Question

Develop the analysis of variance computations for the following completely randomized design. At a = .05, is there a significant difference between the treatment means? Treatment 143 99 108 120 125 115 109 121 96 112 113 117 117 117 128 108 115 106 112 89 104.80 122.84 Ij 120.25 s. 136.79 from the table to compute T (to 2 decimals) 95 113 101.40 152.49 Use the 108 Source of Variation Sum of Squares Degrees Mean Square (to 2 decimals) (to 2 Freedom decimals) F (to 2 decimals) p-value (to 4 decimals) of Treatments 1738.5 Error Total The p-value is - Select your answer -

Answer 1

Date: 01/11/2019 Answer To test the hypothesis is that there is a significant difference between the treatment means at 5% level of significance. The null and alternative hypothesis is : 0 Haat leastone , is unequal The grand mean is, 3 120.25 104.80101.40 = 108.82 The grand mean is 108.82.

The F-test statistics is, By using Excel software, find F-test statistics with the help of following steps: 1) 2) 3) 4) 5) Import the data. Choose ANOVA: Single Factor from Data Analysis in Data menu Choose Input Range and Coluumn in Groups by Choose Output Range Click Ok. ANOVA Source of Variation df SS MS F P-value Between Groups 2 1738.50 869.25 6.33 0.0060 Within Groups 3435.50 25 137.42 Total 27 5174.00 From the Excel output, the F-test statistics is 6.33 The p-value for this test is, From the Excel output, the p value for this test is 0.0060 less than 0.01.

Decision The conclusion is that p-value in this context is less than 0.05 which is 0.0060, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that there is a significant difference between the treatment means. The result is statistically significant.