Lecture 7 - Comparing Two Groups
Howell Chapters 13,14,19,20
Categorizing Research Involving Two Conditions
1) the Extent of Pairing of participants in the two conditions.
2) the nature of the dependent variable, and the type of performance comparisons which it permits.
3) the shape of the distribution of dependent variable scores within groups.
1) Extent of Pairing
A) The Independent Groups Design.
No pairing of participants in the two conditions has been made.
Sample sizes may be different.
Different people are in the two conditions.
Participants may or may not have been randomly assigned.
Problems:
1. Pre-existing differences between the groups
2. Lack of power
B) The Matched Participants Design.
Different people are in the two groups, but they have been matched with respect to a variable that is
correlated with the dependent variable.
98
96
Sample sizes must be equal.
96
A correlation must be computable.
95
Requires a pretest
90
87
Problems:
1. Time & effort required to match
C) The Participants as their Own Controls Design.
The same people are in the two groups – the matched participants design carried to the extreme.
Sample sizes must be equal.
A correlation must be computable.
Order of exposure to treatment conditions may be an issue.
Problems:
1. Carry-over effects from one condition to the next.
D) Clones in the two conditions
Not a possibility in human research, although it’s feasible in research involving nonhumans.
Problems:
1. Cloning of humans is prohibited.
Lecture 7 – Comparing Two Groups - 1
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2) The nature of the dependent variable.
A) Quantitative Interval/Ratio Dependent Variable.
Interval / Ratio aka quantitative data..
The numbers represent quantitative levels performance.
The mean and standard deviation are appropriate summaries of central tendency and variability.
B) Ordinal Dependent Variable.
Order only.
The numbers represent rank ordering of performance only.
We could substitute ranks for the actual observed values without loss of information.
Median is natural measure of central tendency.
C) Categorical Dependent Variable. (Avoid these like the plague.)
Gross performance comparisons only – Typically Good vs. Bad or Success vs Failure
The numbers (if we have actually recorded numbers) represent only a categorical performance distinction.
3) The shape of the distribution of scores within Groups.
This factor is only important for Quantitative Interval / Ratio scaled dependent variables. There are two
possibilities. . .
A) The distributions within groups are US without outliers.
B.) The distributions within groups are skewed or have outliers.
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The Various Tests Comparing Two Research Conditions
by Design and Dependent Variable Characteristics
Memorize this table
Nature of Dependent Variable
Quantitative
(Interval/Ratio)
US Distribution:
Independent Groups t-test
Skewed Distribution:
Mann Whitney U-test
Independent Groups
Type of
Design
Matched Participants
Or
Participants as Own
Controls
Dependent t-test
Ordinal
Categorical
Mann-Whitney Utest
Crosstabs with Chisquare test
Wilcoxon Signed
Ranks Test
McNemar Test
(2 categories only)
We’ll cover all of the above cells.
For each,
1) The situation to which the test is applicable will be described.
2) The null hypothesis and alternative hypotheses tested will be presented.
3) Solving an example problem with SPSS will be shown.
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Independent Groups t Test
Situation:
Research Design is an independent groups design
The dependent variable is interval / ratio, aka quantitative variable.
Key assumption: The distribution of scores within groups is unimodal and symmetric with no outliers.
The interest is on comparing the means of the two conditions.
Hypotheses
H0: Population condition means are equal
H1: Population condition means are not equal.
Test Statistic assuming equal variances in the two populations represented by the two conditions.
Distribution of the equal variances test statistic when the null is true
The T distribution with N1-1 + N2-1 degrees of freedom.
Test Statistic assuming unequal variances in the two populations represented by the two conditions
X1 – X2
t = ------------------S12
S22
-- + --n1
n2
Distribution of the unequal variances test statistic when the null is true
The T distribution with ((S12/n1 + S22/n2)2 / ( (S12/n1)2/(n1-1) + (S22/n2)2/(n2-1)) degrees of freedom.
See Howell, D. (2002). Statistical Methods for Psychology. 5th Ed. Duxbury. P.215.
Lecture 7 – Comparing Two Groups - 4
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Independent Groups t-test example
Two ways of training clerical workers in how to use the Management Information System in an
organization are being compared. Trainees are randomly assigned to either a Lecture Training Course or a
Computer Aided Training Course. After the end of training, a comprehensive exam covering the MIS is
given. Examination score is the dependent variable.
