One-Way Within-Subjects ANOVA Homework 1 Within Subjects ANOVA Problem: adapted
ID: 3374832 • Letter: O
Question
One-Way Within-Subjects ANOVA Homework 1 Within Subjects ANOVA Problem: adapted from Problem #27 on p 429 in Privitera (please use the data in the table below, not those from your textbook) Some studies show that people who think they are intoxicated will show signs of intoxication, even if they did not consume alcohol To test whether this is true, researchers had a group of five adults consume nonalcoholic drinks, which they were told contained alcohol. The participants completed a standard driving test before drinking and then after one nonalcoholic drink and after five nonalcoholic drinks. A standard driving test was conducted in a school parking lot where the participants had to maneuver through traffic cones. The number of cones knocked over during each test was recorded. The following table lists the data for this hypothetical study Driving Test Participant Before Drinking After 1 Drink After 5 Drinks Participant 1 Participant 2 Participant 31 Participant 4 Participant 5 Assignment 1. Conduct the one-way within-subjects ANOVA by working through the steps of hypothesis testing 2. Construct the source table 3 Conduct the post hoc test for each pairwise comparison using the Bonferroni procedure 4 Calculate eta-squared and partial eta-squared for the effect of the number of drinks on driving errorsExplanation / Answer
Here I compute these answers of all question simulteneously in Minitab.
And the output of this is given below;
General Linear Model: Response versus Drink, Participant
Factor Type Levels Values
Drink fixed 3 1, 2, 3
Participant fixed 3 1, 2, 3
Analysis of Variance for Response, using Adjusted SS for Tests
Source DF Seq SS Adj SS Adj MS F P
Drink 2 18.533 18.317 9.158 9.08 0.006
Participant 2 0.717 0.717 0.358 0.36 0.709
Error 10 10.083 10.083 1.008
Total 14 29.333
S = 1.00416 R-Sq = 65.63% R-Sq(adj) = 51.88%
Grouping Information Using Bonferroni Method and 95.0% Confidence
Drink N Mean Grouping
3 5 3.2083 A
2 5 1.2500 B
1 5 0.5417 B
Means that do not share a letter are significantly different.
Bonferroni Simultaneous Tests
Response Variable Response
All Pairwise Comparisons among Levels of Drink
Drink = 1 subtracted from:
Difference SE of Adjusted
Drink of Means Difference T-Value P-Value
2 0.7083 0.6482 1.093 0.9003
3 2.6667 0.6482 4.114 0.0063
Drink = 2 subtracted from:
Difference SE of Adjusted
Drink of Means Difference T-Value P-Value
3 1.958 0.6482 3.021 0.0386
Grouping Information Using Tukey Method and 95.0% Confidence
Drink N Mean Grouping
3 5 3.2083 A
2 5 1.2500 B
1 5 0.5417 B
Means that do not share a letter are significantly different.
Tukey Simultaneous Tests
Response Variable Response
All Pairwise Comparisons among Levels of Drink
Drink = 1 subtracted from:
Difference SE of Adjusted
Drink of Means Difference T-Value P-Value
2 0.7083 0.6482 1.093 0.5396
3 2.6667 0.6482 4.114 0.0054
Drink = 2 subtracted from:
Difference SE of Adjusted
Drink of Means Difference T-Value P-Value
3 1.958 0.6482 3.021 0.0316
Grouping Information Using Bonferroni Method and 95.0% Confidence
Participant N Mean Grouping
2 5 1.9167 A
1 5 1.7083 A
3 5 1.3750 A
Means that do not share a letter are significantly different.
Bonferroni Simultaneous Tests
Response Variable Response
All Pairwise Comparisons among Levels of Participant
Participant = 1 subtracted from:
Difference SE of Adjusted
Participant of Means Difference T-Value P-Value
2 0.2083 0.6482 0.3214 1.000
3 -0.3333 0.6482 -0.5143 1.000
Participant = 2 subtracted from:
Difference SE of Adjusted
Participant of Means Difference T-Value P-Value
3 -0.5417 0.6482 -0.8357 1.000
Grouping Information Using Tukey Method and 95.0% Confidence
Participant N Mean Grouping
2 5 1.9167 A
1 5 1.7083 A
3 5 1.3750 A
Means that do not share a letter are significantly different.
Tukey Simultaneous Tests
Response Variable Response
All Pairwise Comparisons among Levels of Participant
Participant = 1 subtracted from:
Difference SE of Adjusted
Participant of Means Difference T-Value P-Value
2 0.2083 0.6482 0.3214 0.9450
3 -0.3333 0.6482 -0.5143 0.8662
Participant = 2 subtracted from:
Difference SE of Adjusted
Participant of Means Difference T-Value P-Value
3 -0.5417 0.6482 -0.8357 0.6905
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