Question

7. The following presents how long it takes to drain water with each type of channel type Type 1 41.4 43.4 So.0 41.2 Type 2 38.7 48.3 51.1 38.3 Type 3 32.6 33.7 34.8 22.5 Type 4 26.3 30.9 33.3 23.8 Type 5 44.9 47.2 48.5 37.1 a) Using a calculator, obtain the relevant values and construct an ANOVA table. b) At a .05, is there a difference between the means of the posttest scores?

Answer 1

7.

a)

Group 1 41.4 43.4 50 41.2 Group 2 38.7 48.3 51.1 38.3 Group 3 32.6 33.7 34.8 22.5 Group 4 26.3 30.9 33.3 23.8 Group 5 44.9 47.2 48.5 37.1 176.4 44.1 7908.68 6.569 123.6 30.9 3915.74 5.672 96.4999999 114.3 28.575 3321.83 4.309 Sum 176 44.425 7972.51 5.105 78.1875000 Average- 7794.96 St. Dev.- 4.121 50.96 129.44 55.7075 00001 4 4 4 4 4 The total sample size is N -20. Therefore, the total degrees of freedom are dftotal 20-1- 19 Also, the between-groups degrees of freedom are dfbetween 5-1-4, and the within-groups degrees of freedom are

dfwithin = dftotal-dfbetween = 19-4 = 15 First, we need to compute the total sum of values and the grand mean. The following is obtained Xij = 1764 176.4 + 123.64 114.34 177.7-768 2. Also, the sum of squared values is Xi, 7794.96 7908.68 + 3915. T4 3321.83 7972.51 30913.72 2. Based on the above calculations, the total sum of squares is computed as follows 768 SStotal = Xii 30913.72 2.J The within sum of squares is computed as shown in the calculation below ss,ithin ystrithingroups 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 = 〉 S Swithingroups 50.96 ± 129.44 + 96.499999999999 + 55.7075 + 78.187500000001 410.795 Setueen-SSetueen-. 101 1.725 enfkctween4 252.931

MS withinuithin410.795 hdjwithin 15 Now that sum of squares are computed, we can proceed with computing the mean sum of squares eteen1011.725 252.931 MSbetueenetireen between MS withinuithin410.795 hdjwithin 15 Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows F Sbetuen 252.931 within27.386 M Sbet MS The following null and alternative hypotheses need to be tested 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 dfı 4 and dfe 4, therefore, the rejection region for this F-test is R F: F F 3.056

(3) Test Statistics M Sbetween MSwithin27.386 252.931236 (4) Decision about the null hypothesis Since it is observed that F 9.236> Fe 3.056, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p - 0.0006, and since p 0.00060.05, 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 5 population means are equal, at the α 0.05 significance level.

ANOVA Table :

ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	1011.725	4	252.9313	9.235674	0.00057	3.055568
Within Groups	410.795	15	27.38633
Total	1422.52	19

b)

Now we perform Post Hoc test :

Treatments pair	Tukey HSD Q statistic	Tukey HSD P-value	Tukey HSD inferfence
Type1 vs Type2	0.038	0.9000	insignificant
Type1 vs Type3	5.007	0.0212	p<0.05
Type1 vs Type4	5.895	0.0063	p<0.05
Type1 vs Type5	0.162	0.9000	insignificant
Type2 vs Type3	5.045	0.0201	p<0.05
Type2 vs Type4	5.933	0.0060	p<0.05
Type2 vs Type5	0.124	0.9000	insignificant
Type3 vs Type4	0.889	0.9000	insignificant
Type3 vs Type5	5.169	0.0170	p<0.05
Type4 vs Type5	6.058	0.0050	p<0.05