This appears to be a hypothesis test involving inferences concerning two variances (standard deviation is the square root of the variance).
Sample 1: n = 25; variance = sd^2; df = n - 1 = 24
Sample 2: n = 25; variance = sd^2; df = n - 1 = 24
Note: sd = standard deviation.
Test statistic = sample 1 variance / sample 2 variance
You can use the F-distribution at .05 level using the above information for degrees of freedom. This will be your critical value to compare to the test statistic. If the test statistic exceeds the critical value from the table, the null will be rejected in favor of the alternate hypothesis and you can conclude a difference in variances. If the test statistic does not exceed the critical value from the table, then the null is not rejected and you cannot conclude a difference.
I hope this will help.
A sample of 25 concession stand purchases at the October 22 matinee of Bride of Chucky showed a mean purchase of $5.29 with a standard deviation of $3.02. For the October 26 evening showing of the same movie, for a sample of 25 purchases the mean was $5.12 with a standard deviation of $2.14. The means appear to be very close, but not variances. At a = .05, is there a differences in variances?
2 answers
a bumper test, three types of autos were deliberately crashed into a barrier at 5 mph, and the resulting damage (in dollars) was estimated. Five test vehicles of each type were crashed, with the results shown below. Research question: Are the mean crash damages the same for these three vehicles?