Questions: Estimating a Population Question 3, 9.2.17 A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x̄, is found to be 106, and the sample standard deviation, s, is found to be 10. (a) Construct an 80% confidence interval about μ it the sample size, n, is 18. (b) Construct an 80% confidence interval about μ if the sample size, n, is 26. (c) Construct a 90% confidence interval about μ if the sample size, n, is 18. (d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed? (b) Construct an 80% confidence interval about μ if the sample size, n, is 26. Lower bound: 103.4; Upper bound: 108.6 How does increasing the sample size affect the margin of error, E? A. As the sample size increases, the margin of error increases. B. As the sample size increases, the margin of error decreases. C. As the sample size increases, the margin of error stays the same. (c) Construct a 90% confidence interval about μ if the sample size, n, is 18. Lower bound: 101.9; Upper bound: 110.1 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 decreases. C. As the level of confidence increases, the size of the interval stays the same.

Estimating a Population Question 3, 9.2.17

A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x̄, is found to be 106, and the sample standard deviation, s, is found to be 10. 
(a) Construct an 80% confidence interval about μ it the sample size, n, is 18. 
(b) Construct an 80% confidence interval about μ if the sample size, n, is 26. 
(c) Construct a 90% confidence interval about μ if the sample size, n, is 18. 
(d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed? 
(b) Construct an 80% confidence interval about μ if the sample size, n, is 26.

Lower bound: 103.4; Upper bound: 108.6 How does increasing the sample size affect the margin of error, E? 
A. As the sample size increases, the margin of error increases. 
B. As the sample size increases, the margin of error decreases. 
C. As the sample size increases, the margin of error stays the same. 
(c) Construct a 90% confidence interval about μ if the sample size, n, is 18.

Lower bound: 101.9; Upper bound: 110.1 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 decreases. 
C. As the level of confidence increases, the size of the interval stays the same.
Transcript text: Estimating a Population Question 3, 9.2.17 A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, $\overline{\mathrm{x}}$, is found to be 106 , and the sample standard deviation, s , is found to be 10. (a) Construct an $80 \%$ confidence interval about $\mu$ it the sample size, n , is 18. (b) Construct an $80 \%$ confidence interval about $\mu$ if the sample size, n , is 26. (c) Construct a $90 \%$ confidence interval about $\mu$ if the sample size, n , is 18. (d) Could wo have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed? (b) Construct an $80 \%$ confidence interval about $\mu$ if the sample size, n , is 26. Lower bound: 103.4 ; Upper bound: 108.6 How does increasing the sample size affect the margin of error, E? A. As the sample size increases, the margin of error increases. B. As the sample size increases, the margin of error decreases. C. As the sample size increases, the margin of error stays the same. (c) Construct a $90 \%$ confidence interval about $\mu$ if the sample size, n , is 18. Lower bound: 101.9 ; Upper bound: 110.1 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 decreases. C. As the level of confidence increases, the size of the interval stays the same.
failed

Solution

failed
failed

Solution Steps

Step 1: Constructing the 80% Confidence Interval for \( n = 18 \)

To construct the 80% confidence interval for the population mean \( \mu \) when the sample size \( n = 18 \), we use the formula:

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

Where:

  • \( \bar{x} = 106 \) (sample mean)
  • \( s = 10 \) (sample standard deviation)
  • \( n = 18 \) (sample size)
  • \( t \) is the critical value from the t-distribution for \( \alpha = 0.20 \) and \( df = n - 1 = 17 \), which is approximately \( 1.3 \).

Calculating the margin of error:

\[ E = t \frac{s}{\sqrt{n}} = 1.3 \cdot \frac{10}{\sqrt{18}} \approx 3.1 \]

Thus, the confidence interval is:

\[ (106 - 3.1, 106 + 3.1) = (102.9, 109.1) \]

Step 2: Constructing the 80% Confidence Interval for \( n = 26 \)

For the sample size \( n = 26 \), we again use the confidence interval formula:

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

Where:

  • \( n = 26 \)
  • The critical value \( t \) for \( \alpha = 0.20 \) and \( df = 25 \) is approximately \( 1.3 \).

Calculating the margin of error:

\[ E = t \frac{s}{\sqrt{n}} = 1.3 \cdot \frac{10}{\sqrt{26}} \approx 2.6 \]

Thus, the confidence interval is:

\[ (106 - 2.6, 106 + 2.6) = (103.4, 108.6) \]

Step 3: Constructing the 90% Confidence Interval for \( n = 18 \)

For the 90% confidence interval with \( n = 18 \), we use the same formula:

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

Where:

  • The critical value \( t \) for \( \alpha = 0.10 \) and \( df = 17 \) is approximately \( 1.7 \).

Calculating the margin of error:

\[ E = t \frac{s}{\sqrt{n}} = 1.7 \cdot \frac{10}{\sqrt{18}} \approx 4.1 \]

Thus, the confidence interval is:

\[ (106 - 4.1, 106 + 4.1) = (101.9, 110.1) \]

Final Answer

  • 80% confidence interval for \( n = 18 \): \( \boxed{(102.9, 109.1)} \)
  • 80% confidence interval for \( n = 26 \): \( \boxed{(103.4, 108.6)} \)
  • 90% confidence interval for \( n = 18 \): \( \boxed{(101.9, 110.1)} \)
Was this solution helpful?
failed
Unhelpful
failed
Helpful