Ricardo's copy of Two factor ANOVA

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Christian Lawson-Perfect 8 years, 10 months ago

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Ricardo Monge 9 years, 7 months ago

Created this as a copy of Two factor ANOVA.

Name	Status	Author	Last Modified
Two factor ANOVA	Ready to use	Christian Lawson-Perfect	01/06/2016 11:08
José's copy of Two factor ANOVA	Ready to use	Jos Klenner	02/12/2020 15:42
Julie's copy of Two factor ANOVA	Ready to use	Julie Crowley	26/07/2016 12:53
Ricardo's copy of Two factor ANOVA	Ready to use	Ricardo Monge	01/06/2016 09:47
James's copy of Two factor ANOVA	Ready to use	James Eustace	01/06/2016 09:47
YJ's copy of Two factor ANOVA	Ready to use	YJ Kim	01/06/2016 09:47
Andrew's copy of Two factor ANOVA	Ready to use	Andrew Dunbar	01/06/2016 09:47
Christoph's copy of Two factor ANOVA	Ready to use	Christoph Tiedtke	01/06/2016 09:47
Two factor ANOVA - SES	Has some problems	Chris Graham	10/12/2020 00:31
Two factor ANOVA - Mock Test - SES2003	Needs to be tested	Chris Graham	25/02/2019 11:02
Two factor ANOVA	draft	Xiaodan Leng	11/07/2019 05:48
Two factor ANOVA	draft	Xiaodan Leng	11/07/2019 06:27

There are 4 other versions that do you not have access to.

Question statement

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).

{capitalise(factor1_name)}	{capitalise(factor2_name)}
{capitalise(factor1_levels[0])}	{capitalise(factor2_levels[0])}	$\var{r1[0]}$	$\var{r1[1]}$	$\var{r1[2]}$	$\var{r1[3]}$	$\var{r1[4]}$	$\var{r1[5]}$	$\var{r1[6]}$	$\var{r1[7]}$	$\var{r1[8]}$	$\var{r1[9]}$
{capitalise(factor1_levels[0])}	{capitalise(factor2_levels[1])}	$\var{r2[0]}$	$\var{r2[1]}$	$\var{r2[2]}$	$\var{r2[3]}$	$\var{r2[4]}$	$\var{r2[5]}$	$\var{r2[6]}$	$\var{r2[7]}$	$\var{r2[8]}$	$\var{r2[9]}$
{capitalise(factor1_levels[1])}	{capitalise(factor2_levels[0])}	$\var{r3[0]}$	$\var{r3[1]}$	$\var{r3[2]}$	$\var{r3[3]}$	$\var{r3[4]}$	$\var{r3[5]}$	$\var{r3[6]}$	$\var{r3[7]}$	$\var{r3[8]}$	$\var{r3[9]}$
{capitalise(factor1_levels[1])}	{capitalise(factor2_levels[1])}	$\var{r4[0]}$	$\var{r4[1]}$	$\var{r4[2]}$	$\var{r4[3]}$	$\var{r4[4]}$	$\var{r4[5]}$	$\var{r4[6]}$	$\var{r4[7]}$	$\var{r4[8]}$	$\var{r4[9]}$

Carry out a two factor ANOVA on the data to test the following null hypotheses:

{capitalise(measurement_name)} does not depend upon {factor1_name}.
{capitalise(measurement_name)} does not depend upon {factor2_name}.
There is no interaction between {factor1_name} and {factor2_name} in determining {measurement_name}.

Give any introductory information the student needs.

			Name	Type	Generated Value

measurement_name	string	GSR
factor1_name	string	personality
factor1_levels	list	[ "normal", "antisocial" ]
factor2_name	string	stimulus
factor2_levels	list	[ "baseline", "stress" ]
mu1	number	24
mu2_diff	number	3.4
mu3	number	23.5
mu4_diff	number	1.1
sig1	number	2
sig2	number	1.8
num_samples	integer	10

			Name	Type	Generated Value

r1	list	List of 10 items
r2	list	List of 10 items
r3	list	List of 10 items
r4	list	List of 10 items

