Question

Edit View Insert Format Tools Table Window Help Share rt Design Layout References Mailings Review 42, Styles The data below are from an independent-measures experiment comparing three different treatment conditions. Treatment 1 Treatment 2 Treatment3 4 G-24 6 X squared-92 4 T-2 SS-3 T-6 SS-9 T 16 SS6 Use an ANOVA with α-05 to determine whether these data indicate any significant differences afhong treatments. a. b. Use the Tukey's IISD to determine which of the treatments are significantly different from each other at the 05 level.

Answer 1

Answer Date: 17/01/2019 To test the hypothesis is that there is significance difference among treatments conditions at 5% level of significance. a) The null and alternative hypothesis is, Ha: at least one is unequal. The F-test statistics is, find F-test statistics with the helpof following steps: y using Excelsoitware, 1) Import the data. 2) Choose ANOVA: Single Factor from Data Analysis in Data menu. 3) Choose Input Range and Colunn in Groups by 4) Choose Output Range 5) Click Ok.

Anova: Single Factor SUMMARY Groups Treatment 1 Treatment 2 Treatment 3 Count Sum Average Variance 0.5 1.5 4 2 4 16 ANOVA Source of Variation Between Groups Within Groups Total MS P-value F crit 26 18 13 6.50 0.017914 4.256495 From the Excel output, the F-test statistics is6.50 Thep-value for this test is, From the Excel output, the p value for this test is|0.0179 Decision

The conclusion is that p-value in this context is less than 0.05 which is 0.0179, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that there is significance difference among treatments conditions. The result is statistically significant. b) The Tukey HSD value for each of three mean comparisons is, First, compute q then find standard error for computing Tukey HSD statistics The q for Tukey HSD test is, From the q table, look at the independent variable 3 and degrees of freedom within are 9 and at 0.05 significance level and get critical values of the q statistics for the comparisorn is 3.95. The q for Tukey HSD test is 3.95 The critical value of Tukey HSD is

MSwithin HSD = qx Group 2 4 3.95x - 3.95x 0.7071 = 2.79 The critical value of Tukey HSD is 2.79 The mean difference table is, Treatments Mean 1 Mean 2 Difference Treatment 1-Treatment 20.50 Treatment 1-Treatment 30.50 Treatment 2-Treatment 3 1.500 1.50 4.00 4.00 1.00 -3.50 2.50 Decision

The statistical decision is that the mean difference is greater than critical value of Tukey HSD which is 2.79, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that the mean difference of Treatment 1-Treatment 3 is statistically significant. The result is statistically significant.