Econ 41 HW SET 5

# Econ 41 HW SET 5 - Problem Set 5 Econ 41(Winter 2008 Duke...

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Problem Set 5, Econ 41 (Winter 2008) Duke Whang 1. (12.14, page 544) The following ANOVA table is given a. Complete the table. Source of Variation Degrees of Freedom Sum of Squares Mean Square Value of the Test Statistic Between 2 19.2813 Within 89.3677 F = Total 12 b. Using α = 0 . 01, what is your conclusion for the test with the null hypothesis that the means of the three populations are all equal against the alternative hypothesis that the means of the three populations are not all equal? Solution: a. Here is the completed ANOVA table: Source of Variation Degrees of Freedom Sum of Squares Mean Square Value of the Test Statistic Between k - 1 = 2 SSB = 38 . 5626 MSB = SSB/ ( k - 1) = 19 . 2813 Within n - k = 10 SSW = 89 . 3677 MSW = SSW / ( n - k ) = 8 . 93677 F = MSB / MSW = 2 . 16 Total n - 1 = 12 SST = SSB + SSW = 127 . 9303 1. From df b = 2 and df t = 12, we can calculate df w : ( k - 1) + ( n - k ) = ( n - 1) df b + df w = df t 2 + df w = 12 df w = 10 or, of course, we can solve for k from k - 1 = df b = 2 (namely, k = 3) and solve for n from n - 1 = df t = 12 (namely, n = 13), and then write df w = n - k = 13 - 3 = 10. 2. We can solve for SSB = MSB × ( k - 1) = 19 . 2813 × 2 = 38 . 5626 3. From SSB and SSW , we can solve for SST = SSB + SSW = 38 . 5626+89 . 3677 = 127 . 9303 1

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4. From SSW and n - k , we can solve for MSW = SSW/ ( n - k ) = 89 . 3677 / 10 = 8 . 93677 5. From MSB and MSW , we can solve for F = MSB/MSW = 19 . 2813 / 8 . 93677 = 2 . 16 b. We will follow the steps explicitly. 1. These are the null and alternate hypotheses: H 0 : μ 1 = μ 2 = μ 3 H 1 : μ i 6 = μ j for some 1 i < j 3 The null hypothesis H 0 states that all the means are equal. The alternate hypothesis H 1 states that for some pair of variables, the means are not equal; namely it states that not all the means are equal. 2. Since we are testing for equality of means among 3 groups, use the F distribution. 3. Let df n and df d denote the degrees of freedom in the numerator and denominator re- spectively. We have df n = 2 and df d = 10. The critical value for the F 2 , 10 distribution at α = 0 . 01 is given on page C24 of the textbook; it is F c = F 2 , 10 , 0 . 01 = 7 . 56. 4. From the ANOVA table, we know that our test-statistic is F = 2 . 16. 5. Since F 2 . 16 < 7 . 56 F c and the rejection region is to the right of F c (namely, we reject exactly when F > F c ), we fail to reject the null hypothesis (that all three means are equal) at the α = 0 . 01 significance level. 2. (12.17, page 545) The following table give the numbers of classes missed during one semester by 25 randomly selected college students drawn from three different age groups. Below 25 25 to 30 31 and Above 19 9 5 13 6 8 25 11 2 10 14 3 19 5 10 4 9 9 15 3 10 11 16 18 9 a. We are to test if the mean number of classes missed during the semester by all students in each of these three age groups is the same. Write the null and alternative hypotheses. b. What are the degrees of freedom for the numerator and the denominator? c. Calculate SSB, SSW, and SST. d. Show the rejection and nonrejection regions on the F distribution curve for α = 0 . 01.
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• Spring '07
• Guggenberger
• Null hypothesis, Statistical hypothesis testing, significance level

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