Tutorial two
Answers
Exercise One:
With this question you are comparing proportions.
Since the categories are both nominal level i.e. men and women and seen/not seen the
appropriate test is a chi square test.
The Null hypothesis is that there are no differences between men and women as to
whether they’ve seen the movie.
The Alternative hypotheses are
One tailed: men have seen the movie more than women
Two tailed: There is a difference in gender as to whether they’ve seen the movie.
The critical value can be set at .05
Chi-Square Tests
Pears on Chi-Square
Continuity Correctiona
Likelihood Ratio
Fisher's Exact Test
N of Valid Cas es
Value
.586 b
.082
.600
df
1
1
1
Asymp. Sig.
(2-s ided)
.444
.774
.439
Exact Sig.
(2-s ided)
Exact Sig.
(1-s ided)
.642
.392
20
a. Computed only for a 2x2 table
b. 2 cells (50.0%) have expected count less than 5. The minimum expected count is
2.80.
The continuity correction corrects for an overestimate of the chi-square value in a 2x2
table. This is the figure to use when examining for significance. In the table above the
corrected value is .082 with an associated significance level of .774. To be significant,
the Sig. Value needs to be .05 or smaller. In this case .774 is larger than the alpha value
of .05 therefore we can conclude that our result is not significant. I.e. there is no gender
difference in terms of who has seen the movie.
Note however, that two cells have expected counts less than 5, which violates one of the
assumptions of the Chi square. In this case we should use Fisher’s Exact test.
For the second part of the exercise you need to collapse a ratio variable (age) into a
categorical variable. (newage)
In the data window click open the Transform menu, then click on Recode into Different
variables.
Click on the variable (age) and move it onto the box labeled Input
Output
Variable
Type in a new name for this grouped variable e.g. Agegrp.
Type in a label description and then click on Change
Then click on Old and New Values
Under the Old Values section click on Range lowest through ________ and enter 27
In the new Values section enter a value of 1 and then click on Add
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Going back to the Old Values Section click on Range _________ through highest and
enter 28 in the space
In the new values side enter 2 into the value and then Add.
And then continue
A new variable should appear in the data window which can now be used to run a chi
square test to see if age has a difference.
age group * Seen Matrix Crosstabulation
age group
1.00
2.00
Total
Seen Matrix
no
yes
1
10
3.9
7.2
9.1%
90.9%
14.3%
76.9%
5.0%
50.0%
6
3
3.2
5.9
66.7%
33.3%
85.7%
23.1%
30.0%
15.0%
7
13
7.0
13.0
35.0%
65.0%
100.0%
100.0%
35.0%
65.0%
Count
Expected Count
% within age group
% within Seen Matrix
% of Total
Count
Expected Count
% within age group
% within Seen Matrix
% of Total
Count
Expected Count
% within age group
% within Seen Matrix
% of Total
Total
11
11.0
100.0%
55.0%
55.0%
9
9.0
100.0%
45.0%
45.0%
20
20.0
100.0%
100.0%
100.0%
Chi-Square Tests
Pears on Chi-Square
Continuity Correctiona
Likelihood Ratio
Fisher's Exact Test
N of Valid Cas es
Value
7.213 b
4.904
7.739
df
1
1
1
Asymp. Sig.
(2-s ided)
.007
.027
.005
Exact Sig.
(2-s ided)
Exact Sig.
(1-s ided)
.017
.012
20
a. Computed only for a 2x2 table
b. 2 cells (50.0%) have expected count less than 5. The minimum expected count is
3.15.
Since two cells have an expected cell frequency of less than 5 we have to use Fisher’s
exact test the significance factor is .017 which is less than .05 so we can conclude that
there is a significant difference in age as to who has seen the movie. More specifically
that older people, regardless of gender are less ,likely to have seen it.
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Exercise Two
In this case we want to test whether there is a difference in means between the first and
second test scores.
Because the scores are not independent; it is the same person taking the score. The
appropriate test is a paired-samples t-test.
The null hypothesis here is that there is no difference between the scores after social
skills training.
The alternative hypothesis is that there is (or that there is an improvement)
From the Analyze menu click on Compare Means and then Paired Samples T test
Click on the two variables of interest and move them into the paired variables box.
Click OK
Paired Samples Statistics
Pair
1
Mean
24.69
26.88
1st SE tes t s core
2nd SE tes t s core
N
16
16
Std. Deviation
3.400
3.384
Std. Error
Mean
.850
.846
Paired Samples Test
Paired Differences
Mean
Pair
1
1st SE tes t s core 2nd SE tes t s core
-2.19
Std. Deviation
Std. Error
Mean
2.562
.640
95% Confidence
Interval of the
Difference
Lower
Upper
-3.55
-.82
t
-3.416
df
Sig. (2-tailed)
15
.004
In this example you need to look at the last column Sig. This is the probability value.
