Question

2.) The data below are from an independent-measures experiment comparing the effects of insomnia treatments. Treatment 1 is a control group, Treatment 2 meditated before sleeping, and Treatment 3 received a sleeping pill. Note Pay close attention to what information you are given! Treatment1 Treatment 2 Treatment 3 6 G-36 T2 = 8 SS2 -6 T's = 24 SS-6 SS: = 6 a.Complete the ANOVA summary table below to help determine whether these data indicate any significant mean differences among the treatments. ss MS Source Between Treatments Within Treatments Total b. Testing at an alpha level of a 05, is there a significant difference among the treatment means? What is your Fonsol value? Report the results in APA format. Foni- c. Use the Tukey's HSD test to determine which of the treatments are significantdly different from each other. Use the .0S level of significance for all tests and clearly identify which means differ from one another. HSD value- d. Calculate an effect size in terms of proportion of variance account for (n

Answer 1

a.

Source	SS	DF	MS	F
Between Treatments	56	2	28	14
Within Treatments	18	9	2
Total	74	11

T1 = 4, T2 = 8 , T3 = 24

G = 36 , N = 4 * 3 = 12

$\sum$X² = 182

G²/N = 36² / 12 = 108

SST = $\sum$X² - G²/N = 182 - 108 = 74

$\sum$T₁²/n = 4² / 4 = 4

$\sum$T₂²/n = 8² / 4 = 16

$\sum$T₃²/n = 24² / 4 = 144

SS Between = $\sum$T²/n - G²/N = (4 + 16 + 144 ) - 108 = 56

SS Within = 74 - 56 = 18

DF Between = Number of treatments - 1 = 3 - 1 = 2

DF within = N - Number of treatments = 12 - 3 = 9

DF total = N - 1 = 12 - 1 = 11

MS Between = SS Between / DF Between = 56 / 2 = 28

MS Within = SS Within / DF Within = 18 / 9 = 2

F = MS Between / MS Within = 28 / 2 = 14

b.

Numerator df = DF Between = 2

Denominator df = DF within = 9

Critical value of F at df = 2, 9 and significance level of 0.05 is 4.26

As, observed F (14) is greater than the critical value of F, we reject null hypothesis and conclude that there is significant evidence that mean difference among the treatments are not zero.

c.

$HSD = q(\alpha, r, df_w) \sqrt{MS_w / n}$

where $q(\alpha, r, df_w)$ is a critical value of the studentized range for $\alpha$, the number of treatments or samples r, and the within-groups degrees of freedom $df_w$. We get this value from studentized range table.

$MS_w$ is the within groups mean square from the ANOVA table and n is the sample size for each treatment.

From Anova table,

$MS_w$ = 2 , $df_w$ = 9

n = 4, r = 3

For 95% confidence interval , $\alpha$ = 0.05

From studentized range table, $q(0.05, 3, 9)$ = 3.948

So, $q(\alpha, r, df_w) \sqrt{MS_w / n} = 3.948 \sqrt{2 / 4} = 2.79$

HSD = 2.79

M1 = T1 / 4 = 4 / 4 = 1

M2 = T2 / 4 = 8 / 4 = 2

M3 = T3 / 4 = 24 / 4 = 6

M3 - M1 = 6 - 1 = 5

M2 - M1 = 2 - 1 = 1

M3 - M2 = 6 - 2 = 4

As,  M3 - M1 and M3 - M2 are greater than the critical HSD value, Treatment 1 and Treatment 2 means significantly differ from Treatment 3 mean.

d.

$\eta^2$ = SS between / SS total = 56 / 74 = 0.7568