \frac{s}{\sqrt{n}} \] where \( t \) is the critical value from the \( t \)-distribution for \( \alpha = 0.20 \) and \( df = n - 1 = 14 \). The critical value \( t \) is approximately \( 1.3 \).

Calculating the margin of error: \[ E = t \frac{s}{\sqrt{n}} = 1.3 \cdot \frac{10}{\sqrt{15}} \approx 3.5 \]

Thus, the confidence interval is: \[ (108 - 3.5, 108 + 3.5) = (104.5, 111.5) \]

Given:

• Sample size \( n = 27 \)

Using the same formula: \[ \bar{x} \pm t \frac{s}{\sqrt{n}} \] For \( n = 27 \), \( df = 26 \) and the critical value \( t \) is approximately \( 1.3 \).

Calculating the margin of error: \[ E = t \frac{s}{\sqrt{n}} = 1.3 \cdot \frac{10}{\sqrt{27}} \approx 2.5 \]

Thus, the confidence interval is: \[ (108 - 2.5, 108 + 2.5) = (105.5, 110.5) \]

Given:

• Sample size \( n = 15 \)

Using the formula: \[ \bar{x} \pm t \frac{s}{\sqrt{n}} \] For a 99% confidence level, \( \alpha = 0.01 \) and \( df = 14 \). The critical value \( t \) is approximately \( 3.0 \).

Calculating the margin of error: \[ E = t \frac{s}{\sqrt{n}} = 3.0 \cdot \frac{10}{\sqrt{15}} \approx 7.0 \]

Thus, the confidence interval is: \[ (108 - 7.0, 108 + 7.0) = (100.3, 115.7) \]

As the level of confidence increases, the size of the interval increases. This is evident from the comparison of the intervals:

• 80% CI for \( n = 15 \): \( (104.5, 111.5) \)
• 99% CI for \( n = 15 \): \( (100.3, 115.7) \)

For parts (a)-(c), the population is assumed to be normally distributed. If the population were not normally distributed, the confidence intervals could still be computed if the sample size is sufficiently large (typically \( n \geq 30 \)) due to the Central Limit Theorem. However, for small sample sizes, normality is a critical assumption.

Final Answer