Find the 95% confidence interval for the variance of the lengths of pipes if a sample of 16 pipes has a standard deviation of 12 inches.

Question

Quizplus · Accepted Answer

We know that the distribution of sample variances follows a chi-square distribution with (n-1) degrees of freedom, where n is the sample size. Therefore, we can use a chi-square distribution table to find the critical values that correspond to a 95% confidence interval with 15 degrees of freedom (16-1=15).

The critical values are 6.262 and 27.488.

Using the formula for the confidence interval for the variance:

[(n-1)*s^2]/X^2 < 𝜎^2 < [(n-1)*s^2]/Y^2

where n is the sample size, s is the sample standard deviation, X and Y are the critical values from the chi-square distribution table.

Substituting the values we have:

[(16-1)*(12)^2]/27.488 < 𝜎^2 < [(16-1)*(12)^2]/6.262

77.043 < 𝜎^2 < 356.809

So the 95% confidence interval for the variance of the length of pipes is (77.043, 356.809) with a sample of 16 pipes and a standard deviation of 12 inches.

Therefore, the correct choice is B.

Find the 95% Confidence Interval for the Variance of the Lengths