Chi-Square Part II Fenster Chi-Square Part II  Let us see how this works in another example. Attitudes towards Research Attitudes Towards Statistics Favorable Neither favorable nor unfavorable Favorable Unfavorable Row Totals Neither favorable nor unfavorable Unfavorable Col.

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Transcript Chi-Square Part II Fenster Chi-Square Part II  Let us see how this works in another example. Attitudes towards Research Attitudes Towards Statistics Favorable Neither favorable nor unfavorable Favorable Unfavorable Row Totals Neither favorable nor unfavorable Unfavorable Col.

Chi-Square Part II
Fenster
Chi-Square Part II

Let us see how this works in another example.
Attitudes towards Research
Attitudes
Towards
Statistics
Favorable
Neither
favorable nor
unfavorable
Favorable
Unfavorable
Row Totals
9
26
13
48
Neither
favorable nor
unfavorable
19
75
83
177
Unfavorable
16
56
110
182
Col. Totals
44
157
206
407
Chi-Square Part II
It has been argued that people with
favorable attitudes towards research
tend to have favorable attitudes
towards statistics.
 Question: If we knew the attitudes
towards research of a respondent,
can we predict the attitude toward
statistics?

Chi-Square Part II
Step 2
 H1: Knowledge of attitudes toward
research does help us predict
attitudes towards statistics.
 Step 1
 HO: Knowledge of attitudes toward
research does not help us predict
attitudes towards statistics.

Chi-Square Part II






Selecting a significance level: Let’s use
=.05. This gives us a χ2 critical of 9.488.
Your book says the χ2 critical of 9.5.
Step 4: Collect and summarize sample
data.
We will use the chi-square test with 4
degrees of freedom.
Why four? df=(r-1) X (c-1)
We have 3 rows and 3 columns.
so we get df= (3-1) X (3-1)= 2 X 2=4
Chi-Square Part II
If we find a χ2 greater than or equal
to 9.5 we reject the null hypothesis
and conclude that attitudes towards
research can predict attitudes
towards statistics.
 If we find a χ2 less than 9.5 we fail to
reject the null hypothesis and
conclude attitudes towards research
cannot predict attitudes towards
statistics.

Calculation of Expected Frequencies

Expected frequencies= (Row total) X (Column
Total)
Grand Total
Calculation of Expected Frequencies
Cell a – Favorable attitudes towards
both research and statistics.
 (44) X (48) = 5.18
407

Calculation of Expected Frequencies
Cell b – Neither favorable or
unfavorable attitudes towards
research, favorable attitudes towards
statistics.
 (157) X (48) = 18.51
407

Calculation of Expected Frequencies
Cell c –Unfavorable attitudes towards
research, favorable attitudes towards
statistics
 (206) X (48) = 24.29
407

Calculation of Expected Frequencies
Cell d – Favorable attitudes towards
research, neither favorable or
unfavorable attitudes towards
statistics

(44) X (177) = 19.13
407

Calculation of Expected Frequencies
Cell e - Neither favorable or
unfavorable attitudes towards both
statistics and research

(157) X (177) = 68.27
407

Calculation of Expected Frequencies
Cell f – Unfavorable attitudes
towards research, neither favorable
or unfavorable attitudes towards
statistics

(206) X (177) = 89.58
407

Calculation of Expected Frequencies
Cell g – Favorable attitudes towards
research, unfavorable attitudes
towards statistics

(44) X (182) = 19.67
407

Calculation of Expected Frequencies
Cell h - Neither favorable or
unfavorable attitudes towards
research, unfavorable attitudes
towards statistics

(157) X (182) = 70.20
407

Calculation of Expected Frequencies
Cell i – Unfavorable attitudes towards
both research and statistics

(206) X (182) = 92.11
407

So we set up our chi-square table
Cell
f observed f expected
f observed-f
expected
(i.e.,
RESIDUAL
S)
(f observed-f
expected)2
(f observed-f
expected)2/f
expected
a
9
5.18
3.82
14.59
2.81
b
26
18.51
7.49
56.1
3.03
c
13
24.29
-11.29
127.46
5.25
d
19
19.13
-0.13
0.17
0.008
e
75
68.27
6.73
45.29
0.6
f
83
89.58
-6.58
43.29
0.5
g
16
19.67
-3.67
13.46
0.67
h
56
70.20
-14.2
201.64
2.87
i
110
92.11
17.89
320.05
3.5
Total
407
407.00
0.00
20.2
Hypothesis Testing with Chi-Square
Step 5: Making a decision
2
 Χ observed= 20.2
 χ2 critical= 9.488.
 Decision: REJECT HO, and conclude
that attitudes towards research allow
us to predict attitudes towards
statistics.

Hypothesis Testing with Chi-Square
Notes about chi-square:
 (1) Σ (f observed - f expected)=0.
 The RESIDUALS ALWAYS SUM TO
ZERO.
 If
Σ (f observed - f expected) does
not equal zero (within rounding
error), you have made a calculation
error. Recheck your work.

