Two Factor ANOVA March, 07 Contents Two Factor ANOVA Example : peanut butter and jelly Within-cell variance - the denominator of all three F-tests Main effects for rows: Does peanut butter affect taste ratings? Main effect for columns: does jelly affect taste ratings? The interaction: does the effect of peanut butter on taste ratings depend on jelly? The Summary Table Example : peanut butter and ham Example : A x ANOVA on study habits Questions Answers Two Factor ANOVA A two factor ANOVA (sometimes called two-way ANOVA) is a hypothesis test on means for which groups are placed in groups that vary along two factors instead of one. For example, we might want to measure BMI for subjects with different diets AND for different levels of exercise. Or, we might measure the duration of hospital stays for patients undergoing different levels of medication AND different types of physical therapy. In these cases, we re interest in not only how the means vary across each factor (called the main effects), but also how these two dimensions interact with each other (called interactions). Here s how to get to the -factor ANOVA with the flow chart: START HERE Test for ρ = 0 Ch 7. number of correlations correlation (r) measurement scale frequency number of variables χ test frequency Ch 9.5 Test for ρ = ρ Ch 7.4 Means χ test independence Ch 9.9 z-test Ch. Yes Do you know σ? number of means More than number of factors -factor ANOVA Ch No one sample t-test Ch.4 -factor ANOVA Ch 0 independent measures t-test Ch 5.6 Yes independent samples? No dependent measures t-test Ch 6.4
For a two-factor ANOVA, each of the two factors (like diet and exercise ) are comprised of different levels, (like different diets, or different levels of exercise). The complete design involves having subjects placed into all possible combinations of levels for the two factors. For example, if there are two diets and three levels of exercise, there will be x = 6 groups, or cells. Each cell should have the same number of subjects, which called a balanced design. There are ways to deal with unequal sample sizes, but we will not go there. Example : peanut butter and jelly Let s start with a simple example. Suppose you want to study the effects of adding peanut butter and jelly to bread to study how they affect taste preferences. You take 6 students and break them down into 4 groups of 9 in what s called a x design. One group will get just bread, another will get bread with peanut butter, another will get bread with jelly, and one group will get bread with both peanut butter and jelly. We ll run our test with α = 0.05. Each student tastes their food and rates their preference on a scale from (gross) to 0 (excellent). The raw data looks like this: no jelly, no no jelly, peanut jelly, no peanut jelly and peanut peanut butter butter butter butter 8 4 8 6 8 6 9 5 6 4 6 4 6 9 8 4 8 9 9 8 6 4 8 5 5 5 0 9 4 6 4 9 Here are the summary statistics from the experiment: Means no jelly jelly no peanut butter 4. 6. peanut butter 6. 7.6667 SSwc no jelly jelly no peanut butter.5556 50 peanut butter 5.5556 0 Totals grand mean 6. SS total 8.5556 We ve organized statistics in a x matrix for which the factor peanut butter varies across the rows and the factor jelly varies across the columnns. We therefore call peanut butter the row factor and jelly the column factor. The table titled SSwc contains the sums of squared deviation of each score from the mean of for the cell that it belongs to. For example, for the no peanut butter, no jelly category, the mean for that cell is 4., so the SSwc for that cell is: (8 4.) + (6 4.) +... + (5 4.) =.5556 SS total is the sums of squared deviation of each score from the grand mean. Before we run any statistical analyses, it s useful to plot the means with error bars representing the standard errors of the mean. Remember, standard errors can be calculated by first calculating standard deviations from SSwc:
sx = SS wc n And then dividing by by the square root of the sample size to get the standard error of the mean: s x = s x n For example, for the no peanut butter, no jelly category, sx =.5556 9 So =.986 s x =.986 9 = 0.66 Check for yourself that the standard errors of the mean turn out to be: Standard errors no jelly jelly no peanut butter 0.66 0.8 peanut butter 0.5958 0.57 Here s the plot: 8.5 8 no peanut butter peanut butter 7.5 7 6.5 Taste rating 6 5.5 5 4.5 4.5 no jelly jelly Visually inspecting the graph, you can see that the green points are shifted above the red ones. This means that adding peanut butter helps the taste for both the plain bread, and for bread with jelly. This is called a main effect for rows, where the rows correspond the factor peanut butter. Similarly, adding jelly increases the taste rating both for plain bread, and for bread with peanut butter. This is a main effect for columns where columns refer to the factor jelly. Notice also that the lines are roughly parallel. This means that the increase in the taste rating by adding peanut butter is about the same with and without jelly. We therefore say that there is no interaction between the factors of peanut butter and jelly.
