Solve using R Script to compute a 90% confidence interval for the difference between the average running times of films produced by the two companies

Question

Please solve using R Script and show all coding/steps. Thanks in advance!

The following data represent the running times of films produced by two motion-picture companies.

There are 2 companies, A and B.

Company A's Time (minutes) is 103, 94, 110, 87, 98

Company B's Time (minutes) is 97, 82, 123, 92, 175, 88, 118

Construct a 90% confidence interval for σ² _A / σ²_B .

Assume that the running-time differences are approximately normally distributed with unequal variances.

Answer 1

F.05(4,6) = .162255

F.95(4,6) = 4.533677

Thus, as s1^2/s2^2 =0.063655, the 90% confidence interval is 0.063655/4.533677 <= sigma1^2/sigma2^2 <= 0.063655/4.533677, or 0.016246 <= sigma1^2/sigma2^2 <= .453948

Looking at the data, I have a hard time believing that 175 would not be considered an outlier.

In R, if we run

A <- c(103, 94, 110, 87, 98)

B <- c(97, 82, 123, 92, 175, 88, 118)

z <- t.test(A,B,conf.level=.9)

z$parameter

we get a confidence interval for A-B = ( -36.42670 , 11.79813)

The Welch test, a test used with unequal variances, is used for the analysis.