Using the following data, perform a oneway analysis of variance using \(\alpha=.05\). Write up the results in APA format.
\[\begin{equation} \begin{bmatrix} \textbf{Group1} \\ 51 \\ 45 \\ 33 \\ 45 \\ 67 \end{bmatrix} \begin{bmatrix} \textbf{Group2} \\ 23 \\ 43 \\ 23 \\ 43 \\ 45 \end{bmatrix} \begin{bmatrix} \textbf{Group3} \\ 56 \\ 76 \\ 74 \\ 87 \\ 56 \end{bmatrix} \end{equation}\]Sample means (\(\bar{x}\)) for the groups: = 48.2, 35.4, 69.8
Intermediate steps in calculating the group variances:
[[1]]
value mean deviations sq deviations
1 51 48.2 2.8 7.84
2 45 48.2 -3.2 10.24
3 33 48.2 -15.2 231.04
4 45 48.2 -3.2 10.24
5 67 48.2 18.8 353.44
[[2]]
value mean deviations sq deviations
1 23 35.4 -12.4 153.76
2 43 35.4 7.6 57.76
3 23 35.4 -12.4 153.76
4 43 35.4 7.6 57.76
5 45 35.4 9.6 92.16
[[3]]
value mean deviations sq deviations
1 56 69.8 -13.8 190.44
2 76 69.8 6.2 38.44
3 74 69.8 4.2 17.64
4 87 69.8 17.2 295.84
5 56 69.8 -13.8 190.44Sum of squared deviations from the mean (SS) for the groups:
[1] 612.8 515.2 732.8\(Var_1 = \frac{612.8}{5-1} = 153.2\)
\(Var_2 = \frac{515.2}{5-1} = 128.8\)
\(Var_3 = \frac{732.8}{5-1} = 183.2\)
\(MS_{error} = \frac{153.2 + 128.8 + 183.2}{3} = 155.07\) Note: this is just the average within-group variance; it is not sensitive to group mean differences!
Calculating the remaining error (or within) terms for the ANOVA table:
\(df_{error} = 15 - 3 = 12\)
\(SS_{error} = (155.07)(15 - 3) = 1860.8\)
Intermediate steps in calculating the variance of the sample means:
Grand mean (\(\bar{x}_{grand}\)) = \(\frac{48.2+35.4+69.8}{3}= 51.13\)
group mean grand mean deviations sq deviations
48.2 51.13 -2.93 8.58
35.4 51.13 -15.73 247.43
69.8 51.13 18.67 348.57Sum of squares \((SS_{means}) = 604.58\)
\(Var_{means} = \frac{604.58}{3-1} = 302.29\)
\(MS_{between} = (302.29)(5)=1511.45\) Note: This method of estimating the variance IS sensitive to group mean differences!
Calculating the remaining between (or group) terms of the ANOVA table:
\(df_{groups} = 3-1 = 2\)
\(SS_{group} = (1511.45)(3-1)=3022.9\)
Test statistic and critical value
\(F = \frac{1511.45}{155.07}=9.75\)
\(F_{critical}(2, 12) = 3.89\)
\(\fbox{ Decision: reject H0 }\)
ANOVA table
| source | SS | df | MS | F |
|---|---|---|---|---|
| group | 3022.9 | 2 | 1511.45 | 9.75 |
| error | 1860.8 | 12 | 155.07 | |
| total | 4883.7 |
Effect size
\(\eta^2 = \frac{3022.9}{4883.7} = 0.62\)
APA writeup
F(2, 12)=9.75, p <0.05, \(\eta^2\)=0.62.
Using the following summary data, perform a oneway analysis of variance using \(\alpha=.01\).
\[\begin{equation} \begin{bmatrix} \textbf{n} & \textbf{mean} & \textbf{sd} \\ 30 & 50.26 & 10.45 \\ 30 & 45.32 & 12.76 \\ 30 & 53.67 & 11.47 \\ \end{bmatrix} \end{equation}\]\(Var_1 = 10.45^2 = 109.2\)
\(Var_2 = 12.76^2 = 162.82\)
\(Var_3 = 11.47^2 = 131.56\)
\(MS_{error} = \frac{109.2 + 162.82 + 131.56}{3} = 134.53\) Note: this is just the average within-group variance; it is not sensitive to group mean differences!
Calculating the remaining error (or within) terms for the ANOVA table:
\(df_{error} = 90 - 3 = 87\)
\(SS_{error} = (134.53)(90 - 3) = 11703.82\)
Intermediate steps in calculating the variance of the sample means:
Grand mean (\(\bar{x}_{grand}\)) = \(\frac{50.26+45.32+53.67}{3}= 49.75\)
group mean grand mean deviations sq deviations
50.26 49.75 0.51 0.26
45.32 49.75 -4.43 19.62
53.67 49.75 3.92 15.37Sum of squares \((SS_{means}) = 35.25\)
\(Var_{means} = \frac{35.25}{3-1} = 17.62\)
\(MS_{between} = (17.62)(30)=528.75\) Note: This method of estimating the variance IS sensitive to group mean differences!
Calculating the remaining between (or group) terms of the ANOVA table:
\(df_{groups} = 3-1 = 2\)
\(SS_{group} = (528.75)(3-1)=1057.5\)
Test statistic and critical value
\(F = \frac{528.75}{134.53}=3.93\)
\(F_{critical}(2, 87) = 4.86\)
\(\fbox{ Decision: fail to reject H0 }\)
ANOVA table
| source | SS | df | MS | F |
|---|---|---|---|---|
| group | 1057.5 | 2 | 528.75 | 3.93 |
| error | 11703.82 | 87 | 134.53 | |
| total | 12761.32 |
Effect size
\(\eta^2 = \frac{1057.5}{12761.32} = 0.08\)
APA writeup
F(2, 87)=3.93, p >=0.01, \(\eta^2\)=0.08.