The -factor ANOVA quantifies the statistical significance of these three observations: the main main effect for rows, the main effect for columns, and the interaction. Within-cell variance - the denominator of all three F-tests The statistical significance of the main effect for rows, the main effect for columns and the interaction will all be done with F-tests. Just as with the -factor ANOVA, the F-test will be a ratio of variances which are both estimates of the population variance under the null hypothesis. Also, like the -factor ANOVA, the denominator will be an estimate of variance based on the variability within each group, and again we calculate it by summing up the sums of squared error between each score and its own cell s mean: SSwc = (X X cell ) I ve given you SSwc for each cell, so the total SSwc is just the sum of these four numbers: SSwc =.5556 + 5.5556 + 50 + 0 = 7. Just as with the -way ANOVA, the degrees of freedom for SSwc is the total number of scores minus the total number of groups, so df = 9 + 9 + 9 + 9-4 = The variance within cells is SSwc divided by its degrees of freedom: s wc = SS wc dfwc = 7. =.97 Main effects for rows: Does peanut butter affect taste ratings? To quantify the main effect for rows (peanut butter) we average the means across the two levels for columns (jelly). We can show these two means in a third column in the matrix of means: Means no jelly jelly Row Means no peanut butter 4. 6. 5.778 peanut butter 6. 7.6667 6.9445 The difference between these two row means indicates the effect of adding peanut butter after averaging across the two levels of jelly. We ll use these two means to calculate the numerator of the F-test for the main effect for rows. We first calculate SS for rows much like we would if this were a -factor ANOVA. We calculate the sum of squared difference of the row means from the grand mean, scaled by the sample size for the number of scores for each mean. Since there are groups for each mean, the sample size for each row mean is ()(9) = 8. So, SSrow = (nrow)( Xrow X) = SS R = ()(9)(5.778 6.) + ()(9)(6.9445 6.) = 5 The degrees of freedom for SS for rows is equal to the number of rows minus (df = - = ). The variance for rows is SS for rows (SSrows) divided by its degrees of freedom: s rows = SS rows dfrows = 5 = 5 The F-statistic for the main effect for rows is the variance for rows divided by the variance within cells: Frows = s rows s wc = 5.97 = 6.9 Under the null hypothesis, the variance for rows is another estimate of the population variance. So if the null hypothesis is true, this F statistic should be around, on average. A large value of F indicates that the row means vary more than expected under the null hypothesis. If F is large enough, we conclude that this is too unusual for the null hypothesis to be true, so we reject it and conclude that the population means for the two rows are not the same. 4
The critical value for this F can be found in table E, using the degrees of freedom and. dfw df b 4.6. 7.5 5.6 4.5.9 7.5 5.4 4.4.8 7.47 5. The critical value is 4.5. You can use the F-calculator to find that the p-value is 0.074. For our example, since our observed value of F (6.9) is greater than the critical value of F (4.5), we reject the null hypothesis. We can conclude using APA format: There is a significant main effect of peanut butter on taste ratings, F(,) = 6.9, p = 0.074 Main effect for columns: does jelly affect taste ratings? Conducting the F-test on the main effect for columns (jelly) is completely analogous to the main effect for rows. We first calculate the column means by calculating the mean for each of the two colummns. We can show this as a third row in the matrix of means: Means no jelly jelly no peanut butter 4. 6. peanut butter 6. 7.6667 Column Means 5. 7 The difference between these two column means indicates the effect of adding jelly after averaging across the two levels of peanut butter. As for rows, we ll calculate SS for columns, which is the sum of squared difference of the column means from the grand mean, scaled by the sample size for the number of scores for each mean: SS col = (n col )( X col X) = (8)(5. 6.) + (8)(7 6.) = 8.4445 The degrees of freedom for SS for columns is equal to the number of columns minus (df = - = ). The variance for the main effect for columns is SS for columns divided by its degrees of freedom: s col = SS col = df 8.4445 col = 8.4445 The F-statistic for the main effect for columns is the variance for columns divided by the variance within cells: F col = s col s wc = 8.4445.97 = 7.6 The F-calculator shows that the p-value for the main effect for columns (the df s are and ) is 0.07. For our example, since our observed value of F (7.6) is greater than the critical value of F (4.5), we reject the null hypothesis. We can conclude using APA format: There is a significant main effect of jelly on taste ratings, F(,) = 7.6, p = 0.07 The interaction: does the effect of peanut butter on taste ratings depend on jelly? This third F-test determines if there is a statistically significant interaction between our two factors on our dependent variable. For our example, an interaction would occur if the effect of adding peanut butter to the bread added a different amount to the taste rating depending upon whether or not there was jelly (and vice versa). Remember from the graph that the two lines 5
