Question

9.21 The data in the next table resulted from an experiment NW that used a completely randomized design. L09021 Treatment 3 Treatment 1 3.8 Treatment 2 5.4 2.0 1.3 1.2 0.7 2.2 4.1 4.8 SS 3.8 2.3 a. Use statistical software (or the appropriate calculation formulas in Appendix C) to complete the following ANOVA table: Source df SS MSF Treatments Error Total

b. Test the null hypothesis that m1 = m2 = m3, where mi
represents the true mean for treatment i, against the
alternative that at least two of the means differ. Use
a = .05.

Answer 1

Answer Date: 23/07/2019 The one-way ANOVA table is, By using Excel software, find one-way ANOVA table with the help of following steps: 1) Import the data. Choose ANOVA: Single Factor from Data Analysis in Data menu. Choose Input Range and Column in Groups by. 4) Choose Output Range. 5) Click Ok. The one-way ANOVA table Excel output is, df F ANOVA Source of Variation Between Groups Within Groups Total P-value F crit 2.93071 0.104669 4.256495 SS 12.30117 18.888 31.18917 MS 2 6.150583 92.098667 11

To test the hypothesis is that at least two of the means differs at 5% level of significance. The null and alternative hypothesis is, H:11 = 142 = uz Hq: at least two y; is unequal. The F-test statistics is, From the Excel output in part (a), the F-test statistics is 2.93. The p-value for this test is, From the Excel output in part (a), the p value for this test is 0.1047). Decision The conclusion is that p-value in this context is higher than 0.05 which is 0.1047, so the null hypothesis is not rejected at 5% level of significance. There is insufficient evidence to indicate that at least two of the means differs. The result is not statistically significant.