Question

At a local store, 65 female employees were randomly selected and it was found that their mean monthly income was $605 with a standard deviation of $121.50. Seventy-five male employees were also randomly selected and their mean monthly income was found to be $677 with a standard deviation of $168.70. Test the hypothesis that male employees have a higher monthly income than female employees. Use α = 0.01.

Type the conclusion only for the hypothesis test (step 5).

Answer 1

hypothesis

pooled Variance
s2p = ((df1)(s21) + (df2)(s22)) / (df1 + df2) = 94496561.06 / 138 = 684757.69

Standard Error
s(M1 - M2) = √((s2p/n1) + (s2p/n2)) = √((684757.69/65) + (684757.69/75)) = 140.23

test statistic

= (M1 - M2) /s(M1 - M2) = 72 / (140.23)

0.513 4

p value for right tailed test

p(t>0.5134) and degrees of freedom is minimum of n1-1 and n2-1

which is 64

p value is 0.3 046, which is greater than given significance so we can not reject null hypothesis

there is not significant evidence to claim that male employees have a higher monthly income than female.

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