			Name	Type	Generated Value

n	integer	40
f1btss	number	0.03
f1vr	number	0.00
f2btss	number	133.23
f2vr	number	0.22
ivr	number	-35.73
rss	number	21502.70
mrs	number	597.30
mean1	number	25.70
mean2	number	20.40
mean3	number	24.10
mean4	number	22.10

			Name	Type	Generated Value

w1	list	[ 0, 0, 0, 0, 1 ]
w2	list	[ 0, 0, 0, 0, 1 ]
w0	list	[ 0, 0, 0, 0, 1 ]
f1l2	list	List of 20 items
f1l1	list	List of 20 items
f1ssq	list	[ 10821, 10774 ]
tol	number	0.001
btss	number	-21205.92
tss	number	296.78
f1t	list	[ 461, 462 ]
w	list	Nested 3×5 list
f2ssq	list	[ 12476, 9119 ]
f2t	list	[ 498, 425 ]
ssq	list	[ 6619, 4202, 5857, 4917 ]
vvr	list	[ 0.00, 0.22, -35.73 ]
f1ss	number	21595
interactionss	number	-21339.18
tsqovern	list	[ 25.7, 20.4, 24.1, 22.1 ]
mu4	number	22.4
g	number	923
mu2	number	20.6
ss	number	21595
f1tsqovern	list	[ 10626.05, 10672.2 ]
f1rss	number	296.75
f2l1	list	List of 20 items
f2l2	list	List of 20 items
stovern	number	92.3
t	list	[ 257, 204, 241, 221 ]
f1stovern	number	21298.25
f2ss	number	21595
f2stovern	number	21431.45

Automatically regenerate variables?

Name

Data type

Value

(text string)

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Description

The name of the value being measured.

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

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Generated value: `string`

GSR

→ Used by:

Statement
Advice
"Part b)" → "Unnamed gap" - Choice text
"Part b)" → "Unnamed gap" - Choice text

Parts

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Part b) Gap-fill
1. Gap Match choices with answers
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Gap-fill

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 $\mathit{SS}$, $\mathit{MS}$ and $\mathit{VR}$ data to 2 decimal places.

Source	df	SS	MS	VR
{capitalise(factor1_name)}	Gap 0.	Gap 1.	Gap 2.	Gap 3.
{capitalise(factor2_name)}	Gap 4.	Gap 5.	Gap 6.	Gap 7.
Interaction	Gap 8.	Gap 9.	Gap 10.	Gap 11.
Residual	Gap 12.	Gap 13.	Gap 14.	-
Total	Gap 15.	Gap 16.	-	-

The Calculations

Residual Calculations. When you are calculating $\mathit{TSS}$ and $\mathit{BTSS}$, round them both to 2 decimal places, then calculate $\mathit{RSS}$ by taking away these rounded values. The $\mathit{RMS}$ value is then obtained by dividing this $\mathit{RSS}$ value by the residual degrees of freedom.
For {factor1_name}, {factor2_name} and Interaction, calculate the estimations of their variances to 2 decimal places as well. These values go in the $\mathit{MS}$ column.
The $\mathit{VR}$ values are obtained by dividing the first three values in the $\mathit{MS}$ column by the $\mathit{RMS}$ value. Enter the $\mathit{VR}$ values to 2 decimal places in the last column.

Also input the mean values of the factors at their various levels:

	$\overline{x_i}$
{capitalise(factor1_levels[0])}, {factor2_levels[0]}	Gap 17.
{capitalise(factor1_levels[0])}, {factor2_levels[1]}	Gap 18.
{capitalise(factor1_levels[1])}, {factor2_levels[0]}	Gap 19.
{capitalise(factor1_levels[1])}, {factor2_levels[1]}	Gap 20.

Advice

We have two factors: {factor1_name} and {factor2_name}.

{capitalise(factor1_name)} has two levels: {factor1_levels[0]} and {factor1_levels[1]}.

{capitalise(factor2_name)} has two levels: {factor2_levels[0]} and {factor2_levels[1]}.

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 ($\mathit{BTSS}$) and Residual Sum of Squares ($\mathit{RSS}$). Note that the degrees of freedom for this step are $\var{n-4} = \var{n}-4$ as there are $4$ treatments.

