1
Psy 521/621
Lab 4 Activities
t-tests
Learning Objectives:
 Learn to conduct and interpret the t-test for dependent samples
 Learn to conduct and interpret the t-test for independent samples
To accomplish these learning objectives, we will focus on the end-of-lesson exercises in
Lessons 23 and 24 of the Green and Salkind book.
T-test for dependent samples (AKA, the paired-sample t-test)
DATA FILE: Lesson 23 Exercise File 1.sav
Exercise 1 (excerpted from Green & Salkind Lesson 23)
Mike wants to know if overall life stress increases or decreases as working women get
older. He obtains scores on the Index of Life Stress (ILS) from a group of 100 working
women when they are 40 years old, and he gets ILS scores again from 45 of these women
when they are 60 years old. The ILS consists of two scores, interpersonal life stress and
occupational life stress. The two scores combine to form an index of overall life stress.
The SPSS file contains 45 cases, one for each woman, and four variables: interpersonal
stress at 40, occupational stress at 40, interpersonal stress at 60, and occupational stress at
60.
1. Compute scores to obtain a total ILS at age 40 and age 60.
Transformcompute. Target variable = ILS_40. Numeric expression = inter_40 +
occup_40. OK.
Transformcompute. Target variable = ILS_60. Numeric expression = inter_60 +
occup_60. OK.
2. Compute a paired-samples t-test to determine whether overall life stress changes with
age.
Click Analyze Compare MeansPaired-Samples T Test.
Click ILS_40 and ILS_40 will appear as Variable 1 in Current Selections. Hold down the
ctrl button and click ILS_60 and ILS_60. It will appear as Variable 2 in Current
Selections. Both variables should be highlighted in the box above.
Click ► and “ILS_40-ILS_60” will appear in the Paired Variables box. Click OK.
NOTE: If a participant is missing data for either variable, SPSS will not include them in
the analysis.
2
T-Test
Paired Samples Statistics
Pair
1
ILS_40
ILS_60
Mean
151.8444
136.8667
N
45
45
Std. Error
Mean
2.15907
1.48310
Std. Deviation
14.48346
9.94896
Pa ired Sa mples Correlations
N
Pair 1
ILS_40 & ILS_60
45
Correlation
.037
Sig.
.810
Pa ired Sa mples Test
Paired Differences
Pair 1
ILS_40 - ILS_60
Mean
14.9778
Std. Deviation
17.26595
Std. Error
Mean
2.57386
95% Confidence
Interval of the
Difference
Lower
Upper
9.7905
20.1650
t
5.819
df
44
We now have a p-value? What are our two questions?
1) What’s the null hypothesis?
H0: Mean life stress at 40 = mean life stress at 60
2) Did we reject the null?
Yes.
So, what exactly did we find?
Let’s look at our means: the mean for ILS_40 is higher than the mean for ILS_60, so it
looks like life stress decreased with age in this sample.
3. Create a difference variable to show the changes in life stress from 40 to 60 for each
woman. Create a histogram to show these changes graphically.
Transformcompute. Target variable = ILS_chng.
Numeric expression = ILS _60 - ILS _40. In interpreting this variable, positive scores
mean participant’s stress increased with age and negative scores tell you that the
participant’s stress decreased with age. If we computed ILS_ 40 – ILS_60 the
interpretation would be just the opposite (i.e., pos scores = stress decreases, neg scores =
stress increases). Click OK.
Click GraphHistogram. Move ILS_change to Variable. Click OK.
Sig. (2-tailed)
.000
3
FYI: Conducting a one-sample t-test on the ILS_chng variable would give you the same
results as those above.
Graph
7
6
5
4
3
2
Std. Dev = 17.27
1
Mean = -15.0
N = 45.00
0
-50.0
-40.0
-45.0
-30.0
-35.0
-20.0
-25.0
-10.0
-15.0
0.0
-5.0
10.0
5.0
20.0
15.0
30.0
25.0
ILS_CHNG
4. Conduct a paired-samples t-test to determine whether (a) occupational stress, and (b)
interpersonal stress increase or decrease with age.
Occupational Stress
Click Analyze Compare MeansPaired-Samples T Test. Click occup_40 and
occup_40 will appear as Variable 1 in Current Selections. Click occup_60 and occup_60
will appear as Variable 2 in Current Selections. Click ► and occup_40-occup_60 will
appear in the Paired Variables box. Click OK.
Paired Samples Statistics
Mean
Pair
1
Occupational life
stress at age 40
Occupational life
stress at age 60
N
Std. Deviation
Std. Error
Mean
73.64
45
9.547
1.423
61.87
45
6.625
.988
4
Paired Samples Correlations
N
Pair
1
Occupational life stress at
age 40 & Occupational life
stress at age 60
Correlation
45
Sig.
