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 110, and the sample standard deviation, s, is found to be 10. (a) Construct a 90% confidence interval about μ if the sample size, n, is 27. (b) Construct a 90% confidence interval about μ if the sample size, n, is 13. (c) Construct a 70% confidence interval about μ if the sample size, n, is 27. (d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed? A. No, the population needs to be normally distributed. B. No, the population does not need to be normally distributed. C. Yes, the population needs to be normally distributed. D. Yes, the population does not need to be normally distributed.

A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x̄, is found to be 110, and the sample standard deviation, s, is found to be 10.
(a) Construct a 90% confidence interval about μ if the sample size, n, is 27.
(b) Construct a 90% confidence interval about μ if the sample size, n, is 13.
(c) Construct a 70% confidence interval about μ if the sample size, n, is 27.
(d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed?
A. No, the population needs to be normally distributed.
B. No, the population does not need to be normally distributed.
C. Yes, the population needs to be normally distributed.
D. Yes, the population does not need to be normally distributed.

Transcript text: A simple random sample of size $n$ is drawn from a population that is normally distributed. The sample mean, $\bar{x}$, is found to be 110, and the sample standard deviation, $s$, is found to be 10. (a) Construct a $90 \%$ confidence interval about $\mu$ if the sample size, n , is 27. (b) Construct a $90 \%$ confidence interval about $\mu$ if the sample size, n , is 13. (c) Construct a $70 \%$ confidence interval about $\mu$ if the sample size, n , is 27. (d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed? A. No, the population needs to be normally distributed. B. No, the population does not need to be normally distributed. C. Yes, the population needs to be normally distributed. D. Yes, the population does not need to be normally distributed.

Solution

Solution Steps

Step 1: Constructing the 90% Confidence Interval for $ n = 27 $

Given:

Sample mean $ \bar{x} = 110 $
Sample standard deviation $ s = 10 $
Sample size $ n = 27 $

To calculate the 90% 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 = 0.10 $ and $ df = n - 1 = 26 $. The critical value $ t $ is approximately $ 1.71 $.

Calculating the margin of error: \[ \text{Margin of Error} = t \frac{s}{\sqrt{n}} = 1.71 \cdot \frac{10}{\sqrt{27}} \approx 3.28 \]

Thus, the confidence interval is: \[ (110 - 3.28, 110 + 3.28) = (106.72, 113.28) \]

Step 2: Constructing the 90% Confidence Interval for $ n = 13 $

Given:

Sample size $ n = 13 $

Using the same formula: \[ \bar{x} \pm t \frac{s}{\sqrt{n}} \] For $ n = 13 $, $ df = 12 $ and the critical value $ t $ is approximately $ 1.78 $.

Calculating the margin of error: \[ \text{Margin of Error} = t \frac{s}{\sqrt{n}} = 1.78 \cdot \frac{10}{\sqrt{13}} \approx 4.94 \]

Thus, the confidence interval is: \[ (110 - 4.94, 110 + 4.94) = (105.06, 114.94) \]

Step 3: Constructing the 70% Confidence Interval for $ n = 27 $

Given:

Sample size $ n = 27 $

Using the same formula: \[ \bar{x} \pm t \frac{s}{\sqrt{n}} \] For a 70% confidence level, $ \alpha = 0.30 $ and the critical value $ t $ is approximately $ 1.06 $.

Calculating the margin of error: \[ \text{Margin of Error} = t \frac{s}{\sqrt{n}} = 1.06 \cdot \frac{10}{\sqrt{27}} \approx 2.04 \]

Thus, the confidence interval is: \[ (110 - 2.04, 110 + 2.04) = (107.96, 112.04) \]