Assume the sample is taken from a normally distributed population and construct the indicated confidence interval.
-The mean replacement time for a random sample of 12 microwave ovens is 8.6 years with a standard deviation of 5.3 years. Construct the 98% confidence interval for the population variance, $\sigma ^ { 2 }.$
A)(12.5, 101.2)
B)(3.5, 10.1)
C)(2.4, 19.1)
D)(11.8, 86.5)

Question

Quizplus · Accepted Answer

To construct a 98% confidence interval for the population variance $\sigma^2$ when the sample is normally distributed, we use the chi-square distribution. The formula involves the sample variance $s^2$, the sample size $n$, and the chi-square values for the chosen confidence level. The sample variance here is $5.3^2 = 28.09$. The degrees of freedom ($df$) is $n - 1 = 11$. For a 98% confidence interval and 11 degrees of freedom, we look up the chi-square values for the lower and upper tails (1% each since it's two-tailed and 98% confidence). The confidence interval for the variance is given by:$$\left( \frac{(n-1)s^2}{\chi^2_{\alpha/2,df}}, \frac{(n-1)s^2}{\chi^2_{1-\alpha/2,df}} \right)$$Plugging in the values and using the appropriate chi-square values for the 1st and 99th percentiles for 11 degrees of freedom, we get the interval (12.5, 101.2), which matches choice A. This calculation requires looking up the chi-square values in a table or using a calculator/computer, which isn't shown here but leads to the interval in choice A.

Assume the Sample Is Taken from a Normally Distributed Population σ2.\sigma ^ { 2 }.σ2.

Assume the Sample Is Taken from a Normally Distributed Population $\sigma ^ { 2 }.$