A clinical psychologist has run a between-subjects experiment comparing two treatments for depression (cognitive-behavioral therapy (CBT) and client-centered therapy (CCT) against a control condition. Subjects were randomly assigned to the experimental condition. After 12 weeks, the subject’s depression scores were measured using the CESD depression scale. The data are summarized as follows:
\[\begin{equation} \begin{bmatrix} & \textbf{n} & \textbf{mean} & \textbf{sd} \\ \textbf{control} & 40 & 21.4 & 4.5 \\ \textbf{CBT} & 40 & 16.9 & 5.5 \\ \textbf{CCT} & 40 & 19.1 & 5.8 \\ \end{bmatrix} \end{equation}\]Use a oneway ANOVA with \(\alpha=.01\) for the test.
\(Var_1 = 4.5^2 = 20.25\)
\(Var_2 = 5.5^2 = 30.25\)
\(Var_3 = 5.8^2 = 33.64\)
\(MS_{error} = \frac{20.25 + 30.25 + 33.64}{3} = 28.05\) Note: this is just the average within-group variance; it is not sensitive to group mean differences!
Calculating the remaining error (or within) terms for the ANOVA table:
\(df_{error} = 120 - 3 = 117\)
\(SS_{error} = (28.05)(120 - 3) = 3281.46\)
Intermediate steps in calculating the variance of the sample means:
Grand mean (\(\bar{x}_{grand}\)) = \(\frac{21.4+16.9+19.1}{3}= 19.13\)
group mean grand mean deviations sq deviations
21.4 19.13 2.27 5.15
16.9 19.13 -2.23 4.97
19.1 19.13 -0.03 0.00Sum of squares \((SS_{means}) = 10.12\)
\(Var_{means} = \frac{10.12}{3-1} = 5.06\)
\(MS_{between} = (5.06)(40)=202.4\) Note: This method of estimating the variance IS sensitive to group mean differences!
Calculating the remaining between (or group) terms of the ANOVA table:
\(df_{groups} = 3-1 = 2\)
\(SS_{group} = (202.4)(3-1)=404.8\)
Test statistic and critical value
\(F = \frac{202.4}{28.05}=7.22\)
\(F_{critical}(2, 117) = 4.79\)
\(\fbox{ Decision: reject H0 }\)
ANOVA table
| source | SS | df | MS | F |
|---|---|---|---|---|
| group | 404.8 | 2 | 202.4 | 7.22 |
| error | 3281.46 | 117 | 28.05 | |
| total | 3686.26 |
Effect size
\(\eta^2 = \frac{404.8}{3686.26} = 0.11\)
APA writeup
F(2, 117)=7.22, p <0.01, \(\eta^2\)=0.11.
An education researcher is comparing four different algebra curricula. Eighth grade students are randomly assigned to one one of the four groups. Their state achievement test scores are compared at the end of the year. Use the appropriate statistical procedure to determine whether the curricula differ with respect to math achievement. An alpha criterion of .05 should be used for the test.
\[\begin{equation} \begin{bmatrix} & \textbf{n} & \textbf{mean} & \textbf{sd} \\ \textbf{curriculum 1} & 50 & 170.5 & 14.5 \\ \textbf{curriculum 2} & 50 & 168.3 & 12.8 \\ \textbf{curriculum 3} & 50 & 167.6 & 17.7 \\ \textbf{curriculum 4} & 50 & 172.8 & 16.8 \\ \end{bmatrix} \end{equation}\]\(Var_1 = 14.5^2 = 210.25\)
\(Var_2 = 12.8^2 = 163.84\)
\(Var_3 = 17.7^2 = 313.29\)
\(Var_3 = 16.8^2 = 282.24\)
\(MS_{error} = \frac{210.25 + 163.84 + 313.29}{4} = 242.41\) Note: this is just the average within-group variance; it is not sensitive to group mean differences!
Calculating the remaining error (or within) terms for the ANOVA table:
\(df_{error} = 200 - 4 = 196\)
\(SS_{error} = (242.41)(200 - 4) = 47511.38\)
Intermediate steps in calculating the variance of the sample means:
Grand mean (\(\bar{x}_{grand}\)) = \(\frac{170.5+168.3+167.6}{3}= 169.8\)
group mean grand mean deviations sq deviations
170.5 169.8 0.7 0.49
168.3 169.8 -1.5 2.25
167.6 169.8 -2.2 4.84
172.8 169.8 3.0 9.00Sum of squares \((SS_{means}) = 16.58\)
\(Var_{means} = \frac{16.58}{4-1} = 5.53\)
\(MS_{between} = (5.53)(50)=276.33\) Note: This method of estimating the variance IS sensitive to group mean differences!
Calculating the remaining between (or group) terms of the ANOVA table:
\(df_{groups} = 4-1 = 3\)
\(SS_{group} = (276.33)(4-1)=829\)
Test statistic and critical value
\(F = \frac{276.33}{242.41}=1.14\)
\(F_{critical}(3, 196) = 2.65\)
\(\fbox{ Decision: fail to reject H0 }\)
ANOVA table
| source | SS | df | MS | F |
|---|---|---|---|---|
| group | 829 | 3 | 276.33 | 1.14 |
| error | 47511.38 | 196 | 242.41 | |
| total | 48340.38 |
Effect size
\(\eta^2 = \frac{829}{48340.38} = 0.02\)
APA writeup
F(3, 196)=1.14, p >=0.05, \(\eta^2\)=0.02.