-.207
.173
Pa ired Sa mpl es Test
Paired Differences
Mean
Pair
1
Oc cupational life st ress
at age 40 - Occ upational
life stress at age 60
St d. Deviat ion
St d. Error
Mean
12.696
1.893
11.778
95% Confidenc e
Int erval of t he
Difference
Lower
Upper
7.964
15.592
t
6.223
Again, we have a p-value, so what do we ask?
1) What’s the null?
H0: Occ. stress at 40 = occ. stress at 60
2) Did we reject the null? Yes. Occupational stress is higher at age 40 than it is at age 60.
Interpersonal Stress
Click Analyze Compare MeansPaired-Samples T Test. Click inter_40 and inter_40
will appear as Variable 1 in Current Selections. Click inter_60 and inter_60 will appear
as Variable 2 in Current Selections. Click ► and inter_40-inter_60 will appear in the
Paired Variables box. Click OK.
Paired Samples Statistics
Mean
Pair
1
Interpersonal life
stress at age 40
Interpersonal life
stress at age 60
N
Std. Deviation
Std. Error
Mean
78.20
45
11.655
1.737
75.00
45
7.711
1.149
Paired Samples Correlations
N
Pair
1
Interpersonal life stress at
age 40 & Interpersonal
life stress at age 60
Correlation
45
.005
Sig.
.974
df
Sig. (2-tailed)
44
.000
5
Pa ired Sa mples Test
Paired Differences
Mean
Pair
1
Interpersonal life stress at
age 40 - Interpersonal life
stress at age 60
3.20
Std. Deviation
Std. Error
Mean
13.942
2.078
95% Confidence
Interval of the
Difference
Lower
Upper
-.99
7.39
t
1.540
df
Sig. (2-tailed)
44
Our two questions again:
1) What’s the null?
H0: Interp. stress at 40 = interp. stress at 60
2) Did we reject the null? No. Interpersonal stress at age 40 does not appear to be
significantly different than it is at age 60.
5. Let’s see what our results might look like in an APA style write-up.
A paired samples t test was conducted to evaluate whether women’s life stress declined
from age 40 to 60. The results indicated that the mean overall life stress index at age 40
(M = 151.84, SD = 14.48) was significantly greater than the mean overall life stress index
at age 60 (M = 136.87, SD = 9.95), t(44) = 5.82, p < .001. For most women, overall life
stress scores declined over the 20-year period. Two separate paired-samples t tests were
conducted to evaluate whether the change in life stress from age 40 to age 60 was due to
interpersonal stress, occupational stress, or both. The results indicated that interpersonal
stress levels did not differ significantly from age 40 (M = 78.20, SD = 11.66) to age 60
(M = 75.00, SD = 7.71), t(44) = 1.54, p = .13, but that women’s occupational stress
decreased significantly from age 40 (M = 73.64, SD = 9.55) to age 60 (M = 61.87, SD =
6.23), t(44) = 6.22, p < .001. Therefore, the decline in life stress was primarily a function
of the decrease in occupational stress.
Note: means and standard deviations could be displayed in a table rather than in the text,
as long as the text refers the reader to the table.
DATA FILE: Lesson 23 Exercise File 2.sav
Kristy is interested in investigating whether husbands and wives who are having
infertility problems feel equally anxious. Twenty-four couples agree to participate.
Husbands and wives both take the Infertility Anxiety Measure (IAM). Higher scores on
the IAM indicate more anxiety related to infertility. The SPSS data file contains 24 cases,
one for each husband-wife pair, and two variables: husband IAM score and wife IAM
score.
Conduct a paired samples t-test on these data.
Click Analyze Compare MeansPaired-Samples T Test. Click husband and husband
will appear as Variable 1 in Current Selections. Click wife and wife will appear as
.131
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Variable 2 in Current Selections. Click ► and husband-wife will appear in the Paired
Variables box. Click OK.
T-Test
Paired Samples Statistics
Mean
Pair
1
Husband's infertility
anxiety score
Wife's infertility
anxiety score
N
Std. Deviation
Std. Error
Mean
57.46
24
7.337
1.498
62.54
24
12.441
2.540
Paired Samples Correlations
N
Pair
1
Husband's infertility
anxiety score & Wife's
infertility anxiety score
Correlation
24
.822
Sig.
