Problem 1

Using the following summary data, perform a oneway analysis of variance using \(\alpha=.05\). Perform all pairwise contrasts (i.e., group1 vs group2, group2 vs group3, and group1 vs group3). The Bonferroni-corrected alpha should be \(\frac{.05}{3}=.0167\), but you should use \(\alpha_{corrected}=.02\) since that’s the closest value in the book’s table.

\[\begin{equation} \begin{bmatrix} \textbf{n} & \textbf{mean} & \textbf{sd} \\ 20 & 46.87 & 11.6 \\ 20 & 50.92 & 10.33 \\ 20 & 45.82 & 9.18 \\ \end{bmatrix} \end{equation}\]

Solution

ANOVA table

source SS df MS F
group 290 2 145 1.34
error 6185.26 57 108.51
total 6475.26

APA writeup

F(2, 57)=1.34, p >=0.05, \(\eta^2\)=0.04.

Contrasts

Standard error for contrasts: \(StdErr = 2\sqrt{\frac{MS_{error}}{n_1+n_2}} = 2 \sqrt{\frac{108.51}{20 + 20}} = 3.3\)

1 2 3 raw es std err t t critical decision
1 -1 0 -4.05 3.3 -1.23 2.429 fail to reject H0
1 0 -1 1.05 3.3 0.32 2.429 fail to reject H0
0 1 -1 5.10 3.3 1.55 2.429 fail to reject H0

Problem 2

Using the following summary data, perform a oneway analysis of variance using \(\alpha=.05\). Perform all pairwise contrasts. The Bonferroni-corrected alpha should be \(\frac{.05}{3}=.0167\), but you should use \(\alpha_{corrected}=.02\) since that’s the closest value in the book’s table.

\[\begin{equation} \begin{bmatrix} \textbf{n} & \textbf{mean} & \textbf{sd} \\ 10 & 104.9 & 14.7 \\ 10 & 107.4 & 16.5 \\ 10 & 105.8 & 15.4 \\ \end{bmatrix} \end{equation}\]

Solution

ANOVA table

source SS df MS F
group 32.1 2 16.05 0.07
error 6529.5 27 241.83
total 6561.6

APA writeup

F(2, 27)=0.07, p >=0.05, \(\eta^2\)=0.

Contrasts

Standard error for contrasts: \(StdErr = 2\sqrt{\frac{MS_{error}}{n_1+n_2}} = 2 \sqrt{\frac{241.83}{10 + 10}} = 6.95\)

1 2 3 raw es std err t t critical decision
1 -1 0 -2.5 6.95 -0.36 2.552 fail to reject H0
1 0 -1 -0.9 6.95 -0.13 2.552 fail to reject H0
0 1 -1 1.6 6.95 0.23 2.552 fail to reject H0

Problem 3

Using the following summary data, perform a oneway analysis of variance using \(\alpha=.05\). Perform all pairwise contrasts and a corrected \(\alpha\) of .01 for the contrasts.

\[\begin{equation} \begin{bmatrix} \textbf{n} & \textbf{mean} & \textbf{sd} \\ 15 & 8.8 & 3 \\ 15 & 5.6 & 3.4 \\ 15 & 12.2 & 3.3 \\ 15 & 9.9 & 3.2 \\ \end{bmatrix} \end{equation}\]

Solution

ANOVA table

source SS df MS F
group 338.85 3 112.95 10.84
error 583.66 56 10.42
total 922.51

APA writeup

F(3, 56)=10.84, p <0.05, \(\eta^2\)=0.37.

Contrasts

Standard error for contrasts: \(StdErr = 2\sqrt{\frac{MS_{error}}{n_1+n_2}} = 2 \sqrt{\frac{10.42}{15 + 15}} = 1.18\)

1 2 3 4 raw es std err t t critical decision
1 -1 0 0 3.2 1.18 2.71 2.467 reject H0
1 0 -1 0 -3.4 1.18 -2.88 2.467 reject H0
1 0 0 -1 -1.1 1.18 -0.93 2.467 fail to reject H0
0 1 -1 0 -6.6 1.18 -5.59 2.467 reject H0
0 1 0 -1 -4.3 1.18 -3.64 2.467 reject H0
0 0 1 -1 2.3 1.18 1.95 2.467 fail to reject H0

Problem 4

Using the following summary data, perform a oneway analysis of variance using \(\alpha=.05\). Perform all pairwise contrasts using the Bonferroni correction to control the familywise error rate.

\[\begin{equation} \begin{bmatrix} \textbf{n} & \textbf{mean} & \textbf{sd} \\ 30 & 22.8 & 3 \\ 30 & 22.3 & 5.6 \\ 30 & 20.2 & 4.9 \\ \end{bmatrix} \end{equation}\]

Solution

ANOVA table

source SS df MS F
group 114 2 57 2.66
error 1866.73 87 21.46
total 1980.73

APA writeup

F(2, 87)=2.66, p >=0.05, \(\eta^2\)=0.06.

Contrasts

Standard error for contrasts: \(StdErr = 2\sqrt{\frac{MS_{error}}{n_1+n_2}} = 2 \sqrt{\frac{21.46}{30 + 30}} = 1.2\)

1 2 3 raw es std err t t critical decision
1 -1 0 0.5 1.2 0.42 2.392 fail to reject H0
1 0 -1 2.6 1.2 2.17 2.392 fail to reject H0
0 1 -1 2.1 1.2 1.75 2.392 fail to reject H0