You should obtain

	$\overline{x_i}$	$T_i$	$\sum x^2$	$n_i$
{capitalise(factor1_levels[0])}, {factor2_levels[0]}	$\var{mean1}$	$\var{t[0]}$	$\var{ssq[0]}$	$\var{num_samples}$
{capitalise(factor1_levels[0])}, {factor2_levels[1]}	$\var{mean2}$	$\var{t[1]}$	$\var{ssq[1]}$	$\var{num_samples}$
{capitalise(factor1_levels[1])}, {factor2_levels[0]}	$\var{mean3}$	$\var{t[2]}$	$\var{ssq[2]}$	$\var{num_samples}$
{capitalise(factor1_levels[1])}, {factor2_levels[1]}	$\var{mean4}$	$\var{t[3]}$	$\var{ssq[3]}$	$\var{num_samples}$
		$G = \var{g}$	Sum of Squares = $\var{ss}$	$N=\var{n}$

From this we obtain:

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

\[\mathit{TSS} = \sum \sum x^2- \frac{G^2}{\var{n}}=\var{ss}- \frac{\var{g}^2}{\var{n}}=\var{tss} \]

both to 2 decimal places.

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

Second Step

Now ignore the {factor2_name} factor and calculate totals $T_i$ for each level of {factor1_name}. From these totals calculate a variance estimate for {factor1_name} using the same method as before. The degrees of freedom will be one fewer than the number of levels of {factor2_name} and is therefore $1$.

{capitalise(factor1_levels[0])} ({factor2_levels[0]} and {factor2_levels[1]})	$\var{r1[0]}$	$\var{r1[1]}$	$\var{r1[2]}$	$\var{r1[3]}$	$\var{r1[4]}$	$\var{r1[5]}$	$\var{r1[6]}$	$\var{r1[7]}$	$\var{r1[8]}$	$\var{r1[9]}$	$\var{r2[0]}$	$\var{r2[1]}$	$\var{r2[2]}$	$\var{r2[3]}$	$\var{r2[4]}$	$\var{r2[5]}$	$\var{r2[6]}$	$\var{r2[7]}$	$\var{r2[8]}$	$\var{r2[9]}$
{capitalise(factor1_levels[1])} ({factor2_levels[0]} and {factor2_levels[1]})	$\var{r3[0]}$	$\var{r3[1]}$	$\var{r3[2]}$	$\var{r3[3]}$	$\var{r3[4]}$	$\var{r3[5]}$	$\var{r3[6]}$	$\var{r3[7]}$	$\var{r3[8]}$	$\var{r3[9]}$	$\var{r3[0]}$	$\var{r4[1]}$	$\var{r4[2]}$	$\var{r4[3]}$	$\var{r4[4]}$	$\var{r4[5]}$	$\var{r4[6]}$	$\var{r4[7]}$	$\var{r4[8]}$	$\var{r4[9]}$

You should produce the following data from this table:

	$T_i$	$\sum x^2$	$n_i$ (number of observations)
{capitalise(factor1_levels[0])} ({factor2_levels[0]} and {factor2_levels[1]})	$\var{f1t[0]}$	$\var{f1ssq[0]}$	$\var{2*num_samples}$
{capitalise(factor1_levels[1])} ({factor2_levels[0]} and {factor2_levels[1]})	$\var{f1t[1]}$	$\var{f1ssq[1]}$	$\var{2*num_samples}$
	$G = \var{g}$	Sum of Squares = $\var{f1ss}$	$N = \var{n}$

So we can calculate:

\[\text{Variance estimate for }\var{factor1_name} = \sum \frac{T_i^2}{\var{2*num_samples}}- \frac{G^2}{\var{n}}=\frac{\var{f1t[0]}^2}{\var{2*num_samples}}+\frac{\var{f1t[1]}^2}{\var{2*num_samples}}-\frac{\var{g}^2}{\var{n}}=\var{f1btss}\]

Third Step

Repeat step 2 with the factors switched, i.e. use the totals $T_i$ for the {factor2_name} factor levels, ignoring {factor1_name}. This gives a Between Treatments Sum of Squares. Again, the degrees of freedom will be one fewer than the number of levels of {factor1_name}, i.e. $2 - 1 = 1$.