.000
Pa ired Sa mpl es Test
Paired Differenc es
Mean
Pair
1
Husband's infertility
anxiety score - Wife's
infertility anxiet y sc ore
St d. Deviat ion
St d. Error
Mean
7.649
1.561
-5. 083
95% Confidenc e
Int erval of t he
Difference
Lower
Upper
-8. 313
-1. 853
t
df
-3. 256
Let’s identify:
Mean IAM scores for husbands: 57.46
Mean IAM scores for wives: 62.54
What’s the null hypothesis here? H0: husband IAM = wife IAM
Did we reject the null? Yes
Let’s create a boxplot to see what the differences between husband and wife scores on
the IAM look like graphically.
Graphs  Boxplot  Summaries of separate variables. Move Husband and Wife into the
Boxes Represent box. Click OK.
Sig. (2-tailed)
23
.003
7
90
2
80
2
70
60
50
21
40
30
N=
24
Husband's infertilit
24
Wife's infertility a
How would we compute an effect size for this test?
d = mean of the diff scores
SD of the diff scores
alternatively, but equivalently: d = t_
√N
(where N = #pairs)
You can get the values to plug into this equation from the Paired Samples Test output.
What is d for these data?
d = -5.083/7.649 = -.66
OR
d = -3.256/√24 = -.66
Let’s see what an APA results section would look like for this test:
A paired samples t test was conducted to evaluate whether husbands or wives differed on
their levels of infertility anxiety. The results indicated that wives (M = 62.54, SD = 12.44)
were significantly more anxious about infertility than their husbands (M = 57.46, SD =
7.34), t(23) = -3.26, p < .01. A boxplot of the data is presented in Figure XX. The 95%
confidence interval for the mean difference in fertility anxiety between husbands and
wives was -8.31 to -1.35. Finally d = -.66, indicating that mean anxiety score for
husbands is .66 standard deviations below the mean anxiety score for their wives. This is
a medium effect.
8
T-test for independent samples
DATA FILE: Lesson 24 Exercise File 1.sav
Billie wants to test the hypothesis that overweight individuals eat faster than normalweight individuals. She has two assistants sit in a McDonald’s restaurant and identify
individuals who order the Big Mac special (Big Mac, large fries, large Coke) for lunch.
The Big Mackers, as they are affectionately called by the researchers, are classified as
overweight, normal-weight, or neither overweight nor normal-weight. The assistants
identify 10 overweight and 30 normal-weight Big Mackers. The time it takes for the
individuals to complete their Big Mac specials is recorded by the assistants. The SPSS
data file contains two variables: a grouping variable with two levels (overweight = 1,
normal-weight = 2) and time in seconds to eat the meal.
1. Compute an independent-samples t test on these data.
Click AnalyzeCompare MeansIndependent Samples T Test.
Move “Time Spent” to Test Variable box.
Move “Weight” to Grouping Variable. NOTE: ALWAYS GROUP BY CATEGORICAL
VARIABLE.
Click Define Groups. In Group one box, type in 1 and in Group 2 box, type in 2.
Click Continue. Click OK.
T-Test
Group Statistics
Time s pent eating Big
Mac specials in seconds
Weight clas sification
normal weight
over weight
N
10
30
Mean
589.00
698.40
Std. Deviation
42.615
82.949
Std. Error
Mean
13.476
15.144
Independent Samples Test
Levene's Test for
Equality of Variances
F
Time s pent eating Big
Mac specials in seconds
Equal variances
as sumed
Equal variances
not ass umed
2.745
Sig.
.106
t-test for Equality of Means
t
df
Sig. (2-tailed)
Mean
Difference
Std. Error
Difference
95% Confidence
Interval of the
Difference
Lower
Upper
-3.975
38
.000
-109.40
27.522
-165.116
-53.684
-5.397
30.828
.000
-109.40
20.272
-150.754
-68.046
1. What are the t and p values when equal variances are assumed and not assumed?
Assuming equal population variances: t(38) = -3.98, p < .001.
Not assuming equal population variances: t(31) = -5.40, p < .001.
-- Is the t test significant? Yes. We would reject the null hypothesis that mean eating time
for overweight and normal weight individuals is equal.--
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2. a. Mean eating time for overweight individuals = 589 seconds.
b. Standard deviation for normal weight individuals = 82.95 seconds.
c. Results for the Levene’s test of homogeneity: F(1, 38) = 2.75, p > .05. Levene’s test
is not significant. What does that mean? Well, what is our null hypothesis?
H0: population variance of time spent eating Big Mac specials for normal weight
individuals = population variance of time spent eating Big Mac specials for over weight
individuals.