Entering the Data
EXAM CONDIT
79
2
79
1
80
2
71
1
81
2
73
1
76
2
72
1
79
2
78
1
73
2
74
1
75
2
75
1
79
2
72
1
81
2
73
1
74
2
71
1
80
2
70
1
81
2
69
1
84
2
68
1
74
2
73
1
72
2
74
2
Specifying the Analysis:: Analyze -> Compare Means -> Independent Samples t-test
Put the name of the dependent variable into the
“Test Variable(s)” field.
Put the name of the variable whose values define
the groups into the “Grouping Variable:” field.
Then click on the [Define Groups] button and enter
the two values representing the groups.
After that, click on the [OK] button.
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t-test Results
Group Statistics
condit
N
Mean
Std. Deviation
Std. Error Mean
1
14
72.71
3.099
.828
2
15
78.00
3.381
.873
exam
Independent Sam ples Test
Levene' s
Te st for
Eq uality of
Va riance s
F
EX AM Eq ual
variance s 1.9 43
assumed
t-te st for Equa lity o f Me ans
Sig .
t
.17 4
-3.9 78
28
.00 0
-4.9 1
1.2 3
-7.4 4
-2.3 8
-4.0 19
27. 997
.00 0
-4.9 1
1.2 2
-7.4 1
-2.4 1
Eq ual
variance s
not
assumed
df
95%
Co nfide nce
Inte rval of the
Me an
Sig .
Me an
Std . Erro r
(2-t ailed ) Dif feren ce Dif feren ce Lower Up per
Conclusion
Step 1: "Precomparison" of Population Variances
F = 1.943, p = .174. Tests the null hypothesis that the population variances are equal.
If you retain, use the “Equal variances” t that follows.
If you reject, use the “Unequal variances” t that follows.
In this case, p=.174 so retain null that population variances are equal.
Use Equal variances t to compare the means.
Step 2. Comparing Means
Use the "Equal Variances Assumed" t since we’ve retained the hypothesis of equal variances.
t = -3.978, p = .000.
Interpretation: If the population means were equal, the probability of a t as extreme as 3.978 would be .000.
So reject the hypothesis of equal population means. Mean performance was significantly better in
Computer Condition.
Lecture 7 – Comparing Two Groups - 6
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Graphical Representation of the Results
Graphs -> Scatterplot -> Simple
90
80
EXAM
70
60
.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
Training Condition
This graphical representation shows the difference in central tendency between the two groups. It may also
show differences in variability, if there are any.
If sample size is too large, the following display could also be useful.
1
You should develop the habit of creating
one of these two graphs whenever you
conduct the independent groups t-test.
condit
Look for asymmetry, outliers, unequal
variance.
2
65
70
75
80
85
exam
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The Mann-Whitney U-test
(Also called the Wilcoxon Rank Sum Test)
Situation:
A.
B.
The research design employs two independent conditions (no pairing.)
The dependent variable is interval / ratio.
The distribution of scores within groups is skewed.: Time, Income, Charges
or
The research design employs two independent conditions (no pairing.)
The dependent variable is ordinal, i.e., the scores represent only order.
The interest is on comparing the "average value" of the scores in the two conditions.
In fact, however, the Mann-Whitney tests the null hypothesis that the distributions are identical. If
you are willing to assume that they are equally variable, then rejection of the null implies a difference in
central tendency of the two distributions.
Hypotheses
H0: Population distributions are identical
H1: Population distributions are not identical
These hypotheses are not hypotheses of equal “average value”. But they’re typically treated as if they were.
Test Statistic
Mann-Whitney U Statistic or Z-statistic computed from U if sample sizes are large (> 20 or so).
This test is also called the Wilcoxon Rank Sum test, not to be confused with the Wilcoxon Signed Ranks
test.
Distribution of the test statistic when the null hypothesis is true
The U distribution for small sample sizes
The Standard Normal Distribution for large sample sizes.
Lecture 7 – Comparing Two Groups - 8
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Mann-Whitney U-Test Example
Two methods of filing completed insurance forms in an insurance company are being compared. In one
method, a pre-filing stage is completed prior to actually placing the forms in the appropriate cabinet. In the
other method, each individual form is taken directly to its cabinet. Time to complete the filing of 100
forms is used the dependent variable. 30 groups of 100 forms each were filed. (Recall that times are
notorious for being positively skewed.)
