An instructor hy pothesizes that the mean of the final ex am grades in her statistics class is larger for the male students than that it is for the

Question

female students. The data from the final ex am for the last semester are presented here. Test her claim using  = 0.05.

Male Students
36 17 24 39 28
25 38 27 15 21
34 18 22 37 32

Female Students
27 24 33 35 33
25 34 30 29 26
28 23 33 37 25
35 24 23

Answer 1

Find the means 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

Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√n

If only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score. Is it less than .05?

I'll let you do the calculations.