Chapter 22 – Multi-level analysis – differences between more than two conditions (ANOVA)

Exercises

Here is the data file oneway ANOVA caffeine.sav used in the main one-way ANOVA calculations in the chapter.

Exercise 22.1

Calculating one-way unrelated ANOVA

You will need one of these data sets for this exercise

These are fictitious data supposedly collected from an experiment in which participants are given (with their permission) either Red Bull (a high caffeine drink), Diet Coke (moderate caffeine) or decaffeinated Coke (no caffeine, i.e., control group). They are then asked to complete a maze task where they have to trace round a maze to find the exit as quickly as possible.

Carry out a one-way ANOVA analysis on the data either in SPSS, by using a spreadsheet programme or even by hand and make a full report of results. Include the use of a Tukey^b post-hoc test if possible. You should also report effect sizes if you can. If you are calculating by hand you could conduct simple effect t tests between two samples at a time and adjust alpha accordingly.

Show answer

F (2, 34) = 6.661, p = .004 (or < .01), partial eta² = .282, a very large effect size.

Scores in the Red Bull group are significantly higher than scores in the caffeine-free group. This is shown by the Tukey^b test, which says that Red Bull and caffeine-free samples are in different subsets (non-homogenous) or by t tests. The simple effect test between Red Bull and caffeine-free gives t (22) = 3.76 or calculated by the Bonferroni method t (22) = 3.65. Either way this is highly significant (p < .01).

Exercise 22.2

Interpreting SPSS results for a one-way ANOVA

Shown below is the SPSS output after a one-way ANOVA has been performed on data where patients leaving hospital have been treated in three different ways, 1. traditionally (the control group), 2. with extra information (leaflet and video) given as they leave hospital and 3. with this information and a home visit from a health professional. The scores represent an assessment of their quality of recovery after three months. Have a go at answering the multiple-choice questions that appear below.

Test of homogeneity of variances
Score
Levene Statistic	df1	df2	Sig.
5.191	2	36	.010

ANOVA
Score
	Sum of Squares	df	Mean Square	F	Sig.
Between Groups	30.974	2	15.487	5.771	.007
Within Groups	96.615	36	2.684
Total	127.590	38

Score
Tukey B
Type of post-op care	N	Subset for alpha = 0.05
		1	2
Trad. care	13	5.3846
Trad. care + inform.	13	6.7692	6.7692
Trad. care + inform. + visit	13		7.5385
Means for groups in homogeneous subsets are displayed.

Exercise 22.3

The features of post hoc tests

Try the ‘matching’ quiz – match the test with the appropriate description.

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Weblinks

Multi-level analysis – difference between more than two conditions

For online definitions and explanations about ANOVA try:        

HyperStat Online: Introduction to ANOVA (davidmlane.com)

Laerd Statistics guide to one-way ANOVA:

One-way ANOVA in SPSS Statistics – Step-by-step procedure including testing of assumptions. (laerd.com)

For very many helpful pages of information and debates around ANOVA see:

Analysis of Variance and Covariance – The Analysis Factor

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Further Information

The Jonckheere trend test

Chapter 22 of the book describes the Jonckheere trend test and directs the reader here for the means of calculation. Below is a table of fictitious (and very minimal) data upon which we will conduct the test. This will tell us whether there is a significant trend for scores to increase across the three conditions from left to right. Assume that participants have been given information about a fictitious person including one criterion piece concerning the person’s attitude to global warming. In condition 1 participants are told that the person doesn’t care about global warming. In condition 2 no information about the person’s attitude is given to pts. In condition 3 they are told that the person cares a lot about global warming. The scores in the columns are the participant’s rating of how likely they are to like the person.

	Information given about person’s attitude to global warming.
	1. Doesn’t     care	Values to right exceeding	2. No information	Values to right exceeding	3. Cares
Participant
A	3	7	2	4	10
B	5	7	7	3	8
C	6	7	9	2	7
D	3	7	8	2	11
Totals:		28		11

Procedure	Calculation
1. For each score count the number of scores that exceed it to the right. Start at the left hand column.	See the table above; Example: The score of 5 for participant B in the ‘Doesn’t care’ column is exceeded by 7, 9 and 8 in the next column and 10, 8, 7 and 11 in the right hand column, making 7 scores in all.
2. Add the two count columns	See the table above (‘totals’)
3. Add the two totals and call this value X	X = 28 + 11 = 39

OK that was the easy part. Now things are rather tricky when we want to check if our value of Xis significant. There are tables for this test but they only go up to n = 10 in each condition and you have to have the same number in each condition – a rare circumstance.  We need to enter our value of X then into the following equation (take a deep breath):

$$z=\frac{2X-\sum \left(n_i{n}_j\right)-1}{\sqrt{1/18[N^2(2N+3)-3\sum \left(n^2\right)-2\sum \left(n^3\right)]}}$$

n_in_j   means multiply all possible combinations of sample size. If we had sample sizes of 4, 6 and 7 this would mean we found (4×6) + (4×7) + (6×7). In our case here though this is just 4×4 + 4×4 + 4×4 = 48. 

(n²) would be 4² + 6² + 7² but in our case is 3 x 4² = 48

(n³) would be 4³ + 6³ + 7³ but for us it is 3 x 4³  = 132

N is the total sample size i.e. 3 x 4 = 12 Our z value then is

$$z=\frac{78-48-1}{\sqrt{1/18[{12}^2(24+3)-3\times 48-2\times 132]}}$$

= 29/√[1/18 x (3888 – 144 – 264)] = 29/√[3480/18] = 29/√193.33 = 29/13.9 = 2.09 A z value of 2.09 cuts off .0183 of the area of the normal distribution at either end (check in Appendix table 2 of the book) and this means that our overall p is 2 x .0183 for a two-tailed test = .0366 so we have a significant trend!