Hypothesis Testing with Chi-Square

The chi-square test itself cannot tell
us anything about directionality. One
way to get directionality in the chisquare is to look at the (f observed- f
expected) column. We see that
certain cells occur much less
frequently than we would expect.
Hypothesis Testing with Chi-Square

For example cell c (unfavorable attitudes
towards research but favorable attitudes
towards statistics) occurs much less
frequently than we would expect on the
basis of chance.
Analysis of Residuals
Cell
f observed f expected
f observed-f
expected
(i.e.,
RESIDUAL
S)
(f observed-f
expected)2
(f observed-f
expected)2/f
expected
a
9
5.18
3.82
14.59
2.81
b
26
18.51
7.49
56.1
3.03
c
13
24.29
-11.29
127.46
5.25
d
19
19.13
-0.13
0.17
0.008
e
75
68.27
6.73
45.29
0.6
f
83
89.58
-6.58
43.29
0.5
g
16
19.67
-3.67
13.46
0.67
h
56
70.20
-14.2
201.64
2.87
i
110
92.11
17.89
320.05
3.5
Total
407
407.00
0.00
20.2
Hypothesis Testing with Chi-Square


We can also see that three cells that
capture consistency of attitudes between
research and statistics (cell a favorable
attitudes for both, cell e neither favorable
or unfavorable attitudes towards both, cell
i unfavorable attitudes for both) all have a
positive values for (f observed- f
expected).
Those three cells are consistent with the
(unstated and untested) hypothesis that
individuals tend to have similar attitudes
for both research and statistics
Hypothesis Testing with Chi-Square

Only by examining the (f observed- f
expected) can we give any statement
on the directionality of the
relationship. [We could also analyze
the column percentages as we move
across categories of the independent
variable to give us insight on
directionality.]
Hypothesis Testing with Chi-Square

3) In this example, why do we get
statistical significance? We can say
that the cells d, e, f and g do not
contribute to the statistical
significance of the overall
relationship. The individual chisquare values for these four cells are
all very small. The overall
relationship is significant because of
the other cells.
Analysis of Residuals
Cell
f observed f expected
f observed-f
expected
(i.e.,
RESIDUAL
S)
(f observed-f
expected)2
(f observed-f
expected)2/f
expected
a
9
5.18
3.82
14.59
2.81
b
26
18.51
7.49
56.1
3.03
c
13
24.29
-11.29
127.46
5.25
d
19
19.13
-0.13
0.17
0.008
e
75
68.27
6.73
45.29
0.6
f
83
89.58
-6.58
43.29
0.5
g
16
19.67
-3.67
13.46
0.67
h
56
70.20
-14.2
201.64
2.87
i
110
92.11
17.89
320.05
3.5
Total
407
407.00
0.00
20.2
Hypothesis Testing with Chi-Square

Chi-square allows us to decompose
the overall relationship into its
component parts. This
decomposition allows us to assess
whether all categories contribute to
the significance of the overall
relationship.
Hypothesis Testing with Chi-Square
Limitations for χ2
 So far we have stressed the virtues
for χ2 such as weak assumptions,
and a statistical significance test
appropriate for nominal level data.
This is why chi-square is so popular.
 There are two limitations for χ2, one
minor and one major.

Hypothesis Testing with Chi-Square





Minor Limitation
When the expected cell frequency is less
than 5, χ2 rejects the null hypothesis too
easily. (Note: this means the EXPECTED
frequency and NOT the OBSERVED
frequency).
Solution: Use Yates' correction
Yates’ correction
Take the | (f observed- f expected) | -0.5
Hypothesis Testing with Chi-Square
Major Limitation
We have set up a null hypothesis that
there is no relationship between two
variables and have tried to reject this
hypothesis.
We refer to a relationship as being
statistically significant when we have
established, subject to the risk of type I
error, that there is a relationship between
two variables.
But does rejecting the null hypothesis
mean the relationship is significant in the
sense of being a strong or an important




Hypothesis Testing with Chi-Square
Remember significance levels are
dependent upon sample size.
 Let us say that you wanted to
investigate the relationship between
gender and level of tolerance. You
had no money to investigate this
relationship, so you handed out
questionnaires around UML and
found the following:

Hypothesis Testing with Chi-Square
Gender
Attitudes
towards racial
tolerance
High
Males
Females
Row Totals
24
26
50
Low
26
24
50
Column Totals
50
50
100
Hypothesis Testing with Chi-Square





Is there a significant relationship between
gender and attitudes towards racial
tolerance?
Let us use α=.05.
We have one degree of freedom.
χ2 critical=3.8. χ2 observed=0.16.
Since χ2 observed (0.16) < χ2 critical
(3.8), we FAIL to reject the null
hypothesis and conclude that gender does
not help us predict to attitudes towards
racial tolerance.
Now let us say you had an extremely ambitious
study and you found the following relationship
Gender
Attitudes towards
racial tolerance
Males
Females
Row Totals
High
2400
2600
5000
Low
2600
2400
5000
Column Totals
5000
5000
10000
Hypothesis Testing with Chi-Square





Is there a significant relationship between
gender and attitudes towards racial
tolerance?
Let us use α=.05.
We have one degree of freedom.
χ2 critical=3.8, χ2 observed=16.0.
Since χ2 observed (16.0) > χ2 critical
(3.8), we easily reject the null hypothesis
and conclude that gender does help us
predict to attitudes towards racial
tolerance.
Hypothesis Testing with Chi-Square

χ2 is sensitive to the number of cases
in the sample. Even though the
proportions in the cells remain
unchanged, the new χ2 is 100 times
the old chi-square because we have
100 times the number of cases.
Hypothesis Testing with Chi-Square
Corrections for the sample size
problem
 Pearson's contingency coefficient
(You can ask for the Contingency
Coefficient with SPSS CROSSTABS’
output).

Hypothesis Testing with Chi-Square
χ2
χ2 + N
 where N=total number of cases in
sample
 Problem with C: Cannot attain 1.0 in
perfect relationship.
 As the syllabus says, there is no
ideal solution to the sample size
problem with chi-square.

C=