Perform a two-factor ANOVA

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Newcastle University Mathematics and Statistics

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Newcastle University Mathematics and Statistics 9 years, 2 months ago

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Question statement

In a similar fashion to the calculation done in lectures, carry out a two factor ANOVA on the following set of data.

Individuals who are identified as having an antisocial personality disorder may also have reduced physiological responses to anxiety-producing stimuli.

One way of measuring this response is with ”galvanic skin response” (GSR), a measurable reduction in the electrical resistance on the skin. [This is the basis of how a lie detector works.]

The following data represent the results of an experiment to compare the responses of normal and antisocial individuals in regular (baseline) and stress-provoking situations (low score reflects a more anxious individual):

Normal	Baseline	$\var{r1[0]}$	$\var{r1[1]}$	$\var{r1[2]}$	$\var{r1[3]}$	$\var{r1[4]}$	$\var{r1[5]}$	$\var{r1[6]}$	$\var{r1[7]}$	$\var{r1[8]}$	$\var{r1[9]}$
Normal	Stress	$\var{r2[0]}$	$\var{r2[1]}$	$\var{r2[2]}$	$\var{r2[3]}$	$\var{r2[4]}$	$\var{r2[5]}$	$\var{r2[6]}$	$\var{r2[7]}$	$\var{r2[8]}$	$\var{r2[9]}$
Antisocial	Baseline	$\var{r3[0]}$	$\var{r3[1]}$	$\var{r3[2]}$	$\var{r3[3]}$	$\var{r3[4]}$	$\var{r3[5]}$	$\var{r3[6]}$	$\var{r3[7]}$	$\var{r3[8]}$	$\var{r3[9]}$
Antisocial	Stress	$\var{r4[0]}$	$\var{r4[1]}$	$\var{r4[2]}$	$\var{r4[3]}$	$\var{r4[4]}$	$\var{r4[5]}$	$\var{r4[6]}$	$\var{r4[7]}$	$\var{r4[8]}$	$\var{r4[9]}$

The null hypotheses we are testing are:

1. The GSR does not depend upon the personality type.

2. The GSR does not depend upon the stimulus.

3. There is no interaction between personality and stimulus in determining the GSR.

Give any introductory information the student needs.

			Name	Type	Generated Value

ivr	number	9.72
f1vr	number	15.94
f2vr	number	3.26
w2	list	[ 0, 1, 0, 0, 0 ]
w1	list	[ 0, 0, 0, 1, 0 ]
w0	list	[ 1, 0, 0, 0, 0 ]
f1l2	list	List of 20 items
f1l1	list	List of 20 items
n	integer	40
mu3	number	22.5
f1ssq	list	[ 6537, 9490 ]
f1btss	number	133.23
m4	number	22.10
m1	number	20.10
m3	number	20.90
m2	number	15.60
btss	number	241.68
f2btss	number	27.23
rss	number	301.10
tss	number	542.78
f1t	list	[ 357, 430 ]
r4	list	List of 10 items
tol	number	0.01
f2ssq	list	[ 8580, 7447 ]
ssq	list	[ 4079, 2458, 4501, 4989 ]
vvr	list	[ 15.94, 3.26, 9.72 ]
sig1	number	3
sig2	number	2.2
f1ss	number	16027
w	list	Nested 3×5 list
n2	integer	10
interactionss	number	81.22
tsqovern	list	List of 4 items
mu4	number	21.4
mu1	number	20
g	number	787
r2	list	List of 10 items
mu2	number	16.2
ss	number	16027
f1tsqovern	list	[ 6372.45, 9245 ]
r3	list	List of 10 items
f2ss	number	16027
mrs	number	8.36
f1rss	number	409.55
f2l1	list	List of 20 items
f2l2	list	List of 20 items
r1	list	List of 10 items
stovern	number	15725.9
t	list	[ 201, 156, 209, 221 ]
f1stovern	number	15617.45
f2t	list	[ 410, 377 ]
n1	integer	10
f2stovern	number	15511.45
n3	integer	10
n4	integer	10

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
precround(interactionss/mrs,2)

JME function reference

Description

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `number`

9.72

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Ask the student a question, and give any hints about how they should answer this part.

Now complete the following two factor ANOVA table from this data. Input the $SS,\;MS$ and $VR$ data to 2 decimal places.

