Please advise the answer for this question and if there is a

Please advise the answer for this question and if there is a way to complete this using statcrunch or ti83 calc. Thank you!

A nutritionist found that a random sample of sandwiches from two restaurants, labeled A and B, had the sample statistics shown in the table to the bottom right. Construct a 90% confidence interval for sigma 21 / sigma 22 using the definition to the right, where sigma 21 and sigma 22 are the variances of the cholesterol content of sandwiches from A and B, respectively. The 90% confidence interval is ( , ). (Simplify your answer. Round to three decimal places as needed.) When s21 and s22 are the variances of randomly selected, independent samples from normally distributed populations, then a confidence interval for sigma 1 / sigma 2 is as shown below, where FL is the left-tailed critical F-value and FR is the right-tailed critical F-value. S21/S22 1/FR

Solution

DF1=13-1=12

DF2=16-1=15

90% table values

FR= 2.475

FL=0.382

Lower limit = 10.09/(9.57*2.475) = 0.425994533

Upper limit = 10.09/(9.57*0.382) = 2.76004311

90% CI = ( 0.426, 2.760 )