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Two-Way ANOVA in SPSS, Study notes of Statistics for Psychologists

Reject the null hypothesis if p-value ≤ 0.05. Step 4: Construct the ANOVA Table. From the output, FAB = 0.0640 with 6 and 36 degrees of freedom.

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N u m b e r
of Coa ts 1 2 3 4
6
72.0 72.1 75.2 70.4
74.6 76.9 73.8 68.1
67.4 74.8 75.7 72.4
72.8 73.3 77.8 72.4
8
76.9 80.3 80.2 74.3
78.1 79.3 76.6 77.6
72.9 76.6 77.3 74.4
74.2 77.2 79.9 72.9
10
76.3 80.9 79.2 71.6
74.1 73.7 78.0 77.7
77.1 78.6 77.6 75.2
75.0 80.2 81.2 74.4
Batch
Two-Way ANOVA in SPSS STAT 314
Preliminary research on the production of
imitation pearls entailed studying the effect of the
number of coats of a special lacquer applied to an
opalescent plastic bead used as the base of the
pearl on the market value of the pearl. Four
batches of beads (12 beads per batch) were used in
the study, and it is desired to also consider the
batch effect on the market value. The three levels
of lacquer coatings (6, 8, and 10 coats) were fixed
in advance, as were the production formulations
for the four batches. The market value of each
pearl was determined by a panel of experts. The
data are shown in the accompanying table.
Perform the appropriate analysis of variance
procedure (including a profile plot of the means,
hypothesis tests, and multiple comparisons) for
this experiment. Use
α
= 0.05.
1. Each treatment is a combination of the Number of Coats value and the Batch value. Enter the
treatment numbers into one column with the corresponding number of coats, batch number, and
market values into separate variables (see figure, below). Be sure to enter your variables
appropriately.
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pf4
pf5
pf8
pf9
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N u m b e r of Coats 1 2 3 4 6 72.0 72.1 75.2 70. 74.6 76.9 73.8 68. 67.4 74.8 75.7 72. 72.8 73.3 77.8 72. 8 76.9 80.3 80.2 74. 78.1 79.3 76.6 77. 72.9 76.6 77.3 74. 74.2 77.2 79.9 72. 10 76.3 80.9 79.2 71. 74.1 73.7 78.0 77. 77.1 78.6 77.6 75. 75.0 80.2 81.2 74. B a t c h

Two-Way ANOVA in SPSS STAT 314

Preliminary research on the production of imitation pearls entailed studying the effect of the number of coats of a special lacquer applied to an opalescent plastic bead used as the base of the pearl on the market value of the pearl. Four batches of beads (12 beads per batch) were used in the study, and it is desired to also consider the batch effect on the market value. The three levels of lacquer coatings (6, 8, and 10 coats) were fixed in advance, as were the production formulations for the four batches. The market value of each pearl was determined by a panel of experts. The data are shown in the accompanying table. Perform the appropriate analysis of variance procedure (including a profile plot of the means, hypothesis tests, and multiple comparisons) for this experiment. Use α = 0.05.

  1. Each treatment is a combination of the Number of Coats value and the Batch value. Enter the treatment numbers into one column with the corresponding number of coats, batch number, and market values into separate variables (see figure, below). Be sure to enter your variables appropriately.
  1. Now it is time to check the normality assumption [this won’t work if you have m = 1 observation per treatment combination]. Select “Split File” from the “Data” menu so that we can tell SPSS that we want separate Q–Q Plots for each treatment group (see left figure, below). Select “Organize output by groups” and enter “treatment” as the variable that groups are based upon (see right figure, below). Now click “OK”. You can create separate Normal Q–Q Plots to assess the normality of each treatment group (see separate handout on Normal Q–Q Plots) , or you can create a single matrix of Normal Q–Q Plots using the following procedure (a single matrix plot takes up much less space than many separate Normal Q–Q Plots. Select “Rank Cases…” from the “Transform” menu (see left figure, below). Enter “Market Value” as the variable and rank the cases by “Treatment” (see middle figure, below). Click the “Rank Types…” button and be sure that “Normal scores” and “Blom” are selected in the resulting window, then click the “Continue” button (see right figure, below). Now click “OK”. Select “Aggregate…” from the “Data” menu (see left figure, below). Enter “Treatment” in the break variable box and “Market Value” in the summaries of variables box (MEAN) will be selected by default (see middle figure, below). Enter “Market Value” in the summaries of variables box again, then click the “Function…” button and select “Standard Deviation” before clicking the “Continue” button (see right figure, below). Now click “OK”.

