Question

The sugar content of the syrup in canned peaches is normally distributed. A random sample of n=25 cans yields a sample standard deviation of s=6.9 milligrams. Construct a 99% one-sided lower confidence bound for the population variance.

Answer 1

Answer:

99% one-sided lower confidence bound = 26.77

Step-by-step explanation:

We have to calculate a 99% one-sided lower confidence bound for the population variance.

The sample size is n=25.

The degrees of freedom are then:

$df=n-1=25-1=24$

The critical value of the chi-square for this confidence bound is:

$\chi^2_{0.01, \,24}=42.98$

Then, the lower confidence bound can be calculated as:

$LB=\dfrac{(n-1)s^2}{\chi^2_{0.01,24}}=\dfrac{24\cdot(6.9)^2}{42.98}=\dfrac{1,142.64}{42.68}=26.77$