were roughly parallel, indicating that peanut butter had a similar increase in taste ratings with and without jelly. Parallel lines indicate no interaction between the two variables. To run this F-test, we need to calculate a third estimate of the population variance. The formula for the sums of squares for the interaction between rows and columns (SS RxC ) is messy and not very intuitive. Fortunately we don t need it because we can infer it by using the fact that the sums of squared deviation from the grand mean can be broken down into three components: SS total = SSrows + SS cols + SS RxC + SSwc I ve given you SS total and we just calculated SSrows, SS cols and SSwc. So SS RxC = SS total SSrows SS cols SSwc = 8.5556 5 8.4445 7. = The degrees of freedom for the row by column interaction is the number of rows minus times the number of columns minus : (-)(-) =. The variance for the row by column interaction is the SS divided by df: s RxC = SS RxC df RxC = = The F-statistic is: F RxC = s RxC s wc =.97 = 0.5 The F-calculator shows that the p-value for this value of F (the df s are and ) is 0.605. Since our p-value (0.605) is greater than alpha (0.05), we fail to reject the null hypothesis. We can conclude using APA format: There is not a significant interaction between the effects of peanut butter and jelly on taste ratings, F(,) = 0.5, p = 0.605 The Summary Table Like for -factor ANOVAs, the results from -factor ANOVAs are often reported in a summary table like this: SS df MS F F crit p-value Rows 5 5 6.9 4.5 0.074 Columns 8.4445 8.4445 7.6 4.5 0.07 R X C 0.5 4.5 0.605 wc 7..97 Total 8.5556 5 Example : peanut butter and ham Now suppose you want to study the effects of adding peanut butter and ham to bread. The (x) design is like the previous example; you take 6 students and break them down into 4 groups. Again, we ll run our test with α = 0.05. Here are the summary statistics from the experiment: Means no ham ham no peanut butter 4.8889 5.8889 peanut butter 5.6667 4.8889 6
SSwc no ham ham no peanut butter.8889.8889 peanut butter 4 4.8889 Totals grand mean 5. SS total Here s a plot the means with error bars representing the standard errors of the mean. 6. no peanut butter peanut butter 6 5.8 5.6 Taste rating 5.4 5. 5 4.8 4.6 no ham ham The first thing you see is that the lines are not parallel. Looking at the red points (no peanut butter), you can see that adding ham to bread improves its taste. But, if you start with peanut butter on your bread, adding ham makes it taste worse (yuck). This is a classic X shaped interaction, and we ll see below that it s statistically significant. For a main effect for rows we d need the overall rating to change when we add peanut butter. But adding peanut butter increases the rating for bread with no ham but decreases the rating with ham. Overall, the greeen points aren t higher or lower than the red points. It therefore looks like there is no main effect for peanut butter. For a main main effect for columns (ham) we d need the overall rating to change when we add ham. But you can see that the mean of the data point on the left (no ham are no higher or lower than the mean of the data points on the right (with jelly. So we don t expect a main effect for the column factor, ham. Now we ll work through the problem and fill in the summary table at the end: First, the within-cell variance: SSwc = (X X cell ) SSwc =.8889 + 4 +.8889 + 4.8889 = 4.6667 df = 9 + 9 + 9 + 9-4 = s wc = SS wc dfwc = 4.6667 = 0.458 7
For the main effects we can create a new table with both the row and column means: Means no ham ham Row Means no peanut butter 4.8889 5.8889 5.889 peanut butter 5.6667 4.8889 5.778 Column Means 5.778 5.889 Note that the two row means don t differ by much, nor do the two column means. This shows a lack of main effects for both rows and columns. We ll now compute the SS, variance and F-statistic for the main effect for rows. SS R = ()(9)(5.889 5.) + ()(9)(5.778 5.) = 0. With df R = = s rows = SS rows dfrows = 0. = 0. Frows = s rows s wc = 0. 0.458 = 0.4 The F-calculator shows that the p-value is 0.675. Since For our example, since our observed value of F (0.4) is not greater than the critical value of F (4.5), we fail to reject the null hypothesis. We can conclude using APA format: There is not a significant main effect of peanut butter on taste ratings, F(,) = 0.4, p = 0.675 Now for the main effect for columns: SS col = (n col )( X col X) = (8)(5.778 5.) + (8)(5.889 5.) = 0. df = - = s col = SS col = df 0. col = 0. F col = s col s wc = 0. 0.458 = 0.4 The F-calculator shows that the p-value for the main effect for columns (the df s are and ) is 0.675. Because our p-value (0.675) is greater than alpha (0.05), we conclude There is not a significant main effect of ham on taste ratings, F(,) = 0.4, p = 0.675 Finally, for the interaction: SS RxC = SS total SSrows SS cols SSwc = 0. 0. 4.6667 = 7. df = (-)(-) =. s RxC = SS RxC = df 7. RxC = 7. F RxC = s RxC s wc = 7. 0.458 = 5.5 The F-calculator shows that the p-value for this value of F (the df s are and ) is 0.0004. Since our p-value (0.0004) is less than alpha (0.05) We conclude: There is a significant interaction between the effects of 8