Since this value is less than .05 we can conclude that there is a significant difference in
the scores from time one to time two. She can conclude that social skills training for six
weeks improves social skills learning.
Exercise three
There are two things that you need to do here. You need to describe your data and also
decide whether any differences are statistically significant.
It is best to do these things separately.
The first part of this questions asks you to compare means in a table format.
From the Analyze menu click on compare means. And then Means\
The independent variable in this case is occupation
Respiratory and Gastric problems are the dependent variables.
Move the variables across into the appropriate boxes.
On the options menu ensure that SD and number of cases will be calculated.
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Report
Occupation
1
2
Total
Mean
N
Std. Deviation
Mean
N
Std. Deviation
Mean
N
Std. Deviation
Res piratory
Dis orders
1.82
11
1.722
4.00
9
2.550
2.80
20
2.353
Gas tric
Dis orders
4.64
11
2.976
2.33
9
1.658
3.60
20
2.683
For question 2 you want to look just at plumbers and then at bankers. This means that you
will have to split your data file
From the Data view open the Data menu and click on split file option. Click on Compare
Groups and specify the grouping variable i.e. occupation.
Click on OK
Until this option is turned off any tests that will be performed will be performed on the
two groups separately.
The next part asks you to see whether plumbers take more days off due to respiratory or
stomach problems. Since the people involved are the same in both cases i.e. not
independent you need to run a paired samples t test.
See the procedure as above.
This time however the result will be for both plumbers and bankers in one table.
Paired Samples Test
Paired Differences
Occupation
1
2
Std. Deviation
Std. Error
Mean
-2.82
3.763
1.135
-5.35
-.29
-2.484
10
.032
1.67
2.739
.913
-.44
3.77
1.826
8
.105
Mean
Pair
1
Pair
1
Res piratory Disorders
- Gas tric Dis orders
Res piratory Disorders
- Gas tric Dis orders
95% Confidence
Interval of the
Difference
Lower
Upper
t
df
Sig. (2-tailed)
For plumbers there is a significant difference, but not for bankers.
To turn the split data file function off, go back into the Data menu and click on the first
radio button: analyze all cases, do not create groups.
The last part of the question asks you to simply compare days off between bankers and
plumbers.
This will require collapsing days off due to stomach problems and due to respiratory
problems into a new variable.
From the transform menu click on compute
Give a name to the new target variable e.g. Daysoff
Move the respiratory variable over to the numeric expression box,
Then insert the + symbol and move the gastric problem variable over and click OK
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The next step involves comparing the means of the two groups (Bankers and plumbers).
Since the two groups are independent the appropriate test is an independent samples t-test
The null hypothesis is that there are no differences between the two groups
The alternative hypothesis is that there is a difference.
From the Analyze menu click on Compare Means and then Independent Samples t-test.
Move the dependent (Continuous) variable (i.e. daysoff) into the area labeled test
variable.
Move the independent (categorical) variable into the section labeled grouping variable.
Click on define groups and type the numbers used in the data set to code for each group.
I.e. 1 for plumbers and 2 for bankers.
Click on Continue and then OK
Group Statistics
total days off
Occupation
1
2
N
11
9
Mean
6.4545
6.3333
Std. Deviation
3.07778
3.31662
Std. Error
Mean
.92799
1.10554
Independent Samples Test
Levene's Test for
Equality of Variances
F
total days off
Equal variances
ass umed
Equal variances
not as sumed
.049
Sig.
.827
t-tes t for Equality of Means
t
df
Sig. (2-tailed)
Mean
Difference
Std. Error
Difference
95% Confidence
Interval of the
Difference
Lower
Upper
.085
18
.933
.1212
1.43207
-2.88745
3.12987
.084
16.637
.934
.1212
1.44339
-2.92914
3.17157
Levene’s test for equality of variances tests to see whether the variation of the two groups
is the same (an assumption of the t-test). The outcome of this test determines which of the
t-values you need to use.
If the Sig. value is larger than .05 you should use the first line in the table. I.e. equivalent
variances assumed.
If it less than .05 (e.g. .001) this means that the variances of the two groups is not the
same and the t-statistic you need to use is the second one.
To find out if there is a significant difference between the two groups use the Sig (2tailed) column. Use the bottom one since equivalence of variance assumption has not
been violated.
If the value is equal or less than .05 then there is a significant difference in the mean
scores.
Since the difference is above .05 we can assume that there are no significant differences
in number of days taken off between plumbers and bankers.
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