Select “Scatter/Dot…” from the “Legacy Dialogs” submenu of the “Graphs” menu (see left figure, below). Select “Simple Scatter” then click the “Define” button (see middle figure, below). Enter “ExpNormal” in the Y-Axis box, “Market Value” in the X-Axis box, and panel your plots by “Number of Coats” as the rows and “Batch” as the columns (to match the way the data are presented in the problem description) (see right figure, below). Be sure to click the “Titles…” button and enter “Q-Q Plots by Treatment Groups” as Line 1 of your title. Once you click “OK”, you’ll be switched to the output window to see your plots so far (we aren’t done quite yet). Now, double-click the graph matrix to open it in editor mode. From the “Options” menu in editor mode, select “Reference Line from Equation” (see left figure, below). Lines should be added to each plot, and the “Properties” window should open. Change the custom equation so that the equation plotted becomes “y = x” (see middle figure, below). You may now exit out of editor mode as your plots now have the proper diagonal line displayed (see right figure, below). The resulting plots can now be interpreted individually in the same manner as Q–Q plots created by the usual method (Analyze  Descriptive Statistics  Q-Q Plots…) are interpreted.

  1. Select Analyze  General Linear Model  Univariate… (see figure, below).
  2. Select “Market Value” as the dependent variable, and select “Number of Coats” (factor_A) and “Batch Number” (factor_B) as the fixed factors (see figure, below). Click the “Model…” button. In the Univariate:Model window, select the “Custom” option and then the pull-down option in the center for “Main effects”. Select “factor_A” (Number of Coats) and “factor_B” (Batch) and move them to be in the Model. Next, after changing the pull-down option in the center to “Interaction”, select both “factor_A” and “factor_B” and then move them to be in the Model. [ Note: You will not include this “Interaction” component if you have m = 1 observation per treatment combination.] Also, be sure that “Type III” sum of squares and “Include intercept in model” are selected, and then click “Continue” (see left figure, below). Click the “Plots…” button. In the Profile Plots window, select one factor for the horizontal axis and one for the separate lines (the treatment means will be on the vertical axis). Be sure to click “Add” and then click “Continue” (see right figure, above).
  1. Since the lines of the profile plot (below) each follow the same basic pattern with no substantial crossings, we conclude that there are probably no significant interaction effects.
  1. Levene’s Test for Homogeneity of Variances and Normal Q-Q Plots. Step 0: Check Assumptions of Equal Variances (Homogeneity of Variances) and Normality The Levene Statistic p - value = 0.8909 is greater than α = 0.05 (from Step 2), so we fail to reject the null hypothesis that the variances are all equal. Since the points in each plot appear to closely follow the respective diagonal lines, then the populations are approximately normally distributed. Since the variances appear to be equal and the groups seem to be normally distributed (and we have random/independent samples), we may continue with ANOVA.
  2. You should use the output information in the following manner to answer the question. Despite thinking that there are no significant interactions (based on profile plot), we must still formally test for significant interaction between the two factors. Test for significant interaction effects: Step 1: Hypotheses H 0 : There is no interaction between number of coats and batches. Ha : There is an interaction between number of coats and batches. Step 2: Significance Level α = 0. Step 3: Critical Value(s) and Rejection Region(s) Reject the null hypothesis if p - value ≤ 0.05. Step 4: Construct the ANOVA Table From the output, FAB = 0.0640 with 6 and 36 degrees of freedom. p - value = Sig. = 0. Step 5: Conclusion Since p - value = 0.99882834 > 0.05 = α, we fail to reject the null hypothesis. Step 6: State conclusion in words At the α = 0.05 significance level, there is not enough evidence to conclude that number of coats and batches have a significant interaction effect on mean value of the imitation pearls.
  1. Since differences were found in the numbers of coats, we should perform a Tukey-Kramer (Tukey’s W) multiple comparison analysis to determine which of the numbers of coats is best. Using the previous output, here is how such an analysis might appear. Thus, we are 95% confident that 6 coats yields a different (smaller) mean value of the imitation pearls from that when using 8 or 10 coats (these two mean values are similar). This table corresponds to our “underline diagram”. Note that using 8 or 10 coats (since they are similar) will yield the largest mean value of the pearls. Since differences were found in the batches, we should perform a Tukey-Kramer (Tukey’s W) multiple comparison analysis to determine which of the batches is best. Using the previous output, here is how such an analysis might appear. Thus, we are 95% confident that batches 4 and 1 (similar mean values) yield a different (smaller) mean value of the imitation pearls from that when using batches 2 or 3 (similar mean values). This table corresponds to our “underline diagram”. Note that batch method 2 or 3 (since they are similar) will yield the largest mean value of the pearls.