Question
41 random samples of monthly electric bill amounts are selected from a normally distributed population. The samples have a mean of $108 and a standard deviation of $5. Construct a 98% confidence interval for the population standard deviation.
A) $1.89 < σ < $2.75
B) $3.14 < σ < $9.02
C) $3.96 < σ < $6.72
D) $4.23 < σ < $6.14
A) $1.89 < σ < $2.75
B) $3.14 < σ < $9.02
C) $3.96 < σ < $6.72
D) $4.23 < σ < $6.14
Answers
GPT-4o mini
To construct a confidence interval for the population standard deviation using the Chi-Square distribution, we can use the formula:
\[
\left( \sqrt{\frac{(n-1)s^2}{\chi^2_{\alpha/2}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}} \right)
\]
Where:
- \( n \) is the sample size.
- \( s \) is the sample standard deviation.
- \( \chi^2_{\alpha/2} \) and \( \chi^2_{1-\alpha/2} \) are the critical values from the Chi-Square distribution for \( n-1 \) degrees of freedom.
In this case:
- \( n = 41 \)
- \( s = 5 \)
- The confidence level is 98%, so \( \alpha = 0.02 \) and \( \alpha/2 = 0.01 \).
The degrees of freedom \( n-1 \) is \( 41 - 1 = 40 \).
Next, we look up the critical values in the Chi-Square distribution table:
1. For \( \chi^2_{0.01} \) with \( 40 \) degrees of freedom.
2. For \( \chi^2_{0.99} \) with \( 40 \) degrees of freedom.
Using a Chi-Square table or calculator:
- \( \chi^2_{0.01}(40) \approx 63.691 \)
- \( \chi^2_{0.99}(40) \approx 19.493 \)
Now we apply these values to the confidence interval formula:
1. Calculate the lower limit:
\[
\sqrt{\frac{(41-1)(5^2)}{63.691}} = \sqrt{\frac{40 \cdot 25}{63.691}} = \sqrt{\frac{1000}{63.691}} \approx \sqrt{15.69} \approx 3.96
\]
2. Calculate the upper limit:
\[
\sqrt{\frac{(41-1)(5^2)}{19.493}} = \sqrt{\frac{40 \cdot 25}{19.493}} = \sqrt{\frac{1000}{19.493}} \approx \sqrt{51.25} \approx 7.16
\]
Thus, the 98% confidence interval for the population standard deviation is approximately:
\[
3.96 < \sigma < 7.16
\]
The interval that best fits our results is **C) \( 3.96 < \sigma < 6.72 \)**.
\[
\left( \sqrt{\frac{(n-1)s^2}{\chi^2_{\alpha/2}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}} \right)
\]
Where:
- \( n \) is the sample size.
- \( s \) is the sample standard deviation.
- \( \chi^2_{\alpha/2} \) and \( \chi^2_{1-\alpha/2} \) are the critical values from the Chi-Square distribution for \( n-1 \) degrees of freedom.
In this case:
- \( n = 41 \)
- \( s = 5 \)
- The confidence level is 98%, so \( \alpha = 0.02 \) and \( \alpha/2 = 0.01 \).
The degrees of freedom \( n-1 \) is \( 41 - 1 = 40 \).
Next, we look up the critical values in the Chi-Square distribution table:
1. For \( \chi^2_{0.01} \) with \( 40 \) degrees of freedom.
2. For \( \chi^2_{0.99} \) with \( 40 \) degrees of freedom.
Using a Chi-Square table or calculator:
- \( \chi^2_{0.01}(40) \approx 63.691 \)
- \( \chi^2_{0.99}(40) \approx 19.493 \)
Now we apply these values to the confidence interval formula:
1. Calculate the lower limit:
\[
\sqrt{\frac{(41-1)(5^2)}{63.691}} = \sqrt{\frac{40 \cdot 25}{63.691}} = \sqrt{\frac{1000}{63.691}} \approx \sqrt{15.69} \approx 3.96
\]
2. Calculate the upper limit:
\[
\sqrt{\frac{(41-1)(5^2)}{19.493}} = \sqrt{\frac{40 \cdot 25}{19.493}} = \sqrt{\frac{1000}{19.493}} \approx \sqrt{51.25} \approx 7.16
\]
Thus, the 98% confidence interval for the population standard deviation is approximately:
\[
3.96 < \sigma < 7.16
\]
The interval that best fits our results is **C) \( 3.96 < \sigma < 6.72 \)**.