Entering the data
TIME METHOD
130
1
150
1
190
1
130
1
135
1
140
1
90
1
110
1
120
1
105
1
115
1
120
1
115
1
120
1
125
1
150
2
140
2
130
2
135
2
220
2
300
2
150
2
130
2
145
2
160
2
140
2
120
2
180
2
155
2
165
2
Specifying the Analysis
Analyze -> Nonparametric tests -> Legacy Dialogs -> 2 Independent Samples . . .
Lecture 7 – Comparing Two Groups - 9
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NPar Tests
Descriptive Statistics
N
Mean
Std. Deviation
Minimum
Maximum
time
30
143.8333
39.55972
90.00
300.00
method
31
1.4516
.56796
.00
2.00
Mann-Whitney Test
Ranks
method
time
N
Mean Rank
Sum of Ranks
1.00
15
10.40
156.00
2.00
15
20.60
309.00
Total
30
Ranking is such that smallest
score gets rank of 1. So it
appears that the times are
generally larger in the Direct
condition.
Test Statisticsa
time
Mann-Whitney U
Wilcoxon W
Z
Asymp. Sig. (2-tailed)
Exact Sig. [2*(1-tailed Sig.)]
36.000
156.000
-3.184
.001
.001b
The program prints p-values for two
test statistics . . .
1) A Z-statistic (Asymp. Sig.)
2) An exact test statistic (Exact Sig)
a. Grouping Variable: method
b. Not corrected for ties.
Conclusion
If the null hypothesis of equal distribution locations were true, the probability of a U as extreme as 36 or a Z
as extreme as 3.184 would be .001.
So reject the hypothesis of equal locations.
Time to file took longer using the Direct Method, #2.
Lecture 7 – Comparing Two Groups - 10
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Graphical Presentation of Results
400
Long positive tail
300
200
TIME
100
0
.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
Filing Method
As was the case with the independent groups t-test, it may help you communicate your findings by creating
a scatterplot of scores vs. group. Note the positive skew of the scores within each group, especially the
group on the right – the Direct group.
EXPLORE’s box-and-whisker plots could also be used here.
Lecture 7 – Comparing Two Groups - 11
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CROSSTABS - Two way Chi-square Test
Situation:
The research design employs two independent conditions (no pairing.)
The dependent variable is categorical.
The interest is on comparing proportions within each category of the dependent variable across
groups.
Examples here will use a dichotomous dependent variable.
Hypotheses
H0: Population Proportions are Equal
H1: Population Proportions are Not Equal.
Test Statistic
Two-way Chi-square Statistic
X2 = (O-E)2/E
Distribution of the test statistic if the null hypothesis is true
The Chi-square distribution with degrees of freedom = (No. of DV categories - 1) * (No. of Groups - 1)
Relationship to other presentations of Chi-square
This presentation is not the traditional presentation of Chi-square. Traditionally, chi-square analysis is
presented to either
1) compare observed with expected frequencies in categories of a single variable – a test called the one-way
chi-square, or
2) to test the independence of two variables – the two-way chi-square.
The presentation here is a variation on 2 above. It turns out that the test comparing equality of
proportions across two groups is equivalent to the test of independence of two variables in which
those variables are the classification variable and the grouping variable.
I prefer the emphasis on comparing proportions across groups presented here because that’s how we use the
two-way chi-square in actual practice.
Lecture 7 – Comparing Two Groups - 12
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Two-way Chi-square Example
Two methods of presenting information to juries are being compared. In the first, character assassination
of the defendant plays a major role in the prosecution. In the second, the prosecution focuses on the facts
of the case without attempting to malign the defendant. Thirty different simulated cases are presented to
college 30 different simulated juries made up of college students over the course of an academic year. The
simulated juries met and reached a unanimous decision concerning guilt or innocence. The dependent
variable is jury verdict.
For the data here.
VERDICT=1 means guilty; VERDICT=0 mean innocent.
PROSMETH=1 means character assassination; PROSMETH=2 means facts only.
VERDICT PROSMETH
1
1
1
1
1
1
0
1
1
1
1
1
1
1
0
1
1
1
1
1
1
1
1
1
1
1
1
1
0
1
The data
VERDICT PROSMETH
0
2
0
2
0
2
1
2
0
2
1
2
1
2
0
2
1
2
1
2
0
2
1
2
1
2
0
2
1
2
Specifying the Analysis: Analyze -> Descriptive Statistics -> Crosstabs
Dependent variable defines the
Rows.
Independent variable defines
the columns.
Click here and request the chisquare statistic.