peanut butter and ham on test scores, F(,) = 5.5, p = 0.0004 The final summary table looks like this: SS df MS F F crit p-value Rows 0. 0. 0.4 4.5 0.675 Columns 0. 0. 0.4 4.5 0.675 R X C 7. 7. 5.5 4.5 0.0004 wc 4.6667 0.458 Total 5 Example : A x ANOVA on study habits This third example is an two-factor ANOVA on study habbits. You manipulate two factors, music and study location. for the factor music, you have students either study in silence, or listen to the music of their choice. For the factor study location, you have groups of students study in one of places: coffee shop, home and library. You find 5 students for each of the 4 groups. You generate the following summary statistics: Means coffee shop home library no music 7.68 75.64 65. music 6. 70.88 60.084 SSwc coffee shop home library no music 996.44 609.76 9688.64 music 8454.64 0880.64 865.0847 Totals grand mean 67.758 SS total 586.767 Here s a plot the means with error bars representing the standard errors of the mean. 9
80 no music music 75 70 Exam score 65 60 55 coffee shop home library study location You ll probably first notice that the red points are higher than the green points. This is a main effecct for the row factor, music, for which listening to music leads to lower scores. Next you ll notice that the scores vary across the column factor, study location, with the best scores for students studying at home. Since there are three locations of study, we re looking for a difference across means - just like a -factor ANOVA with three levels. Here, we see that there does look like there is a main effect for study location. Now we ll work through the problem and fill in the summary table at the end: The within-cell variance is calculated as in the examples above by adding up SSwc across the 6 cells: SSwc = (X X cell ) SSwc = 996.44 + 8454.64 + 609.76 + 0880.64 + 9688.64 + 865.0847 = 559.988 df = n total k = 50 6 = 44 s wc = SS wc dfwc = 559.988 44 = 70.555 For the main effects, again we ll create a new table with both the row and column means: Means coffee shop home library column means no music 7.68 75.64 65. 7.77 music 6. 70.88 60.084 64.95 row means 67.4 7.6 6.648 Next we ll compute the F-statistic for the main effect of rows (music) Note that since there are three columns, there are three means contributing to each row mean, the sample size for each row mean is (5)() = 75. SSrow = (nrow)( Xrow X) = (75)(7.77 67.758) + (75)(64.95 67.758) = 75.9 0
With df = - =. s rows = SS rows dfrows = 75.9 = 75.9 Frows = s rows s wc = 75.9 70.555 = 4.7 The F-calculator shows that the critical value is.9 and the p-value is 0.0. Our observed value of F (4.7) is greater than the critical value of F (.9), we reject the null hypothesis. We can conclude using APA format: There is a significant main effect of music on test scores, F(,44) = 4.7, p = 0.0 Now for the main effect for columns (study location): SS col = (n col )( X col X) = (50)(67.4 67.758) + (50)(7.6 67.758) + (50)(6.648 67.758) = 84.66 df = - = s col = SS col = df 84.66 col = 4.68 F col = s col s wc = 4.68 70.555 =.84 The F-calculator shows that the p-value for the main effect for columns (the df s are and 44) is 0.07. Because our p-value (0.07) is less than alpha (0.05), we conclude There is a significant main effect of study location on taste ratings, F(,44) = 4.7, p = 0.0 Finally, for the interaction: SS RxC = SS total SSrows SS cols SSwc = 586.767 75.9 84.66 559.988 = 60.9757 df = (-)(-) =. s RxC = SS RxC = df 60.9757 RxC = 0.4878 F RxC = s RxC s wc = 0.4878 70.555 = 0.5 The F-calculator shows that the p-value for this value of F (the df s are and 44) is 0.705. Since our p-value (0.705) is greater than alpha (0.05) We conclude: There is not a significant interaction between the effects of music and study location on taste ratings, F(,44) = 0.5, p = 0.705 The final summary table looks like this: SS df MS F F crit p-value Rows 75.9 75.9 4.7.9 0.0 Columns 84.66 4.68.84.06 0.07 R X C 60.9757 0.4878 0.5.06 0.705 wc 559.988 44 70.555 Total 586.767 49 Questions Your turn again. Here are 0 random practice questions followed by their answers.
) For a 499 project you measure the effect of kinds of musical groups and 4 kinds of grad students on the warmth of republicans. You find republicans for each group and measure warmth for each of the x4=8 groups. You generate the following table of means: Means grad students: grad students: grad students: grad students: 4 musical groups: 5.709 8.7877 0.6754 5.908 musical groups: 7.5.5 8.569 4.8408 and the following table for SSwc: SSwc grad students: grad students: grad students: grad students: 4 musical groups: 8.9805 869.68 69.455 7.895 musical groups: 405.066 494.8676 894.755 5.8769 The grand mean is 5.646 and SS total is 4687.0678. Make a plot of the data with error bars representing standard errors of the mean. Using an alpha value of α=0.05, test for a main effect of musical groups, a main effect of grad students and an interaction on the warmth of republicans. ) Your