Question

Researchers believe that the population variance of female student GPAs at UD is greater than the population variance of male student GPAs at UD. Use a 5% level of significance and the following data from the combined Class Dataset to test this idea (this is a question from section11.2, but WHY?):

Female GPAS:	2.9	3.5	3.7	3.5	3.6	3.5	3.2	3.9	3.2
Male GPAs:	2.9	2.7	3.2	3.3	3.5	3.4	3.5	3.2

Enter your Summary Statistics here (rounded off to the 2^nd decimal place, if necessary):

	Female	Male
X:
s:
n:

      Q1: Given the researchers’ stated belief, which group should be considered                 Group #1, Males or Females? WHY?

      Q2: What if the researchers simply wanted to know if the two population                    variances were different? In that case how would you determine which                      group should be called Group #1?

      Q3: Write out your Null (H₀) and Alternative Hypotheses (H_A).

       

        H0:

        HA:

      Q4: What kind of a TS will you calculate for this pair of hypotheses?
             Z₀? T₀? F₀? χ02?

      Q5: Calculate the value of the TS you would use to run this hypothesis                       test, complete with degrees of freedom, if necessary.

      Q6: Write out the full RR that you would use to run this hypothesis test.

      Q7: Write out the full Decision and Conclusion that your TS and RR leads                  you to make.

      Q8: Calculate the PValue for your TS (and if you’re working from one of the                probability tables, you might have a statement like: “0.01< PV <0.05”).

Answer 1

For Females:

∑x = 31

∑x² = 107.5

n1 = 9

Mean , x̅1 = Ʃx/n = 31/9 = 3.4444

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(107.5-(31)²/9)/(9-1)] = 0.3005

For males:

∑x = 25.7

∑x² = 83.13

n2 = 8

Mean , x̅2 = Ʃx/n = 25.7/8 = 3.2125

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(83.13-(25.7)²/8)/(8-1)] = 0.2850

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Q1: The group with higher standard deviation is considered as group 1

Here group #1 will be female.

Q2:

Here also group #1 will be female.

Q3:

Null and alternative hypothesis:

Hₒ : σ₁² = σ₂²

H₁ : σ₁² > σ₂²

Q4:

Test statistic to be F test.

Q5:

Test statistic:

F = s₁² / s₂² = 0.3005² / 0.285² = 1.1111

df₁ = n₁-1 = 8

df₂ = n₂-1 = 7

Q6:

Upper tailed critical value, FU = F.INV.RT(0.05, 8, 7) = 3.7257

Rejection region: Reject Ho if F > 3.7257.

Q7:

Decision:

Fail to reject Ho.

Conclusion:

There is not enough evidence to conclude that the population variance of female student GPAs at UD is greater than the population variance of male student GPAs at UD

Q8:

P-value = F.DIST.RT(1.1111, 8, 7) = 0.4512

0.40 < p-value < 0.50