Click here and request Column
percentages
Lecture 7 – Comparing Two Groups - 13
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Results
Ca se Pr oces sing Summ ary
Ca ses
Va lid
N
VE RDICT Ju ry verdict * PRO SME TH P rosecution Met hod 30
Missing
Pe rcent
100 .0%
N
0
To tal
Pe rcent
.0%
N
30
Pe rcent
100 .0%
VE RDICT Jury v e rdict * PRO SME TH P rose cution Me thod Cros stabulation
PROSM ETH
Pro secu tion M etho d
VE RDICT
0 Not Co unt
Jury verd ict Gu ilty % within PRO SME TH P rose cution Met hod
1
Co unt
Gu ilty % within PRO SME TH P rose cution Met hod
To tal
Co unt
% within PRO SME TH P rose cution Met hod
1 M alig n
def enda nt
3
2 Facts
onl y
7
To tal
10
20. 0%
46. 7%
33. 3%
12
8
20
80. 0%
53. 3%
66. 7%
15
15
30
100 .0%
100 .0%
100 .0%
Chi-Square Tests
Pe arson Chi-Squa re
Asy mp.
Ex act
Ex act
Sig .
Sig .
Sig .
Va lue df (2-sided ) (2-sided ) (1-sided )
2.4 00 b 1
.12 1
Co ntinu ity Co rrect ion a
1.3 50
1
.24 5
Likeliho od Ra tio
2.4 51
1
.11 7
2.3 20
1
.12 8
Fisher's Exac t Test
Lin ear-b y-Lin ear A ssoci ation
.24 5
N o f Val id Ca ses
The chi-square
statistic tests the
significance of
the difference
between these
percentages.
.12 3
30
a. Co mput ed on ly fo r a 2x 2 tab le
b. 0 c ells (.0%) have expe cted coun t less than 5. Th e mi nimu m
exp ecte d cou nt is 5.00.
Conclusion
Whew! This procedure prints a plethora of chi-square values.
Conventional wisdom is to use the Pearson continuity correction chi-square for tables bigger than 2 x 2
and to use the Fisher’s Exact test for 2 x 2 tables.
If you have a 2x2 table, refer to the Fisher’s Exact test p-value, .245 here.
Here, referring to the Fisher’s exact test, if the null hypothesis of equal population proportions were equal,
the probability of a difference in proportions as large as the obtained difference would be .245. Do not
reject the null hypothesis of equal population proportions.
The evidence we have does not suggest that prosecution method makes a difference in % of guilty verdicts.
Lecture 7 – Comparing Two Groups - 14
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Tests for Correlated Groups Designs
Correlated Groups t-Test
Situation:
The research design employs two conditions in which participants are paired – either the matched
participants design or the participants as their own controls design.
The dependent variable is quantitative (interval / ratio).
The distribution of scores within groups is unimodal and symmetric.
The interest is on comparing the means of the two conditions.
Hypotheses
H0: Population Means are Equal
H1: Population Means are Not Equal.
Test Statistic
Correlated Groups t-test aka dependent samples t-test
D is the mean of difference
scores.
SD is the standard
deviation of difference
scores.
In the above, r is the Pearson correlation between the paired scores.
Distribution of t when null hypothesis is true.
The T distribution with N-1 degrees of freedom.
Relationship to independent groups t-test.
When r=0, this test is essentially equivalent to the independent groups t-test.
Since the factor involving r is subtracted in the denominator, the more positive the value of r, the smaller the
denominator, and thus the larger the value of t.
This means that all other things being equal, the dependent t will be larger than the independent samples t.
Thus, for example, for equivalent sample sizes, using participants as their own controls creates a more
powerful test of differences between means than using an independent subjects design.
Lecture 7 – Comparing Two Groups - 15
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Correlated t-test Example
Two product package designs are being compared. Participants are shown two packages, one after the
other. They rate each package on a scale of 1 to 11, with 11 representing most favorable. Half the
participants are shown A first then B. The other half are shown B first, then A. The dependent variable is
product rating.
Entering the data.
Note that each set of ratings is put in a different column.
Pairing is maintained across columns – 6 is paired with 3, 7 with 5, etc.
ARATING BRATING
6
7
5
8
4
7
6
8
7
8
9
5
6
7
6
3
5
4
8
1
3
4
7
7
7
8
2
5
6
3
Specifying the Analysis: Analyze -> Compare means -> Paired samples t
Test question:
Describe how data are entered for the independent groups t.
