Statistics
posted by Gummy .
2. In 2009 a random sample of 70 unemployed people in Alabama showed an average weekly benefit of $199.65. In Mississippi, for a random sample of 65 the number was $187.93. Assume population standard deviations of $32.48 and $26.15 respectively.
a. Using the 5% level of significance, test whether
the two means are different.
b. Assume the pvalue for this test is .0439.
c. If the level of significant used was 10% (or 1%),
would the Ho be rejected?

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.
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