Question

(1) A psychologist would like to examine the relative effectiveness of 3 treatments for mild phobias. A sample of N=12 individuals who display a moderate fear of spiders is obtained. These individuals are randomly assigned (with n=4) to each of the 3 treatments. The dependent variable is a measure of reported fear of spiders after treatment. The data are as follows:

(Treatment 1) 0 4 0 4

(Treatment 2) 1 5 4 2

(Treatment 3) 8    5   6    9

a. Test whether there are any differences among the effectiveness of 3 treatments, test with α=.05(4 step hypothesis testing)

b. If the effectiveness of 3 treatments are different, use Tukey’s HSD to find which treatments are different from each other. State which treatments are different.

Answer 1

Date: 13/05/2019 Answer a) To test the hypothesis is that there are differences among the effectiveness of three treatments at 5% level of significance Step 1 The null and alternative hypothesis is H, : at least one 1s unequal Step 2 The F-test statistics is, By using Excel software, find F-test statistics with the help of following steps: 1)Import the data 2) Choose ANOVA: Single Factor from Data Analysis in Data menu. 3) Choose Input Range and Column in Groups by 4) Choose Output Range.

5) Click Ok Anova: Single Factor SUMMARY Groups Count Sum Average Variance 2 5.333333 3 3.333333 7 3.333333 Treatment 1 Treatment 2 Treatment 3 4 4 4 12 28 ANOVA Source of Variation Between Groups Within Groups Total P-value Fcrit MS 56 36 92 28 2 0.014666 4.256495 4 From the Excel output, the F-test statistics is 7.00 Step 3

The p-value for this test is, From the Excel output, thep value for this test is | 0.014 Step 4 Decision The conclusion is that p-value in this context is less than 0.05 which is 0.0147, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that there are differences among the effectiveness of three treatments. 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 comparison is 3.95 The q for Tukey HSD test is 3.95 The critical value of Tukey HSD is, MS within. N Group 3.95x 3.95x 4 3.95 The critical value of Tukey HSD is 3.95 The mean difference table is

Mean 1 Mean 2 |Difference Treatments Treatment 1- Treatment2 Treatment 1- Treatment 3 Treatment 2- Treatment 3 -1 -5 -4 3 2 3 Decision The statistical decision is that the mean difference is greater than critical value of Tukey HSD which is-3.95, 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 and Treatment 2- Treatment 3 Status is statistically significant. The result is statistically significant.