Lecture 7 – Comparing Two Groups - 16
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Describe how data are entered for the correlated groups t.
Lecture 7 – Comparing Two Groups - 17
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Results
Pa ired Samples S tatis tics
Pa ir ARATING Rating of P ackag e A
1
BRATING Rating of P ackag e B
Std .
Std .
Error
De viatio n Me an
1.3 5
.35
Me an N
6.6 0 15
4.8 7
15
2.2 3
.58
Pa ired S amples Correl ations
Pa ir 1
N
ARATING Ra ting of Pa ckage A &
15
BRATING Ra ting of Pa ckage B
Co rrelat ion
.88 0
Sig .
.00 0
SPSS prints the Pearson r between the paired scores. We’re not really interested in it, except to make sure
that it’s positive. But it’s there if you need it.
Pa ired S amples Test
Pa ired Differences
95 %
Co nfide nce
Std . Int erval of th e
Std .
Error Dif feren ce
Me an De viatio n Me an Lo wer Up per
Pa ir 1
ARATING Rating of Pa ckag e A
- B RATI NG Ratin g of P acka ge B
1.7 3
1.2 2
.32
1.0 6
2.4 1
t
df
5.4 90
14
Sig .
(2-tailed )
.00 0
Conclusion
If the null were true, the probability of a t as extreme as 5.490 would be .000. So reject the null.
The data suggest that in the population, people would favor the A package.
Lecture 7 – Comparing Two Groups - 18
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The Wilcoxon Signed Ranks Test
Situation:
The research design employs two conditions in which participants are paired – either the matched
participants design or the participants as their own controls design.
The dependent variable is ordinal, such as ranks.
The interest is on comparing the central tendency of the two conditions, specifically the distribution of
differences between paired values.
Hypotheses
H0: Equal central tendency of the two populations.
H1: Unequal central tendency.
Test Statistic
Wilcoxon Signed Ranks test.
1.
2.
3.
4.
5.
Compute the absolute value of the difference between each pair of scores.
Assign ranks to the absolute values of the differences.
Put a negative sign in front of the rank of each negative difference.
Sum the positively signed ranks and sum the negatively signed ranks.
The test statistic is the smaller of the absolute values of the two sums.
Lecture 7 – Comparing Two Groups - 19
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Wilcoxon Signed Ranks example
From Howell, David. (2002). Statistical Methods for Psychology. 5th Ed. Duxbury. P. 714.
The data are from a study of the effects of running on systolic blood pressure. The blood pressure of 8
persons is measured. Then each is put on a 6-month running program. After 6 months, blood pressure is
measured again.
Step 1
Step 2
Step 3
The data
Step 4
ID
1
2
3
4
5
6
7
8
BEFORE
130
170
125
170
130
130
145
160
Number of cases read:
AFTER DIFF ABSDIFF
120
163
120
135
143
136
144
120
8
-10
-7
-5
-35
13
6
-1
-40
10
7
5
35
13
6
1
40
Rank
RABSDIFF
5
4
2
7
6
3
1
8
Number of cases listed:
-5
-4
-2
-7
6
3
-1
-8
Sum of positive ranks = 9
Sum of negative ranks = 27
8
Specifying the Analysis
Analyze -> Nonparametric tests -> Legacy Dialogs -> 2 Related Samples
Click on the names of the two columns to be compared.
Lecture 7 – Comparing Two Groups - 20
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NPar Tests: Wilcoxon Signed Ranks Test
Ra nks
N
AFTER BE FORE
Me an Rank
Su m of Ranks
Ne gative
Ra nks
6
4.5 0
27 .00
Po sitive Ranks
2b
4.5 0
9.0 0
Tie s
0c
To tal
8
a
a. AFTER < BE FORE
b. AFTER > BE FORE
c. BE FORE = A FTER
Test Statisticsa
after - before
Z
-1.260b
Asymp. Sig. (2-tailed)
.208
Exact Sig. (2-tailed)
.250
Exact Sig. (1-tailed)
.125
Point Probability
.027
a. Wilcoxon Signed Ranks Test
b. Based on positive ranks.
Do not reject the null. The difference in BPs is not statistically significant.
Lecture 7 – Comparing Two Groups - 21
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McNemar Test for change in dichotomous variables
Situation:
Participants as their own controls design was employed – typically in a before-after design.
Data are categorical
The interest is on comparing the proportions before and after.
