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Question
Given the following sample data| Item | Group 1 | Group 2 | Group 3 | Group 4 |
| 1 | 20.9 | 28.2 | 17.8 | 21.2 |
| 2 | 27.2 | 26.2 | 15.9 | 23.9 |
| 3 | 26.6 | 21.6 | 18.4 | 19.5 |
| 4 | 22.1 | 29.7 | 20.2 | 17.4 |
| 5 | 25.3 | 30.3 | 14.1 | |
| 6 | 30.1 | 25.9 | ||
| 7 | 23.8 |
Based on the computations for the within- and between-sample variation, develop the ANOVA table and test the appropriate null hypothesis using alpha= 0.05. Use the p-value approach.
A) Since p-value = 0.0678 > 0.05 reject H0 and conclude that at least two population means are different.
B) Since p-value = 0.000136 < 0.05 reject H0 and conclude that at least two population means are different.
C) Since p-value = 0.0678 > 0.05 accept H0 and conclude that all population means are the same.
D) Since p-value = 0.000136 < 0.05 accept H0 and conclude that all population means are the same.
Answer
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B) The null hypothesis, the specification not being met, was not rejected when in fact it was not being met, this is a Type I error.
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A) Because t = -2.69 is less than -2.3456, do not reject H0
Conclude that 2012 average out-of-pocket prescription drug expenses are not lower than the 2002 average.
B) Because t = -2.69 is less than -2.3456, reject H0
Conclude that 2012 average out-of-pocket prescription drug expenses are lower than the 2002 average.
C) Because t = -1.69 is less than -0.8712, reject H0
Conclude that 2012 average out-of-pocket prescription drug expenses are lower than the 2002 average.
D) Because t = -1.69 is less than -0.8712, do not reject H0
Conclude that 2012 average out-of-pocket prescription drug expenses are not lower than the 2002 average.
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A) p-value = 0.0872 > 0.025; therefore do not reject H0
B) p-value = 0.0422 > 0.005; therefore do not reject H0
C) p-value = 0.0314 < 0.105; therefore reject H0
D) p-value = 0.0121< 0. 0805; therefore reject H0
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