advsor asks you to measure the effect of kinds of psych 5 students and 4 kinds of facial expressions on the recognition of exams. You find 9 exams for each group and measure recognition for each of the x4=8 groups. You generate the following table of means: expres- facial sions: Means expres- facial sions: expres- facial sions: expres- facial sions: 4 psych 5 students: 9.6 8.884 5.55 6.745 psych 5 students: 6.5774.858.9679.96 and the following table for SSwc: expres- facial sions: SSwc expres- facial sions: expres- facial sions: expres- facial sions: 4 psych 5 students: 58.098 046.547 899.896 5.749 psych 5 students: 496.6466 886.57 408.0069 68.59 The grand mean is 0.79 and SS total is 559.8989. Make a plot of the data with error bars representing standard errors of the mean. Using an alpha value of α=0.05, test for a main effect of psych 5 students, a main effect of facial expressions and an interaction on the recognition of exams. ) I go and measure the effect of kinds of poo and 4 kinds of monkeys on the smell of potatoes. You find 0 potatoes for each group and measure smell for each of the x4=8 groups. You generate the following table of means: Means monkeys: monkeys: monkeys: monkeys: 4 poo: 87.4 84.56 88.4095 84.9765 poo: 8.5465 8.9855 87.45 85.5
and the following table for SSwc: SSwc monkeys: monkeys: monkeys: monkeys: 4 poo: 48.0966 464.487 506.5 488.08 poo: 45.40 498.5705 647.584 44.5555 The grand mean is 85.5447 and SS total is 46.448. Using an alpha value of α=0.05, test for a main effect of poo, a main effect of monkeys and an interaction on the smell of potatoes. 4) Your advsor asks you to measure the effect of kinds of teenagers and 4 kinds of examples on the farts of statistics problems. You find 8 statistics problems for each group and measure farts for each of the x4=8 groups. You generate the following table of means: Means examples: examples: examples: examples: 4 teenagers: 64.786 6.56 64.05 6.988 teenagers: 65.85 6.475 60.4 6.08 and the following table for SSwc: SSwc examples: examples: examples: examples: 4 teenagers: 04.08 8.6 65.405 9.984 teenagers: 55.648.8 5.66.8705 The grand mean is 6.97 and SS total is 980.8584. Using an alpha value of α=0.05, test for a main effect of teenagers, a main effect of examples and an interaction on the farts of statistics problems. 5) I go and measure the effect of kinds of fingers and kinds of photoreceptors on the happiness of personalities. You find personalities for each group and measure happiness for each of the x=6 groups. You generate the following table of means: Means photoreceptors: photoreceptors: photoreceptors: fingers: 64.75 70.7568 65.7986 fingers: 6.84 6.588 6.409 and the following table for SSwc: SSwc photoreceptors: photoreceptors: photoreceptors: fingers: 4.7444 85.4 40.49 fingers: 96.404 4.79 79.768 The grand mean is 64.96 and SS total is 64.985. Make a plot of the data with error bars representing standard errors of the mean. Using an alpha value of α=0.0, test for a main effect of fingers, a main effect of photoreceptors and an interaction on the happiness of personalities. 6) Let s measure the effect of kinds of sponges and kinds of elbows on the education of Americans. You find 9
Americans for each group and measure education for each of the x=6 groups. You generate the following table of means: Means elbows: elbows: elbows: sponges: 74.66 79.6484 7.9 sponges: 77.76 75.56 8.4047 and the following table for SSwc: SSwc elbows: elbows: elbows: sponges: 84.4889 5.65 4.45 sponges: 644.684 96.844 70.90 The grand mean is 76.896 and SS total is 547.096. Make a plot of the data with error bars representing standard errors of the mean. Using an alpha value of α=0.0, test for a main effect of sponges, a main effect of elbows and an interaction on the education of Americans. 7) You measure the effect of kinds of eggs and kinds of monkeys on the health of teams. You find teams for each group and measure health for each of the x=4 groups. You generate the following table of means: Means monkeys: monkeys: eggs: 57.49 56.45 eggs: 56.887 5.4 and the following table for SSwc: SSwc monkeys: monkeys: eggs: 45.975 9.899 eggs: 84.007 800.00 The grand mean is 55.9764 and SS total is 68.56. Using an alpha value of α=0.0, test for a main effect of eggs, a main effect of monkeys and an interaction on the health of teams. 8) You measure the effect of kinds of video games and kinds of otter pops on the taste of fathers. You find 7 fathers for each group and measure taste for each of the x=4 groups. You generate the following table of means: Means otter pops: otter pops: video games: 78.847 8.0406 video games: 80.094 80.959 and the following table for SSwc: 4
