Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation ?.
Assume that the population has a normal distribution. Round the confidence interval limits to the same number of decimal places as the sample standard deviation.
-The mean replacement time for a random sample of 20 washing machines is 9.4 years and the standard deviation is 2.5 years. Construct a 99% confidence interval for the standard deviation, ?, of the replacement times of all washing machines of this type.
A) $1.8 \mathrm { yr }

Question

Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation ?.
Assume that the population has a normal distribution. Round the confidence interval limits to the same number of decimal places as the sample standard deviation.
-The mean replacement time for a random sample of 20 washing machines is 9.4 years and the standard deviation is 2.5 years. Construct a 99% confidence interval for the standard deviation, ?, of the replacement times of all washing machines of this type. 
A) $1.8 \mathrm { yr } < \sigma < 4.2 \mathrm { yr }$
B) $1.8 \mathrm { yr } < \sigma < 3.9 \mathrm { yr }$
C) $1.7 \mathrm { yr } < \sigma < 4.8 \mathrm { yr }$
D) $1.8 \mathrm { yr } < \sigma < 5.3 \mathrm { yr }$

Quizplus · Accepted Answer

The confidence interval for the population standard deviation is given by:$$\frac{(n-1)s^2}{\chi^2_{\alpha/2, n-1}} < \sigma^2 < \frac{(n-1)s^2}{\chi^2_{1-\alpha/2, n-1}}$$where s is the sample standard deviation, n is the sample size, and \alpha is the level of significance.For a 99% confidence level, \alpha = 0.01 and \chi^2_{0.005, 19} = 30.144 and \chi^2_{0.995, 19} = 8.907.Therefore, the confidence interval is:$$\frac{(20-1)2.5^2}{30.144} < \sigma^2 < \frac{(20-1)2.5^2}{8.907}$$$$1.8^2 < \sigma^2 < 4.2^2$$$$1.8 < \sigma < 4.2$$

Use the Given Degree of Confidence and Sample Data to Find