Practice for the Mid-Term
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SPSS DATA ANALYSIS: You will only be tested in SPSS on chisquare, t-test, and analysis of variance. All of the questions will
use the socialsurvey.sav data set.
1. Open the socialsurvey.sav data set in SPSS Data Editor.
Suppose you think that people who like to go out like to go to a
variety of different events, but some people are just stay-athomes. Test the hypothesis at the p <.05 level of confidence that
there is an association between having visited an art gallery or
museum in the last year (the variable (#48) is called “Visit Art
Museum or Gallery in Last Yr (visitart)” and having attended a
sporting event in the last year (the variable (#47) is called
“Attended Sports Event in Last Year (attsprts)”
Is the obtained relationship in the expected direction (art gallery
visitors went to more sports events, relative to their numbers,
than nonvisitors? Is the relationship statistically significant? Give
the value of the test statistic and its associated probability level.
Create a new Word document and type Problem # 1. Below that
copy and paste your SPSS output. Save the document as
“SPSSOutputYourLastName.doc.”
Chi-Square: Appropriate Test for the Impact of
a Nominal Level IV on another Nominal Level
IV
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To answer this question you will run a Chi-Square
Analysis. You will be able to figure this out by first
considering the level of measurement for the two
variables. In the “Variable View” window look at the
two variables, and note that each is measured not in
terms of ranking or numeric values but in terms of
discrete categories (whether they did or did not
attend a sports event, and whether they did or did not
visit a museum or gallery). For two nominal scale
variables like this, the correct statistic to analyze their
relationship would be Chi-square.
Running a Chi-Square Analysis in
SPSS
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Now that you have decided to run a Chi-square test, go to
Analyze/Descriptives/Crosstabs
Move the Visited Art Galleries variable into the Columns box
(you do this because the column variable is the one you are
treating as independent, and in your hypothesis you have asked if
there is a significant association between going to art galleries and
going to sporting events). In the case of two variables like this
there probably is no causal relationship, so which variable is the
column variable is more or less arbitrary, but if your hypothesis
were about the effect of say, gender, then gender is the obvious
independent variable and would have to go in the column box)
Move the Attended Sport Events variable into the Rows box
Under Cells, click Observed, Expected, Row, Column and
Total and click Continue (you do this so that you will have all
the information you need to see if the direction of the relationship
you predicted is in fact correct. In this case you want to know, is
there an association between visiting art museums and going to
sports events that is greater than what you might expect by
chance)
How to Make Your Decisions
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Under Statistics, click Chi-square, then Continue, and
OK. (Chi-square is your test statistic)
To confirm your hypothesis, you need for the obtained
value of Chi-square to be significant at the .05 level.
SPSS will print out the exact probability level for you.
You need for p to be less than .05
However, if after getting a significant Chi-square you
look over the table of observed versus expected counts
and note that the trend is going in the wrong direction
(that is, museum goers are less likely than expected to
go to sports events) then although you can say that
there may be an association, you have not established
that more art visits are associated with more sporting
event attendance.
Comparing Obtained (Count) to Expected (Expected
Count) to Determine Direction of Relationship between
the Two Variables
On the right is what your output
should look like for checking
out the direction of the
relationship. Note that within
the group of art visitors that
went to a sports event, the
counter is higher than the
expected count (orange),
while within the group of nonvisitors to art museums, the
count of those who went to
sports events is lower than the
expected count (green). (The
expected count for a cell is
obtained in your SPSS output
by multiplying the column
total by the row marginal (see
arrows).
So your obtained results are in the
direction you were expecting.
Now it remains to be seen if
this is a statistically significant
relationship
Is the Obtained Value of Chi-square
Significant?
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Is the observed positive relationship between visiting a museum or gallery
and attending a sports event statistically significant beyond the .05 level?
According to the output, you have a chi-square of 79.414 (df = 1) which is
significantly less likely than the .05 probability level. If it were not
significant, the value in the fourth column below would read .051 or larger.
So it’s fair to say that you have confirmed the hypothesis there is a
significant association between attending visiting art museums and visiting
sporting events, and that the trend is for the relationship to be positive
Chi-Square Tests
Pearson Chi-Square
Continuity Correction a
Likelihood Ratio
Fisher's Exact Test
Linear-by-Linear
Association
N of Valid Cases
Value
79.414b
78.474
80.584
79.361
df
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1
1
1
Asymp. Sig.
(2-sided)
.000
.000
.000
Exact Sig.
(2-sided)
Exact Sig.
(1-sided)
.000
.000
.000
1487
a. Computed only for a 2x2 table
b. 0 cells (.0%) have expected count less than 5. The minimum expected count is
282.17.
