Question

Statistics Psychology ANOVA tablesPlease help me understand ANOVA tables better.

GENERAL INSTRUCTION FOR ANOVA (Analysis of Variance) SHOW ALL WORK: STATE A BASIC NULL AND RESEARCH HYPOTHESIS FOR EACH PROBLEM O USE .05 LEVEL OF SIGNIFICANCE FOR EACH PROBLEM CALCULATE EFFECT SIZE FOR INSTRUCTOR PROBLEM AND PROBLEM #3 AND RECEIVE ONE POINT EXTRA CREDIT FOR EACH. PROBLEM #1 INSTRUCTOR: Treatment for an injury (scores for pain) Control (no tx) Pain Medication Physical Therapy a o *****Note: A score of 0 is no pain and a score of 10 is intense pain. Complete an ANOVA table and state the conclusion of the statistical test

Answer 1

The following table is obtained: Group 1 Group 2 Group 3 Sum = Average = 131 St. Dev. = 1.414 1.225 SS = n = 5 The total sample size is N = 15. Therefore, the total degrees of freedom are: d ftotal = 15 - 1 = 14 Also, the between-groups degrees of freedom are dfbetween = 3-1 = 2, and the within-groups degrees of freedom are: dfwithin = d ftotal - dfbetween = 14 - 2 = 12 First, we need to compute the total sum of values and the grand mean. The following is obtained Xij = 40 + 15 + 25 = 80 Also, the sum of squared values is x} = 328 +49 +131 = 508 Based on the above calculations, the total sum of squares is computed as follows Sus = { 13 - + (x3) – 50 ml – 31.333 The within sum of squares is computed as shown in the calculation below: SS within = SS withingroups = 8 + 4 + 6 = 18 The between sum of squares is computed directly as shown in the calculation below: The between sum of squares is computed directly as shown in the calculation below:

Now that sum of squares are computed, we can proceed with computing the mean sum of squares: SSbetween 63.333 M S between = 31.667 dfbetween ween S Swithin 18 M Swithin = dfwithin 12 = 1.5 Now that sum of squares are computed, we can proceed with computing the mean sum of squares: M Sbetween = S Sbetween d freteneen 63.333 = 31.667 2 SS within 18 M Swithin = dfvithin 19 = 1.5 121.5 Finally, with having already calculated the mean sum of squares, the F- statistic is computed as follows: M Sbetween MS within 31.667 - = 21.111 1.5 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: M1 = M2 = M3 Ha: Not all means are equal The above hypotheses will be tested using an F-ratio for a One-Way ANOVA. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the degrees of freedom are dfi = 2 and df2 = 2, therefore, the rejection region for this F-test is R = {F:F> Fe = 3.885) (3) Test Statistics F _M Sbetween M Swithin = 31.667_, ' = 21.111 1.5 (4) Decision about the null hypothesis Since it is observed that F = 21.111 > Fe = 3.885, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0.0001, and since p= 0.0001 <0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that not all 3 population means are equal, at the a = 0.05 significance level.

A nova somłe Treatment les Ersor with Total Table af ss 63.353 12 18 14 81.333 sa E 31.664 21 w 1.5