SSwc otter pops: otter pops: video games: 5.546.59 video games:.999 7.007 The grand mean is 80.476 and SS total is.89. Using an alpha value of α=0.05, test for a main effect of video games, a main effect of otter pops and an interaction on the taste of fathers. 9) For some reason you measure the effect of kinds of spaghetti and 4 kinds of dinosaurs on the warmth of grandmothers. You find 7 grandmothers for each group and measure warmth for each of the x4=8 groups. You generate the following table of means: Means dinosaurs: dinosaurs: dinosaurs: dinosaurs: 4 spaghetti: 5.69 56.84 64.4486 55.964 spaghetti: 5.7 59.957 5.6 57.5457 and the following table for SSwc: SSwc dinosaurs: dinosaurs: dinosaurs: dinosaurs: 4 spaghetti: 45.0559 049.89.969 85.87 spaghetti: 84.954 89.55 57.856 84.788 The grand mean is 56.684 and SS total is 566.954. Make a plot of the data with error bars representing standard errors of the mean. Using an alpha value of α=0.05, test for a main effect of spaghetti, a main effect of dinosaurs and an interaction on the warmth of grandmothers. 0) You want to measure the effect of kinds of brains and kinds of cartoon characters on the clothing of beers. You find 0 beers for each group and measure clothing for each of the x=6 groups. You generate the following table of means: charac- cartoon ters: Means charac- cartoon ters: charac- cartoon ters: brains: 5.4 4.6 4.669 brains:.507 7.875 4.40 and the following table for SSwc: charac- cartoon ters: SSwc charac- cartoon ters: charac- cartoon ters: brains:.09 9.486 77.0077 brains: 89.7686 7.856 07.566 The grand mean is 4.98 and SS total is 467.956. Using an alpha value of α=0.05, test for a main effect of brains, a main effect of cartoon characters and an interaction on the clothing of beers. 5
Answers 6
) Within cell: SSwc = 8.9805 + 405.066 + 869.68 + 494.8676 + 69.455 + 894.755 + 7.895 + 5.8769 = 404.654 dfwc = 04 ()(4) = 96 s wc = 404.65 96 = 4. Rows: musical groups mean row = 5.709+8.7877+0.6754+5.908 4 mean row = 7.5+.5+8.569+4.8408 4 = 5.7 = 6.0 SS R = (4)()(5.7 5.646) + (4)()(6.0 5.646) = 4. df R = = s R = 4. = 4. F R (, 96) = 4. 4. = 0.4 Columns: grad students mean col : 5.709+7.5 mean col : 8.7877+.5 mean col : 0.6754+8.569 mean col 4 : 5.908+4.8408 = 6.608 = 6.589 = 4.46 = 5.808 SS C = ()()(6.608 5.646) +()()(6.589 5.646) +()()(4.46 5.646) +()()(5.808 5.646) = 7.9 df C = 4 = s C = 7.9 = 4.067 F C (, 96) = 4.067 4. = 0.574 Interaction: SS RXC = 4687.07 (404.654 + 4. + 7.9) = 558. df RXC = ( )(4 ) = s RXC = 558.05 = 86.068 F RXC (, 96) = 86.068 4. = 4.478 7
SS df MS F F crit p-value Rows 4. 4. 0.4.94 0.56 Columns 7.9 4.067 0.57.7 0.66 R X C 558.05 86.068 4.4.7 0.0059 wc 404.654 96 4. Total 4687.0678 0 There is not a significant main effect for musical groups (rows) on the warmth of republicans, F(,96) = 0.40, p = 0.56. There is not a significant main effect for grad students (columns), F(,96) = 0.570, p = 0.66. There is a significant interaction between musical groups and grad students, F(,96) = 4.40, p = 0.006. 45 warmth of republicans 40 5 0 musical groups musical groups 4 grad students 8
) Within cell: SSwc = 58.098 + 496.6466 + 046.547 + 886.57 + 899.896 + 408.0069 + 5.749 + 68.59 = 089.4 dfwc = 5 ()(4) = 44 s wc = 089. 44 = 75.78 Rows: psych 5 students mean row = 9.6+8.884+5.55+6.745 4 mean row = 6.5774+.858+.9679+.96 4 = 7.569 =.09 SS R = (4)(9)(7.569 0.79) + (4)(9)(.09 0.79) = 5.59 df R = = s R = 5.59 = 5.59 F R (, 44) = 5.59 75.78 = 5. Columns: facial expressions mean col : 9.6+6.5774 mean col : 8.884+.858 mean col : 5.55+.9679 mean col 4 : 6.745+.96 =.95 = 0.5 = 9.066 = 8.9708 SS C = ()(9)(.95 0.79) + ()(9)(0.5 0.79) + ()(9)(9.066 0.79) + ()(9)(8.9708 0.79) = 9.4 df C = 4 = s C = 9.4 =.08 F C (, 44) =.08 75.78 =.744 Interaction: SS RXC = 559.9 (089.4 + 5.59 + 9.4) = 75.9 df RXC = ( )(4 ) = s RXC = 75.96 = 58.6405 F RXC (, 44) = 58.6405 75.78 = 0.779 9
SS df MS F F crit p-value Rows 5.59 5.59 5..9 0.000 Columns 9.4.08.74.67 0.65 R X C 75.96 58.6405 0.78.67 0.5069 wc 089.4 44 75.78 Total 559.8989 5 There is a significant main effect for psych 5 students (rows) on the recognition of exams, F(,44) = 5.00, p = 0.000. There is not a significant main effect for facial expressions (columns), F(,44) =.740, p = 0.6. There is not a significant interaction between psych 5 students and facial expressions, F(,44) = 0.780, p = 0.507. recognition of exams 0 5 0 psych 5 students psych 5 students 5 4 facial expressions 0
) Within cell: SSwc = 48.0966 + 45.40 + 464.487 + 498.5705 + 506.5 + 647.584 + 488.08 + 44.5555 = 776.878 dfwc = 60 ()(4) = 5 s wc = 776.8 5 = 4.8476 Rows: poo mean row = 87.4+84.56+88.4095+84.9765 4 mean row = 8.5465+8.9855+87.45+85.5 4 = 86.4 = 84.747 SS R = (4)(0)(86.4 85.5447) + (4)(0)(84.747 85.5447) = 0.777 df R = = s R = 0.777 = 0.777 F R (, 5) = 0.777 4.8476 = 4. Columns: monkeys mean col : 87.4+8.5465 mean col : 84.56+8.9855 mean col : 88.4095+87.45 mean col 4 : 84.9765+85.5 = 85.48 = 8.774 = 87.8755 = 85.0458 SS C = ()(0)(85.48 85.5447) +()(0)(8.774 85.5447) +()(0)(87.8755 85.5447) +()(0)(85.0458 85.5447) = 5.794 df C = 4 = s C = 5.798 = 7.5979 F C (, 5) = 7.5979 4.8476 = 4.78 Interaction: SS RXC = 46.4 (776.877 + 0.777 + 5.798) = 84.746 df RXC = ( )(4 ) = s RXC = 84.746 = 8.488 F RXC (, 5) = 8.488 4.8476 =.69
SS df MS F F crit p-value Rows 0.777 0.777 4..9 0.0446 Columns 5.798 7.5979 4.7.66 0.005 R X C 84.746 8.488.4.66 0.49 wc 776.877 5 4.8476 Total 46.448 59 There is a significant main effect for poo (rows) on the smell of potatoes, F(,5) = 4.00, p = 0.045. There is a significant main effect for monkeys (columns), F(,5) = 4.70, p = 0.004. There is not a significant interaction between poo and monkeys, F(,5) =.40, p = 0.5.
