James Desreumaux, VP of Human Resources of Maritime Boat Manufacturing (MBM), is reviewing the employee training programs of MBM factories.His staff randomly selected personnel files for 10 workers in Nova Scotia, and determined that their mean training time was 25 hours and that the standard deviation was 5 hours.Assume that training times are normally distributed.The 90% confidence interval for the population variance of training times is ___.
A)11.83 to 83.33
B)2.37 to 16.67
C)2.66 to 13.51
D)13.30 to 67.67
E)15.00 to 68.00

Question

Quizplus · Accepted Answer

We are looking for the confidence interval for the population variance. We know that:

- Sample size (n) = 10
- Sample mean (x̄) = 25
- Sample standard deviation (s) = 5
- Confidence level = 90%

To find the confidence interval for population variance, we will use the chi-square distribution. The formula for the chi-square distribution is:

χ² = (n-1)s² / σ²

where χ² is the chi-square value, s is the sample standard deviation, σ² is the population variance, and n is the sample size.

We can rearrange the formula to get the confidence interval for σ²:

( n - 1 ) s² / χ²₂₋₀.₁₀  ≤  σ²  ≤  ( n - 1 ) s² / χ²₀.₁₀

where χ²₂₋₀.₁₀ and χ²₀.₁₀ are the chi-square values for the upper and lower tails of the distribution with 9 degrees of freedom (n-1). We can find these values using a chi-square table or calculator. For a 90% confidence level, the chi-square values are:

χ²₀.₁₀ = 3.247
χ²₂₋₀.₁₀ = 16.92

Substituting the values in the formula, we get:

(10-1)×5²/16.92 ≤ σ² ≤ (10-1)×5²/3.247

Simplifying:

13.30 ≤ σ² ≤ 67.67

Therefore, the 90% confidence interval for the population variance of training times is 13.30 to 67.67. The answer is D.