problem set 5 answers

# problem set 5 answers - Homework5 Answers Exercise 12.14(a...

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Homework5 Answers Exercise 12.14 (a) ANOVA table Source of var d. of f. Sum of squares Mean Squares Value of test stat Between k-1=2 SSB=38.5626 MSB =SSB/(k-1) =19.2813 Within n-k=10 SSW=89.3677 MSW =SSW/(n-k) =8.9368 F=MSB/MSW =2.16 Total k-1+n-k =n-1=12 SST=127.9303 (b) Step 1 Since we want to test if the three means of the populations are equal, the null hypothesis will be given by: 3 2 1 0 : μ μ μ = = H : 1 H At least one of the means is different Step 2 Since we are testing for equality of means among three groups, use the F distribution. Step 3 Degrees of freedom for the numerator=2 Degrees of freedom for the denominator=10 Thus, for α =0.01, the critical value is given by F c =7.56 (this value comes from finding F c in P(F> F c )=0.01 where F(2,10)). Step 4 From the ANOVA table, we have F=2.16. Step 5 Since 2.16<7.56, (observed value of the test statistic<critical value) we do not reject the null hypothesis. Conclude that the means of the three populations are equal. Exercise 12.17 (a) Since we want to test if the three means of the populations are equal, the null hypothesis will be given by: 3 2 1 0 : μ μ μ = = H : 1 H Not all three population means are equal. 1

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Given that we want to test if more than 2 population means are equal, we will use the One-Way ANOVA test to test this hypothesis. (b) The degrees of freedom of the numerator are given by k-1 and the ones for the denominator are n-k. Thus, since k=3 (we have 3 samples), and 25 6 9 10 3 2 1 = + + = + + = n n n n , degrees of freedom of the numerator=3-1=2 and degrees of freedom of the denominator=25-3=22.
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• Spring '07
• Guggenberger
• Null hypothesis, critical value, american league

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