4) Within cell: SSwc = 04.08 + 55.648 + 8.6 +.8 + 65.405 + 5.66 + 9.984 +.8705 = 78.047 dfwc = 64 ()(4) = 56 s wc = 78.05 56 =.959 Rows: teenagers mean row = 64.786+6.56+64.05+6.988 4 mean row = 65.85+6.475+60.4+6.08 4 = 6.647 = 6.6775 SS R = (4)(8)(6.647 6.97) + (4)(8)(6.6775 6.97) = 5.566 df R = = s R = 5.566 = 5.566 F R (, 56) = 5.566.959 = 0.4 Columns: examples mean col : 64.786+65.85 mean col : 6.56+6.475 mean col : 64.05+60.4 mean col 4 : 6.988+6.08 = 65.88 = 6.469 = 6.86 = 6.06 SS C = ()(8)(65.88 6.97) + ()(8)(6.469 6.97) + ()(8)(6.86 6.97) + ()(8)(6.06 6.97) = 40.09 df C = 4 = s C = 40.09 = 46.7697 F C (, 56) = 46.7697.959 =.5 Interaction: SS RXC = 980.858 (78.046 + 5.566 + 40.09) = 5.78 df RXC = ( )(4 ) = s RXC = 5.78 = 7.909 F RXC (, 56) = 7.909.959 =.86
SS df MS F F crit p-value Rows 5.566 5.566 0.4 4.0 0.597 Columns 40.09 46.7697.5.77 0.05 R X C 5.78 7.909.8.77 0.90 wc 78.046 56.959 Total 980.8584 6 There is not a significant main effect for teenagers (rows) on the farts of statistics problems, F(,56) = 0.400, p = 0.50. There is a significant main effect for examples (columns), F(,56) =.50, p = 0.05. There is not a significant interaction between teenagers and examples, F(,56) =.80, p = 0.90. 4
5) Within cell: SSwc = 4.7444 + 96.404 + 85.4 + 4.79 + 40.49 + 79.768 = 9.697 dfwc = ()() = 6 s wc = 9.64 6 = 6.85 Rows: fingers mean row = 64.75+70.7568+65.7986 mean row = 6.84+6.588+6.409 = 66.9768 = 6.947 SS R = ()()(66.9768 64.96) + ()()(6.947 64.96) = 55.869 df R = = s R = 55.869 = 55.869 F R (, 6) = 55.869 6.85 =.85 Columns: photoreceptors mean col : 64.75+6.84 mean col : 70.7568+6.588 mean col : 65.7986+6.409 = 6.6087 = 67.75 = 64.048 SS C = ()()(6.6087 64.96) + ()()(67.75 64.96) + ()()(64.048 64.96) = 7.9 df C = = s C = 7.907 = 6.9604 F C (, 6) = 6.9604 6.85 = 9.7465 Interaction: SS RXC = 64.9 (9.696 + 55.869 + 7.907) = 8.499 df RXC = ( )( ) = s RXC = 8.499 = 90.7496 F RXC (, 6) = 90.7496 6.85 = 5.945 5
SS df MS F F crit p-value Rows 55.869 55.869.85 6.84 < 0.000 Columns 7.907 6.9604 9.75 4.78 0.000 R X C 8.499 90.7496 5.9 4.78 0.0057 wc 9.696 6 6.85 Total 64.985 There is a significant main effect for fingers (rows) on the happiness of personalities, F(,6) =.850, p = 0.000. There is a significant main effect for photoreceptors (columns), F(,6) = 9.750, p = 0.000. There is a significant interaction between fingers and photoreceptors, F(,6) = 5.90, p = 0.006. 75 happiness of personalities 70 65 fingers fingers 60 photoreceptors 6
6) Within cell: SSwc = 84.4889 + 644.684 + 5.65 + 96.844 + 4.45 + 70.90 = 068.989 dfwc = 4 ()() = 08 s wc = 069 08 = 94.57 Rows: sponges mean row = 74.66+79.6484+7.9 mean row = 77.76+75.56+8.4047 = 75.07 = 78.474 SS R = ()(9)(75.07 76.896) + ()(9)(78.474 76.896) = 84.9056 df R = = s R = 84.9056 = 84.9056 F R (, 08) = 84.9056 94.57 =.0 Columns: elbows mean col : 74.66+77.76 mean col : 79.6484+75.56 mean col : 7.9+8.4047 = 76.0608 = 77.4505 = 77.64 SS C = ()(9)(76.0608 76.896) + ()(9)(77.4505 76.896) + ()(9)(77.64 76.896) = 40.908 df C = = s C = 40.908 = 0.454 F C (, 08) = 0.454 94.57 = 0.7 Interaction: SS RXC = 547. (068.9889 + 84.9056 + 40.908) = 05.9 df RXC = ( )( ) = s RXC = 05.94 = 56.467 F RXC (, 08) = 56.467 94.57 = 5.588 7
SS df MS F F crit p-value Rows 84.9056 84.9056.0 6.88 0.0846 Columns 40.908 0.454 0. 4.8 0.809 R X C 05.94 56.467 5.59 4.8 0.0049 wc 068.9889 08 94.57 Total 547.096 There is not a significant main effect for sponges (rows) on the education of Americans, F(,08) =.00, p = 0.085. There is not a significant main effect for elbows (columns), F(,08) = 0.0, p = 0.80. There is a significant interaction between sponges and elbows, F(,08) = 5.590, p = 0.005. education of Americans 85 80 75 sponges sponges 70 elbows 8
7) Within cell: SSwc = 45.975 + 84.007 + 9.899 + 800.00 = 67.897 dfwc = 9 ()() = 88 s wc = 67.9 88 = 69.56 Rows: eggs mean row = 57.49+56.45 mean row = 56.887+5.4 = 56.84 = 55.5 SS R = ()()(56.84 55.9764) + ()()(55.5 55.9764) = 68.894 df R = = s R = 68.894 = 68.894 F R (, 88) = 68.894 69.56 = 0.99 Columns: monkeys mean col : 57.49+56.887 mean col : 56.45+5.4 = 57.0654 = 54.8874 SS C = ()()(57.0654 55.9764) + ()()(54.8874 55.9764) = 09.09 df C = = s C = 09.09 = 09.09 F C (, 88) = 09.09 69.56 =.5694 Interaction: SS RXC = 68.5 (67.897 + 68.894 + 09.09) = 4.6905 df RXC = ( )( ) = s RXC = 4.6905 = 4.6905 F RXC (, 88) = 4.6905 69.56 = 0.64 SS df MS F F crit p-value Rows 68.894 68.894 0.99 6.9 0.5 Columns 09.09 09.09.57 6.9 0.5 R X C 4.6905 4.6905 0.6 6.9 0.469 wc 67.897 88 69.56 Total 68.56 9 There is not a significant main effect for eggs (rows) on the health of teams, F(,88) = 0.990, p = 0.. 9
There is not a significant main effect for monkeys (columns), F(,88) =.570, p = 0.4. There is not a significant interaction between eggs and monkeys, F(,88) = 0.60, p = 0.47. 0
8) Within cell: SSwc = 5.546 +.999 +.59 + 7.007 = 8.7458 dfwc = 68 ()() = 64 s wc = 8.7458 64 = 0.99 Rows: video games mean row = 78.847+8.0406 mean row = 80.094+80.959 = 80.47 = 80.56 SS R = ()(7)(80.47 80.476) + ()(7)(80.56 80.476) = 0.088 df R = = s R = 0.088 = 0.088 F R (, 64) = 0.088 0.99 = 0.7 Columns: otter pops mean col : 78.847+80.094 mean col : 8.0406+80.959 = 79.467 = 8.478 SS C = ()(7)(79.467 80.476) + ()(7)(8.478 80.476) = 68.76 df C = = s C = 68.76 = 68.76 F C (, 64) = 68.76 0.99 = 4.76 Interaction: SS RXC =.89 (8.7458 + 0.088 + 68.76) = 4.674 df RXC = ( )( ) = s RXC = 4.674 = 4.674 F RXC (, 64) = 4.674 0.99 = 84.49 SS df MS F F crit p-value Rows 0.088 0.088 0.7.99 0.545 Columns 68.76 68.76 4.76.99 < 0.000 R X C 4.674 4.674 84..99 < 0.000 wc 8.7458 64 0.99 Total.89 67 There is not a significant main effect for video games (rows) on the taste of fathers, F(,64) = 0.70, p = 0.545.