3. Let’s compute an effect size for this statistic:
d = Mean Difference/SDpooled
SDpooled = Sq. Root of (N1 – 1)SD21 + (N2 – 1)SD22
N1 + N2 – 2
=Sq. Root of (9*1817) + (29*6881)
=Sq. Root of 16353 +199549
=Sq. Root of 215888/10 + 30 – 2
= 75.37
d = -109.4/ 75.37 = -1.45
Alternatively, you can compute d with this formula: d = t * √[(N1 + N2) / (N1*N2)]
4. Create a boxplot to summarize the results graphically
Graphs Boxplot  Summaries for groups of cases. Move “time spent” to the Variable
box and “Weight” to the Category Axis box.
Time spent eating Big Mac specials in seconds
1000
900
800
700
600
500
400
N=
10
normal weight
30
over weight
Weight classification
5. Here is a sample APA results section:
An independent-samples t test was conducted to evaluate the hypothesis that overweight
individuals tend to eat faster than normal weight individuals. Overweight individuals did
10
eat significantly more quickly (M = 589, SD = 42.62) than normal weight individuals (M
= 698, SD = 82.95), t(38) = -3.98 , p < .001, d = -.24.
DATA FILE: Lesson 24 Exercise File 2
Fred wants to determine whether mainstreaming special needs children into regular
classrooms will help or hurt the academic achievement of the non-special needs children.
Two 7th grade teachers in two similar junior high schools have allowed him to use their
classrooms as samples. One school has recently integrated special needs children in
regular classrooms, while the other has not. Fred is supplied with the student’s raw scores
on a standardized achievement test at the beginning of the year and the end of the year.
The SPSS data file contains 40 cases and three variables: 1) integrat with 2 levels (1 =
classroom has not been integrated, and 2 = classroom has been integrated); 2) pretest,
which are raw scores on the standardized test at the beginning of the year, and 3) posttest,
which are raw scores on the standardized test at the end of the year.
6. Create a dependent variable by computing a difference score between pretest and
posttest achievement scores.
Transformcompute. Target variable = ach_change. Numeric expression = posttest pretest. (Note: When you subtract posttest from pretest, positive scores tell you that the
participant’s score (achievement) increased over the school year and negative scores tell
you that the participant’s score (achievement) decreased over the school year). Click OK.
7. Conduct an independent-samples t test to evaluate the effect of integration on academic
achievement.
Click AnalyzeCompare MeansIndependent Samples T Test. Move Ach_change to
Test Variable. Move Integrat to Grouping Variable. Click Define Groups. In Group one
box, type 1 and in Group 2 box, type 2. Click Continue. Click OK.
T-Test
Group Statistics
ach_change
Classroom integration
not integrated
integrated
N
15
25
Mean
1.5333
9.6000
Std. Deviation
7.40527
16.54287
Std. Error
Mean
1.91203
3.30857
Independent Samples Test
Levene's Test for
Equality of Variances
F
ach_change
Equal variances
as sumed
Equal variances
not ass umed
4.098
Sig.
.050
t-test for Equality of Means
t
df
Sig. (2-tailed)
Mean
Difference
Std. Error
Difference
95% Confidence
Interval of the
Difference
Lower
Upper
-1.778
38
.083
-8.06667
4.53779
-17.25294
1.11961
-2.111
35.852
.042
-8.06667
3.82133
-15.81779
-.31555
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8. What are the results of Levene’s test?
F(1, 38) = 4.10, p = .05. Levene’s test is significant, so we reject the null hypothesis that
the variances are equal across the grouping variable. In other words, we would conclude
that the variance in ach_chng for those in the integrated classroom is not equal to the
variance in ach_chng for those in the non-integrated classroom.
9. Given the significant Levene’s test, we will report the results where equal variances are
not assumed: t(35.85) = -2.11, p = .04.
10. Let’s put together a sample APA results section and compute a new effect size.
An independent-samples t test was conducted to evaluate the effect of integration on the
achievement scores of regular students. Regular students in integrated classrooms
achieved significantly greater gains in achievement scores (M = 1.53, SD = 7.41) than
students in non-integrated classrooms (M = 9.60, SD = 16.54), t(35.85) = -2.11 , p = .04,
η2 = .10.
As an alternative to d, you could compute η2
η2 = _____t2_______ = ____(-2.11)2_____ = __4.452___ = .10
t2 + (N1 + N2 – 2) (-2.11)2 + (15+25-2) 4.452 + 38
η2 is interpreted as the proportion of variance in the dependent variable that can be
accounted for by group membership. So in this case we would say that 10% of the
variance in the change in achievement scores is accounted for by which classroom a
student was in (integrated v. non-integrated). The interpretation for η2 is like that of R2,
which you will learn more about later.
Rules of thumb for η2: small = .01
medium = .06
large = .14