{capitalise(factor2_levels[0])} ({factor1_levels[0]} and {factor1_levels[1]})	$\var{r1[0]}$	$\var{r1[1]}$	$\var{r1[2]}$	$\var{r1[3]}$	$\var{r1[4]}$	$\var{r1[5]}$	$\var{r1[6]}$	$\var{r1[7]}$	$\var{r1[8]}$	$\var{r1[9]}$	$\var{r2[0]}$	$\var{r3[1]}$	$\var{r3[2]}$	$\var{r3[3]}$	$\var{r3[4]}$	$\var{r3[5]}$	$\var{r3[6]}$	$\var{r3[7]}$	$\var{r3[8]}$	$\var{r3[9]}$
{capitalise(factor2_levels[1])} ({factor1_levels[0]} and {factor1_levels[1]})	$\var{r2[0]}$	$\var{r2[1]}$	$\var{r2[2]}$	$\var{r2[3]}$	$\var{r2[4]}$	$\var{r2[5]}$	$\var{r2[6]}$	$\var{r2[7]}$	$\var{r2[8]}$	$\var{r2[9]}$	$\var{r4[0]}$	$\var{r4[1]}$	$\var{r4[2]}$	$\var{r4[3]}$	$\var{r4[4]}$	$\var{r4[5]}$	$\var{r4[6]}$	$\var{r4[7]}$	$\var{r4[8]}$	$\var{r4[9]}$

You should obtain the following data from this table:

	$T_i$	$\sum x^2$	$n_i$ (number of observations)
{capitalise(factor2_levels[0])} ({factor1_levels[0]} and {factor1_levels[1]})	$\var{f2t[0]}$	$\var{f2ssq[0]}$	$\var{2*num_samples}$
{capitalise(factor2_levels[1])} ({factor1_levels[0]} and {factor1_levels[1]})	$\var{f2t[1]}$	$\var{f2ssq[1]}$	$\var{2*num_samples}$
	$G = \var{g}$	Sum of Squares = $\var{f2ss}$	$N = \var{n}$

So we can calculate:

\[ \text{Variance estimate for }\var{factor2_name} = \sum \frac{T_i^2}{\var{2*num_samples}}- \frac{G^2}{\var{n}}=\frac{\var{f2t[0]}^2}{\var{2*num_samples}}+\frac{\var{f2t[1]}^2}{\var{2*num_samples}}-\frac{\var{g}^2}{\var{n}}=\var{f2btss}\]

Fourth Step

Now determine a Sum of Squares for Interaction by subtracting the sums of squares obtained for {factor1_name} (Step 2) and {factor2_name} (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:

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

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 $\mathit{VR}$ values are obtained by dividing the $\mathit{RMS}$ value into the $\mathit{MS}$ values for the factors and the interaction.

Source	df	SS	MS	VR	Decision
{capitalise(factor1_name)}	$1$	$\var{f1btss}$	$\var{f1btss}$	$\var{f1vr}$	{dec(f1vr)} that {measurement_name} is independent of {factor1_name}.
{capitalise(factor2_name)}	$1$	$\var{f2btss}$	$\var{f2btss}$	$\var{f2vr}$	{dec(f2vr)} that {measurement_name} is independent of {factor2_name}.
Interaction	$1$	$\var{interactionss}$	$\var{interactionss}$	$\var{ivr}$	{dec(ivr)} that {factor1_name} and {factor2_name} are independent in terms of {measurement_name}.
Residual	$\var{n-4}$	$\var{rss}$	$\var{mrs}$	-
Total	$\var{n-1}$	$\var{precround(f1btss+f2btss+interactionss+rss,2)}$	-	-

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Christian Lawson-Perfect 8 years, 10 months ago

Ricardo Monge 9 years, 7 months ago

Error in variable testing condition

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Generated value: `string`

Parts

Gap-fill

The Calculations

First Step

Second Step

Third Step

Fourth Step

The ANOVA Table

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Ricardo's copy of Two factor ANOVA

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Christian Lawson-Perfect 8 years, 10 months ago

Ricardo Monge 9 years, 7 months ago

Error in variable testing condition

Error

Warning

Generated value: string

Parts

Gap-fill

The Calculations

First Step

Second Step

Third Step

Fourth Step

The ANOVA Table

Generated value: `string`