The mean replacement time for a random sample of 12 cd players is 8.6 years with a standard deviation of 3.5years. Construct the 98% confidence interval for the population variance, $$\sigma ^ { 2 }$$ . Assume the data are normallydistributed
A) (5.4, 44.1)
B) (2.3, 6.6)
C) (1.6, 12.6)
D) (5.1, 37.7)

Question

Quizplus · Accepted Answer

To construct a 98% confidence interval for the population variance 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 desired level of confidence. The sample variance here is $s^2 = 3.5^2 = 12.25$. The degrees of freedom ($df$) for the chi-square distribution is $n - 1 = 12 - 1 = 11$. The 98% confidence interval for the variance is calculated using the chi-square values that correspond to the upper and lower tails of the 98% confidence level. The formula for the confidence interval is:$$\left( \frac{(n-1)s^2}{\chi^2_{\alpha/2, df}}, \frac{(n-1)s^2}{\chi^2_{1-\alpha/2, df}} \right)$$Given the values and the chi-square distribution table (or calculator), the interval calculated matches choice (A) (5.4, 44.1), which indicates the range within which the true population variance is expected to lie with 98% confidence.

The Mean Replacement Time for a Random Sample of 12 σ2\sigma ^ { 2 }σ2

The Mean Replacement Time for a Random Sample of 12 $\sigma ^ { 2 }$