Consider the following information. SSTR = 6750 H₀: μ₁ =μ₂ =μ₃ =μ₄ SSE = 8000 H_a: At least one mean is different The null hypothesis is to be tested at the 5% level of significance.The null hypothesis A) should be rejected. B) should not be rejected. C) was designed incorrectly. D) cannot be tested.

Question

Quizplus · Accepted Answer

The null hypothesis is that all means are equal. The alternative hypothesis is that at least one mean is different. The F-test for one-way ANOVA can be used to test the null hypothesis. The test statistic is calculated as F = (SSTR / dfB) / (SSE / dfE), where SSTR is the sum of squares due to treatment, SSE is the sum of squares due to error, dfB is the degrees of freedom for treatment, and dfE is the degrees of freedom for error.

The critical value of F for a one-way ANOVA with k treatments and n observations per treatment at the 5% level of significance is F(k-1, kn-k) = F(2,20) = 3.10.

Using the given information, SSTR = 6750, SSE = 8000, dfB = 3-1 = 2, and dfE = 24-3 = 21. Thus, F = (6750 / 2) / (8000 / 21) = 7.73. Since 7.73 > 3.10, we reject the null hypothesis and conclude that at least one mean is different. Therefore, the answer is A, the null hypothesis should be rejected.

Consider the Following Information. ​

Consider the Following Information.