Difference of Means Test for Two
Levels of the IV: T Test
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For the exam, we will only consider the t test for independent
samples in doing an SPSS application. A t test is used when you
have only two groups on the IV (two categories of a nominal-level
independent variable, such as gender), and interval or ratio level
for the DV
There are three varieties of t test in your SPSS list of options
(independent, dependent or pairedsuch as pre-post comparisons,
and single sample (such as comparing a sample mean to an
assumed population mean or other known parameter)
Here’s a sample test question: Test the hypothesis that people
with a college degree ((#59) “College Degree” (degree2) watch
fewer hours of TV (the variable (#35) “Hours per Day Watching TV
(tvhours)”) than people without a college degree. Test the
hypothesis at the .01 level of confidence. Report the test statistic,
df, probability level and the means for the two levels of the
independent variable. Use the Levene test to determine which
statistic you should report
Determine the Fit between Levels of
Measurement of IV and DV and the Statistical
Test
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Look at the variables in Variable View. You see that
the IV, college degree, has two categories and that it
is a nominal level variable (ignore the “ordinal” tag
that is attached to all of the variables in the data file;
it is not true of all of them) because there are only
two discrete categories, have the degree and don’t
have the degree
Similarly, look at the DV, hours spent watching TV.
This is at least an interval level measure.
Therefore, this question is suitable for the t test,
which is for testing the impact of a two-level nominal
level IV on variation in an interval or better DV
(alternatively, testing whether the means on two
“groups” (levels of the IV) differ significantly on the
DV (are drawn from different populations)
Running a t Test for Independent
Samples in SPSS
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In Data Editor, go to Analyze/ Compare Means/
Independent Samples T-Test
Move the College Degree (degree2) variable into the
Group Variable box and click on the Define button to
assign values to levels of the variable. Use the values
from the Variable View which assign 0 to Group 1 (no
degree) and 1 to Group 2 (college degree)
Move the Hours Watching per Day Watching TV
variable into the Test Variable(s) box
Under Options, set the confidence interval to 99% and
click Continue, and then OK
Use your output to answer the question
Finding Answers to the Question in
Your SPSS T Test Output
First, check to see if the variances between the two groups in the IV are
significantly different based on Levene’s statistic: they are (note the probability
level which shows the statistic fell into the critical region). This determines
which value of t you will report. The tests shows that you can’t assume equal
variances (they are significantly different) so you have to use the value of t
which is calculated assuming unequal variances: 12.275
Independent Samples Test
Levene's Test for
Equality of Variances
F
Hours Per Day
Watching TV
Equal variances
assumed
Equal variances
not assumed
66.470
Sig .
.000
t-test for Eq uality of Means
t
df
Sig . (2-tailed)
Mean
Difference
Std. Error
Difference
99% Confidence
Interval of the
Difference
Lower
Upper
8.861
1484
.000
1.18
.134
.839
1.528
12.275
1149.777
.000
1.18
.096
.935
1.433
Levene’s test shows signfiicant difference of
variances between two levels of the IV
Answering Your Question
Group Statistics
Hours Per Day
Watching TV
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Colleg e Degree
No College degree
Colleg e degree
N
1140
346
Mean
3.17
1.99
Std. Deviation
2.392
1.217
Std. Error
Mean
.071
.065
Here are
the group
means
Here’s a way to write the answer: A test of the hypothesis that
people with a college degree would differ from people without a
college degree on number of hours spent per day watching TV
indicated that there was a significant difference in number of
hours spent watching TV (t (unequal variances) = 12.275, df =
1149.777, p < .0005) . (Note that since your test is one-tailed (you
predicted a direction, you have to “cut the probability in half” since SPSS only
reports the two-tailed test). Persons with a college degree spent an
average of 1.99 hours per week watching TV while persons
without a college degree spent an average of 3.17 hours per
day watching TV.
Univariate Analysis of Variance: an Appropriate Test of
the Impact of a 3- or More Level IV on an Interval or
Ratio Level DV: Sample Problem
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A Sample Problem for the Midterm: Test the
hypothesis that religious preference of respondent
(the variable (#27) “Religious Preference (relig)”)
has a significant impact on hours per day watching
TV (the variable (#35) “Hours per Day Watching TV
(tvhours)”) Make a decision in advance to reject the
null hypothesis (and confirm the research
hypothesis) if the obtained value of the test
statistic falls into the .01 confidence region
Sample Problem, cont’d
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Write up your results as if you were writing for a journal. Include
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The value of the test statistic
The degrees of freedom
The level of significance associated with the test statistic
Report the effect size (amount of variance accounted for, the partial eta
squared)
The statistical power associated with your test
Report the results of the test for equality of variances
Report the means for each condition (level of the variable “religious
preference”).
Run post hoc tests using Sheffe to see if there are significant pair-wise
differences in mean TV hours watched among the levels of the
independent variable (religious preference) and report which ones are
significant.
If there is anything in your printout that suggests that the Sheffe tests
might not be appropriate, run the more appropriate type of post-hoc
test. Make an assessment as to the importance of the observed
relationship between religious preference and hours watching TV based
on the effect size
What Kind of Variables Do I Have?
