Explanation Given info: The table shows the index scores for a sample of states from four regions of the United States. The level of significance is 0.10. Calculation: The hypotheses are given below: Null hypothesis: H 0 : There is no difference between means. Alternative hypothesis: H a : There is difference between means. The critical value of Scheffe test is obtained by multiplying the critical values of the one-way ANOVA test by k − 1 . From the Exercise 11, the critical value of ANOVA is 2.28. Here k − 1 =4-1 =3 Thus, the critical value of Scheffe test is, C V Scheffe = 2.28 × 3 = 6.84 Rejection region: The null hypothesis would be rejected if test value > 6.84 . Consider x ¯ 1 , x ¯ 2 , x ¯ 3 and x ¯ 4 represents the means of Northeast, Midwest, South and West. Step-by-step procedure to obtain the test mean and standard deviation using the MINITAB software: Choose Stat > Basic Statistics > Display Descriptive Statistics . In Variables enter the columns Northeast, Midwest, South and West . Choose option statistics , and select Mean, Standard deviation and N total . Click OK . Output using the MINITAB software is given below: The formula for comparing the means is, ( x ¯ a − x ¯ b ) 2 S S w ∑ ( n i − 1 ) ( 1 n a + 1 n b ) Here, The sample sizes n 1 = 6 , n 2 = 8 , n 3 = 11 , n 4 = 9 The sample means are x ¯ 1 = 62.550 , x ¯ 2 = 62.400 , x ¯ 3 = 61.673 and x ¯ 4 = 62.878 The sample variances are s 1 2 = 0.699 , s 2 2 = 0.760 , s 3 2 = 0.910 and s 4 2 = 0.412 The formula to find S S w = ∑ ( n i − 1 ) s i 2 S S w = ∑ ( n i − 1 ) s i 2 = ( 6 − 1 ) 0.699 + ( 8 − 1 ) 0.760 + ( 11 − 1 ) 0.910 + ( 9 − 1 ) 0.412 = 3.495 + 5.320 + 9.100 + 3.296 = 21.211 Comparison between the means x ¯ 1 and x ¯ 2 : The formula to compare the means x ¯ 1 and x ¯ 2 is, ( x ¯ 1 − x ¯ 2 ) 2 S S w ∑ ( n i − 1 ) ( 1 n 1 + 1 n 2 ) That is, ( x ¯ 1 − x ¯ 2 ) 2 S S w ∑ ( n i − 1 ) ( 1 n 1 + 1 n 2 ) = ( 62.550 − 62.400 ) 2 21.211 ( 6 − 1 ) + ( 8 − 1 ) + ( 11 − 1 ) + ( 9 − 1 ) ( 1 6 + 1 8 ) = 0.0225 21.211 5 + 7 + 10 + 8 ( 0.1667 + 0.125 ) = 0.0225 0.7070 × 0.2917 = 0.0225 0.2062 = 0.11 Thus, the value of the comparison between means x ¯ 1 and x ¯ 2 is 0.11. Conclusion: The test value is 0.11. Here, the test value is lesser than the critical value. That is, 0.11 < 6.84 . Thus, the null hypothesis is not rejected. Hence, there is no significant difference between the means x ¯ 1 and x ¯ 2 . Comparison between the means x ¯ 1 and x ¯ 3 : The formula to compare the means x ¯ 1 and x ¯ 3 is, ( x ¯ 1 − x ¯ 3 ) 2 S S w ∑ ( n i − 1 ) ( 1 n 1 + 1 n 3 ) That is, ( x ¯ 1 − x ¯ 3 ) 2 S S w ∑ ( n i − 1 ) ( 1 n 1 + 1 n 3 ) = ( 62.550 − 61.673 ) 2 21.211 ( 6 − 1 ) + ( 8 − 1 ) + ( 11 − 1 ) + ( 9 − 1 ) ( 1 6 + 1 11 ) = 0.7691 21.211 5 + 7 + 10 + 8 ( 0.1667 + 0.0909 ) = 0.7691 0.7070 × 0.2576 = 0.0225 0.1821 = 0.12 Thus, the value of the comparison between means x ¯ 1 and x ¯ 3 is 0.12. Conclusion: The test value is 0.12. Here, the test value is lesser than the critical value. That is, 0.12 < 6.84 . Thus, the null hypothesis is not rejected. Hence, there is no significant difference between the means x ¯ 1 and x ¯ 3 . Comparison between the means x ¯ 1 and x ¯ 4 : The formula to compare the means x ¯ 1 and x ¯ 4 is, ( x ¯ 1 − x ¯ 4 ) 2 S S w ∑ ( n i − 1 ) ( 1 n 1 + 1 n 4 ) That is, ( x ¯ 1 − x ¯ 4 ) 2 S S w ∑ ( n i − 1 ) ( 1 n 1 + 1 n 4 ) = ( 62

Elementary Statistics: Picturing the World (6th Edition)

6th Edition

ISBN: 9780321911216

Author: Ron Larson, Betsy Farber

Publisher: PEARSON

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Determine which means have a significant difference by using the Scheffe test.

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Elementary Statistics: Picturing the World (6th Edition)

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