Question

3. (9 pts.) A random sample of 10 purchases of a fixed grocery list was taken at Albert's and the standard deviation of the grocery bills was $1.84. A random sample of 10 purchases of the same fixed grocery list was taken at Miller's and the standard deviation of the grocery bills was $1.40. The grocery bills for this fixed list are normally distributed for both stores. Find a 90% confidence interval for the ratio of the variances of the grocery bills at the two stores. Also interpret your results in terms of the original problem.

Answer 1

For n₁ - 1, n₂ - 1 = 9, 9 degrees of freedom, we have from F distribution tables here:

Therefore, the confidence interval for the ratio of variances here is computed as:

$= \frac{s_1^2}{s_2^2F_{0.95}} < \frac{\sigma_1^2}{\sigma_2^2} < \frac{s_1^2}{s_2^2F_{0.05}}$

$= \frac{1.84^2}{1.4^2*3.1789}< \frac{\sigma_1^2}{\sigma_2^2} < \frac{1.84^2}{1.4^2*0.3146}$

$= 0.5434 < \frac{\sigma_1^2}{\sigma_2^2} < 5.4906$

This is the required confidence interval here.

The interpretation for the confidence interval here is that we are 90% confident that the ratio of true variances lies in the above obtained confidence interval here.