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First consult the Variable View in SPSS Data Editor to
find out what kinds of variables these are (look under
the value labels). You will see that religious
preference is a nominal level variable with five
categories, and tv hours is a ratio level variable. The
appropriate analysis for studying the effect of a
nominal level variable with more than two levels
(categories) on an interval or ratio level variable is an
analysis of variance (ANOVA). So you make the
decision that you will run an ANOVA and treat
religious preference as the IV and hours watching TV
as the DV
Running ANOVA in SPSS
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Go to Analyze/ General Linear Model / Univariate
Move the Religious Preference variable into the Fixed Factor(s)
Window (this is where “fixed” IVs go)
Move the Hours Per Day Watching TV Variable into the Dependent
Variable box (you are saying TV hours watched is “dependent” on
religious preference)
Don’t make any changes under Model, Contrasts, or Plots
Under Options, move Overall, Relig to the Display Means window
Also under Options/Display, select descriptive statistics, estimates
of effect size, observed power, and homogeneity tests. You know
to ask for these because the question asked you to provide them.
This will give you the mean tv hours according to religion, the
effect size (how much variance in tv hours you can explain with
religion), how much power you had to detect a difference if there
is one, and whether or not your levels of the IV have different
variances and thus you need to do the alternative tests (Tamhane,
not Sheffe, for example)
Finally under Options set the significance level to .01 and click
Continue
More ANOVA in SPSS
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Click the Post Hoc Button and move relig into the Post
Hoc Tests for window
Under equal variances assumed select Scheffe (you
will use this test if the group variances do not differ
significantly according to the Lehane test; Lehane test
will show up on your output)
Under equal variances not assumed select Tamhane
T2 test (you will use this test if the group variances
are significantly different according to the Lehane
test)
Click Continue and then OK
Consult your output to answer the question
Getting the Answers from Your SPSS
Output: F, Significance, eta square,
Power
Here is the overall F statistic and its associated level of
signifcance. It is not significant according to the .01 level
you set up because the obtained value is larger than .01,
so you can’t reject the null hypothesis
Tests of Between-Subjects Effects
Dependent Variable: Hours Per Day Watching TV
Source
Corrected Model
Intercept
RELIG
Error
Total
Corrected Total
Type III Sum
of Squares
53.308b
2741.557
53.308
7393.922
19921.000
7447.230
df
4
1
4
1478
1483
1482
Mean Square
13.327
2741.557
13.327
5.003
F
2.664
548.020
2.664
Sig .
.031
.000
.031
Partial Eta
Squared
.007
.270
.007
Noncent.
Parameter
10.656
548.020
10.656
Observed
a
Power
.522
1.000
.522
a. Computed using alpha = .01
DF
b. R Squared = .007 (Adjusted R Sq uared = .004)
Here are the partial eta squared (percent of variance
in DV explained by the IV) and the power estimate
More Answers from the SPSS output: Means
on the DV by Level of the IV; Equality of
Variances Test
Descriptive Statistics
Dependent Variable: Hours Per Day Watching TV
Religious Preference
Protestant
Catholic
Jewish
None
Other
Total
Mean
2.90
2.75
2.45
3.42
2.62
2.90
Std. Deviation
2.061
2.133
1.410
3.472
2.104
2.242
Levene's Test of Equality of Error Variancesa
Dependent Variable: Hours Per Day Watching TV
F
6.850
df1
4
df2
1478
Sig .
.000
Tests the null hypothesis that the error variance of
the dependent variable is equal across g roups.
a. Design: Intercept+RELIG
N
947
332
31
139
34
1483
Here are the group means which
show how average hours
watching TV varied as a function
of religious preference. This
table also gives you a numerical
breakdown of the religious
preference categories
Here is the Levene test of equality of
variances between levels of the IV, which
in this case is significant beyond the .001
level. This means the variances can’t be
assumed to be equal, so instead of the
Sheffe post hoc tests you use a test like
the Tamlane for unequal group variances
Post-hoc Comparisons when Equal
Variances Can’t be Assumed
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If the overall effect is not
significant the post-hoc
pairwise comparisons probably
won’t be (for example
comparing Catholics to
Protestants) but let’s look at
the output anyhow. Look at
the table in your output called
Post Hoc Tests, Religious
Preference, Multiple
Comparison (the bottom half).
We have already established
that we have to use the post
hoc tests which assume
unequal group variances, so
we will look only at the
Tamhane tests in the bottom
half of the table. Look at the
column called Significance and
you will see that none of the
tests yields a value of
significance less than .05
Significance levels
Answer to the Question: What
your “Results” Section Would Say
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Answer: To test the hypothesis that religious preference has a
significant impact on hours spent watching TV, a one-way analysis of
variance was conducted. The obtained value of F (4, 1478) of 2.664
was not significant at the .01 level. The effect size (partial eta squared)
was .007. Power to detect the effect was .522. The Levene test for the
equality of variances among the levels of the independent variable
(religious preference) found that the variances were significantly
different (F = 6.850, p < .001), suggesting that an alternative post hoc
test for pair-wise differences of means should be used. The mean TV
hours watched by religious preference were: Jewish, 2.45; “other,”
2.62; Catholic, 2.75; Protestant, 2.90; and “none,” 3.42. Tamhane
tests of post-hoc differences indicated that there were no significant
differences in TV hours watched between any levels of the independent
variable. However, power to detect a between-groups effect was low
(.522) despite the large sample size, so the issue might be revisited in
a new study with more power to detect a small effect.