Source	df	SS	MS	VR
Personality	Gap 0.	Gap 1.	Gap 2.	Gap 3.
Stimulus	Gap 4.	Gap 5.	Gap 6.	Gap 7.
P-S interaction	Gap 8.	Gap 9.	Gap 10.	Gap 11.
Residual	Gap 12.	Gap 13.	Gap 14.	-
Total	Gap 15.	Gap 16.	-	-

The Calculations.

1. Residual Calculations. When you are calculating $TSS$ and $BTSS$ round them both to 2 decimal places, then calculate $RSS$ by taking away these rounded values. The $RMS$ value is then obtained by dividing this $RSS$ value by the residual degrees of freedom.

2. For Personality, Stimulus and Interaction, calculate the estimations of their variances to 2 decimal places as well. These values go in the $MS$ column.

3. The $VR$ values are obtained by dividing the first three values in the $MS$ column by the $RMS$ value. Enter the $VR$ values to 2 decimal places in the last column.

Also input the mean values of the factors at their various levels, input your answers to 2 decimal places:

	$\overline{x}_i$
Normal, Baseline	Gap 17.
Normal, Stress	Gap 18.
Antisocial, Baseline	Gap 19.
Antisocial, Stress	Gap 20.

Advice

We have two factors: Personality and Stimulus.

Personality has two levels: Normal and Antisocial

Stimulus has two levels: Baseline and Stress.

First Step:

First regard the different treatment combinations as a set of independent samples and analyse as for a one-way analysis with unrelated measurements. From this analysis, we obtain the Total Sum of Squares, the Between Treatments Sum of Squares ($BTSS$) and Residual Sum of Squares ($RSS$). Note that the degrees of freedom for this step are 36=40-4 as there are 4 treatments.

You should obtain:

	$\overline{x}_i$	$T_i$	$\sum x^2$	$n_i$
Normal, Baseline	$\var{m1}$	$\var{t[0]}$	$\var{ssq[0]}$	10
Normal, Stress	$\var{m2}$	$\var{t[1]}$	$\var{ssq[1]}$	10
Antisocial, Baseline	$\var{m3}$	$\var{t[2]}$	$\var{ssq[2]}$	10
Antisocial, Stress	$\var{m4}$	$\var{t[3]}$	$\var{ssq[3]}$	10
		$G=\;\var{g}$	Sum of Squares=$\var{ss}$	$N=40$

From this we obtain:

\[ BTSS =\sum \frac{T_i^2}{10}- \frac{G^2}{40}=\frac{\var{t[0]}^2}{10}+\frac{\var{t[1]}^2}{10}+\frac{\var{t[2]}^2}{10}+\frac{\var{t[3]}^2}{10}-\frac{\var{g}^2}{40}=\var{btss}\]

\[TSS = \sum \sum x^2- \frac{G^2}{40}=\var{ss}- \frac{\var{g}^2}{40}=\var{tss}\] both to 2 decimal places.

\[RSS = TSS-BTSS=\var{tss}-\var{btss}=\var{rss}\]

Second Step.

Now ignore the Stimulus factor and calculate totals $T_i$ for each level of Personality. From these totals calculate a variance estimate for personality using the same method as before. The degrees of freedom will be one fewer than the levels of Stimulus and is therefore 1.

Normal (Baseline and Stress)	$\var{r1[0]}$	$\var{r1[1]}$	$\var{r1[2]}$	$\var{r1[3]}$	$\var{r1[4]}$	$\var{r1[5]}$	$\var{r1[6]}$	$\var{r1[7]}$	$\var{r1[8]}$	$\var{r1[9]}$	$\var{r2[0]}$	$\var{r2[1]}$	$\var{r2[2]}$	$\var{r2[3]}$	$\var{r2[4]}$	$\var{r2[5]}$	$\var{r2[6]}$	$\var{r2[7]}$	$\var{r2[8]}$	$\var{r2[9]}$
Antisocial (Baseline and Stress)	$\var{r3[0]}$	$\var{r3[1]}$	$\var{r3[2]}$	$\var{r3[3]}$	$\var{r3[4]}$	$\var{r3[5]}$	$\var{r3[6]}$	$\var{r3[7]}$	$\var{r3[8]}$	$\var{r3[9]}$	$\var{r3[0]}$	$\var{r4[1]}$	$\var{r4[2]}$	$\var{r4[3]}$	$\var{r4[4]}$	$\var{r4[5]}$	$\var{r4[6]}$	$\var{r4[7]}$	$\var{r4[8]}$	$\var{r4[9]}$

You should have the following data from this table:

	$T_i$	$\sum x^2$	$n_i$ (number of observations)
Normal, (Baseline and Stress)	$\var{f1t[0]}$	$\var{f1ssq[0]}$	20
Antisocial, (Baseline and Stress)	$\var{f1t[1]}$	$\var{f1ssq[1]}$	20
	$G=\;\var{g}$	Sum of Squares=$\var{f1ss}$	$N=40$

So we can calculate:

\[\mbox{Variance estimate for personality} =\sum \frac{T_i^2}{20}- \frac{G^2}{40}=\frac{\var{f1t[0]}^2}{20}+\frac{\var{f1t[1]}^2}{20}-\frac{\var{g}^2}{40}=\var{f1btss}\]

Third Step.

Repeat step 2 with the factors switched, i.e. use the totals $T_i$ for the Stimulus factor levels ignoring Personality.This gives a Between Treatments Sum of Squares. Again the degrees of freedom will be 2-1=1.

Baseline (Normal and Antisocial)	$\var{r1[0]}$	$\var{r1[1]}$	$\var{r1[2]}$	$\var{r1[3]}$	$\var{r1[4]}$	$\var{r1[5]}$	$\var{r1[6]}$	$\var{r1[7]}$	$\var{r1[8]}$	$\var{r1[9]}$	$\var{r2[0]}$	$\var{r3[1]}$	$\var{r3[2]}$	$\var{r3[3]}$	$\var{r3[4]}$	$\var{r3[5]}$	$\var{r3[6]}$	$\var{r3[7]}$	$\var{r3[8]}$	$\var{r3[9]}$
Stress (Normal and Antisocial)	$\var{r2[0]}$	$\var{r2[1]}$	$\var{r2[2]}$	$\var{r2[3]}$	$\var{r2[4]}$	$\var{r2[5]}$	$\var{r2[6]}$	$\var{r2[7]}$	$\var{r2[8]}$	$\var{r2[9]}$	$\var{r4[0]}$	$\var{r4[1]}$	$\var{r4[2]}$	$\var{r4[3]}$	$\var{r4[4]}$	$\var{r4[5]}$	$\var{r4[6]}$	$\var{r4[7]}$	$\var{r4[8]}$	$\var{r4[9]}$

You should have the following data from this table:

	$T_i$	$\sum x^2$	$n_i$ (number of observations)
Baseline (Normal and Antisocial)	$\var{f2t[0]}$	$\var{f2ssq[0]}$	20
Stress (Normal and Antisocial)	$\var{f2t[1]}$	$\var{f2ssq[1]}$	20
	$G=\;\var{g}$	Sum of Squares=$\var{f2ss}$	$N=40$

So we can calculate:

\[ \mbox{Variance estimate for stimulus} =\sum \frac{T_i^2}{20}- \frac{G^2}{40}=\frac{\var{f2t[0]}^2}{20}+\frac{\var{f2t[1]}^2}{20}-\frac{\var{g}^2}{40}=\var{f2btss}\]

Step 4.

Now determine a Sum of Squares for Interaction by subtracting the sums of squares obtained for Personality (Step 2) and Stimulus (step 3) from the overall Between Treatments Sum of squares obtained in Step 1. The degrees of freedom is also obtained by subtraction and is 1.

This gives:

\[\mbox{Variance estimate for the interaction}= \var{btss}-\var{f1btss}-\var{f2btss} = \var{interactionss}\]

The Anova Table.

We now have all the terms required to construct the ANOVA table and hence test the null hypothesis relating to each factor and to the interaction. Note that the $VR$ values are obtained by dividing the $RMS$ value into the $MS$ values for the factors and the interaction.

Source	df	SS	MS	VR	Decision
Personality	1	$\var{f1btss}$	$\var{f1btss}$	$\var{f1vr}$	{dec(f1vr)} that GSR is independent of personality.
Stimulus	1	$\var{f2btss}$	$\var{f2btss}$	$\var{f2vr}$	{dec(f2vr)} that GSR is independent of the stimulus.
P-S interaction	1	$\var{interactionss}$	$\var{interactionss}$	$\var{ivr}$	{dec(ivr)} that personality and stimulus are independent in terms of GSR.
Residual	36	$\var{rss}$	$\var{mrs}$	-
Total	39	$\var{precround(f1btss+f2btss+interactionss+rss,2)}$	-	-

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Perform a two-factor ANOVA

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Newcastle University Mathematics and Statistics 9 years, 2 months ago

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