Hypotheses
H0: Proportions changing from one category to the other are equal..
H1: Unequal proportions.
Test Statistic
McNemar test.
Test Statistic
A chi-square from the following 2x2 table
Before 0
1
1
A
C
After
0
B
D
In the table, A is the count of the number of persons who changed from 0 to 1.
D is the count of the number of persons who changed from 1 to 0.
The statistic is X2 = (A-D)2 / (A+D).
Distribution when null is true.
The Chi-square distribution with 1 degree of freedom
Lecture 7 – Comparing Two Groups - 22
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McNemar test Example
The example is from Siegel, S. & Castellan, N. John, Jr. (1988). Nonparametric Statistics for the Behavioral
Sciences. 2nd Ed. McGraw-Hill. p. 75.
Preference for Ronald Reagan vs. Jimmy Carter was measured just prior to their 1980 debate and just after
the debate. Participants were asked to indicate which of the two they preferred.
The data. For both variables, 0=Reagan preferred; 1=Carter preferred.
BEFORE
AFTER
1
1
1
1
1
1
1
1
1
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
0
0
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Specifying the Analysis
Analyze -> Nonparametric tests -> Legacy Dialogs -> 2 Related Samples
Uncheck the default Wilcoxon signed ranks test.
Check the McNemar Test.
The Results
NPar Tests
McNemar Test
Crosstabs
Tes t Sta tistic s b
BE FORE & A FTER
75
N
BE FORE & AFTER
BE FOR
E
0
1
Exact S ig.
(2-tailed )
AFTER
0
1
27
7
13
28
a
.26 3
a. Bin omia l dist ribut ion u sed.
b. McNem ar Te st
Seven persons switched from Reagan before to Carter after while 13 switched from Carter before to Reagan
after. But the p value is > .05, so the difference in no. of persons switching was not significant.
Lecture 7 – Comparing Two Groups - 23
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Examples.
1. Are Males or Females better able to fake when so instructed?
in 2014
Nguyen (2001) gave participantsSkipped
a Big Five
Questionnaire twice – once under instructions to respond
honestly, again under instructions to fake good. Who was better able to fake good – men or women?
Examining the data . . .
The Male data are on the top.
The Male distribution looks slightly positively skewed, suggesting that both the t-test (on the assumption
that the male data are not TOO skewed) and the Mann-Whitney test (on the assumption that they male data
may too skewed) should be conducted.
Lecture 7 – Comparing Two Groups - 24
2/8/2016
First the t-test.
Next the Mann-Whitney U-test
Lecture 7 – Comparing Two Groups - 25
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Finally, a second graphical comparison of the two distributions
So both tests suggest that male and female fake about the same amount when so instructed.
Lecture 7 – Comparing Two Groups - 26
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2. Are males or females more likely to fake when not instructed.
Biderman and Nguyen (2008) gave participants a Big Five questionnaire twice with instructions to respond
honestly in both administrations. On the second administration, participants were told that the best scores
would win a small prize. Did males or females fake more?
Examining the data and comparing the distributions in one step
The top distribution is of female scores.
Both distributions are positively skewed. So the Mann-Whitney U-test should be conducted.
So, males and females fake about
equally as much when given an
incentive to fake.
Lecture 7 – Comparing Two Groups - 27
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A final note on stating the results of comparisons of two groups
Compare the following two descriptions of the same result . ..
Study evaluates blood transfusion rates for hip, knee surgeries
Overweight and obese patients were significantly less likely to receive blood transfusions during hip and knee
replacement procedures, researchers with Henry Ford Hospital found in an analysis of data on 2,399 surgeries
performed from 2011 to 2013. Patients in the obese body-mass index cohort received blood transfusions in
21.9% of hip replacements and 8.3% of knee replacements, while patients in the normal BMI cohort received
transfusions in 34.8% of hip procedures and 17.3% of knee procedures. The study was presented at a meeting
of the International Society for Technology in Arthroplasty. BeckersASC.com (10/5)
vs
Heavier hip, knee replacement patients have lower blood
transfusion rates: 4 insights
Written by Anuja Vaidya | October 05, 2015
0
inShare
A new study, by researchers at Henry Ford Hospital in Detroit, found that overweight or obese patients were less likely to require
blood transfusions when undergoing hip and knee replacement surgery.
Researchers evaluated 2,399 patients, of which 1,503 underwent knee replacement and 896
Lecture 7 – Comparing Two Groups - 28
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