statistics

Part 1)
A physical therapists wishes to determine whether an exercise program increases flexibility. He measures the flexibility (in inches) of 12 randomly selected subjects both before and after an intensive eight-week training program and obtains the following data:
Before: 18.5, 21.5, 16.5, 21, 20, 15, 19.75, 15.75, 18, 22, 15, 20.5
After: 19.25, 21.75, 16.5, 20.5, 22.25, 16, 19.5, 17, 19.25, 19.5, 16.5, 20
Test the claim that the flexibility before the exercise program is less than the flexibility after the exercise program at a significance level of 0.01. Note: the data is approximately normal with no outliers.

Part 2)
Using the data from the previous problem, construct a 95% confidence interval about the population mean difference. Interpret your result.

Any help would be greatly appreciated. Thank you for your time.

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  1. Find the mean first = sum of scores/number of scores

    Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.

    Standard deviation = square root of variance

    Use this data with the information in my response to your previous post.

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