There is a significant main effect for otter pops (columns), F(,64) = 4.760, p = 0.000. There is a significant interaction between video games and otter pops, F(,64) = 84.0, p = 0.000.
9) Within cell: SSwc = 45.0559 + 84.954 + 049.89 + 89.55 +.969 + 57.856 + 85.87 + 84.788 = 40.449 dfwc = 56 ()(4) = 48 s wc = 40.4 48 = 90.55 Rows: spaghetti mean row = 5.69+56.84+64.4486+55.964 4 mean row = 5.7+59.957+5.6+57.5457 4 = 57.468 = 55.775 SS R = (4)(7)(57.468 56.684) + (4)(7)(55.775 56.684) = 9.85 df R = = s R = 9.85 = 9.85 F R (, 48) = 9.85 90.55 = 0.44 Columns: dinosaurs mean col : 5.69+5.7 mean col : 56.84+59.957 mean col : 64.4486+5.6 mean col 4 : 55.964+57.5457 = 5.98 = 58.857 = 58.54 = 56.756 SS C = ()(7)(5.98 56.684) + ()(7)(58.857 56.684) + ()(7)(58.54 56.684) + ()(7)(56.756 56.684) = 7.50 df C = 4 = s C = 7.50 = 90.500 F C (, 48) = 90.500 90.55 =.00 Interaction: SS RXC = 566.95 (40.449 + 9.85 + 7.50) = 55.7 df RXC = ( )(4 ) = s RXC = 55.7 = 75.09 F RXC (, 48) = 75.09 90.55 =.9408
SS df MS F F crit p-value Rows 9.85 9.85 0.44 4.04 0.50 Columns 7.50 90.500.8 0.40 R X C 55.7 75.09.94.8 0.57 wc 40.449 48 90.55 Total 566.954 55 There is not a significant main effect for spaghetti (rows) on the warmth of grandmothers, F(,48) = 0.440, p = 0.50. There is not a significant main effect for dinosaurs (columns), F(,48) =.000, p = 0.40. There is not a significant interaction between spaghetti and dinosaurs, F(,48) =.940, p = 0.6. warmth of grandmothers 70 65 60 55 spaghetti spaghetti 50 4 dinosaurs 4
0) Within cell: SSwc =.09 + 89.7686 + 9.486 + 7.856 + 77.0077 + 07.566 = 5.6805 dfwc = 60 ()() = 54 s wc = 5.68 54 = 6.5496 Rows: brains mean row = 5.4+4.6+4.669 mean row =.507+7.875+4.40 = 4.70 = 5.6 SS R = ()(0)(4.70 4.98) + ()(0)(5.6 4.98) = 4.676 df R = = s R = 4.676 = 4.676 F R (, 54) = 4.676 6.5496 = 0.7 Columns: cartoon characters mean col : 5.4+.507 mean col : 4.6+7.875 mean col : 4.669+4.40 = 4.05 = 6.005 = 4.555 SS C = ()(0)(4.05 4.98) + ()(0)(6.005 4.98) + ()(0)(4.555 4.98) = 8.06 df C = = s C = 8.06 = 9.0 F C (, 54) = 9.0 6.5496 =.90 Interaction: SS RXC = 467.95 (5.6806 + 4.676 + 8.06) = 7.5696 df RXC = ( )( ) = s RXC = 7.5696 = 5.7848 F RXC (, 54) = 5.7848 6.5496 = 5.467 5
SS df MS F F crit p-value Rows 4.676 4.676 0.7 4.0 0.40 Columns 8.06 9.0.9.7 0.066 R X C 7.5696 5.7848 5.46.7 0.0069 wc 5.6806 54 6.5496 Total 467.956 59 There is not a significant main effect for brains (rows) on the clothing of beers, F(,54) = 0.70, p = 0.40. There is not a significant main effect for cartoon characters (columns), F(,54) =.900, p = 0.064. There is a significant interaction between brains and cartoon characters, F(,54) = 5.460, p = 0.007. 6