Question

Question Details You may need to use the appropriate technology to answer this question. The following data are from a completely randomized design. In the following calculations, use a 0.05. Treatment Treatment Treatment 2 3 1 63 82 68 46 73 55 54 88 60 37 61 45 50 76 57 X 123.33 138.00 92.67 (a) Use analysis of variance to test for a significant difference among the means of the three treatments. State the null and alternative hypotheses. OH: Not all the population means are equal. Ho 2 3 H: Not all the population means are equal. O Ha: At least two of the population means are equal. H: At least two of the population means are different. O Ho H1 2H H 23 Find the value of the test statistic. (Round your answer to two decimal places.) Find the p-value. (Round your answer to three decimal places.) p-value State your conclusion. Reject Ha. There is sufficient evidence to conclude that the means of the three treatments are ne Do not reject Ho. There is sufficient evidence to conclude that the means of the three treatments

Answer 1

Answer Date: 17/06/2019 To test the hypothesis is that there is a significant difference among the means of the three treatments at 5% level of significance a) The null and alternative hypothesis is at least one is unequa H 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 Column in Groups by Choose Output Range Click Ok

Anova: Single Factor SUMMARY Average Variance Count Groups Treatment 1 Treatment 2 Treatment 3 Sum 50 123.3333 4 200 304 76 138 4 228 57 92.66667 ANOVA Source of Variation Between Groups Within Groups Total F crit MS df F P-value SS 724 6.135593 0.020846281 4.256495 2 1448 1062 118 2510 11 From the Excel output, the F-test statistics is 6.14 The p-value for this test is, From the Excel output, thep value for this test is 0.021

Decision The conclusion is that p-value in this context is less than 0.05 which is 0.021, hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that there is a significant difference among the means of the three treatments. The result is statistically significant so the null b) The Fisher's LSD value is, First, compute t critical value then find Fisher's LSD value The t critical value is, The sample size is small and two-tailed test. Look in the column headed a = 0.05 and the row headed in the t distribution table by using degree of freedom within is 9 and the t critical value is 2.262 The critical value of Fisher's LSD is

1 MSW x nA 1 LSD t, DFWx 1 2.262 x 18 x 4 1 _ 4 17.38 The critical value of Fisher's LSD is 17.38 The mean difference table is Difference Absolute Treatments Mean 1 Mean 2 -26 Treatment 1-Treatment 2 50 76 26 -7 57 Treatment 1-Treatment 3 50 7 19 Treatment 2-Treatment 3 76 57 19 Decision The statistical decision is that the mean difference is greater than critical value of Fisher's LSD which is 17.38, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that there is significant difference between the means for Treatment 1-Treatment 2 and Treatment 2-Treatment 3 is statistically significant. The result is statistically significant