Transcript Document

Chi Square

Classifying yourself as studious or not.
Yes
58
No
42
Total
100
Are they significantly different?
Studious
Read ahead
Yes
No
Total
12
18
30
No
46
24
70
Total
58
42
100
Yes
Does reading ahead make a difference? Independence?
One variable


Choice of PSYA01 Section
L01
25
L02
40
L03
15
L30
36
Total
116
Is this more than a chance difference?
(O  E ) 2
 
E
2
O = the observed frequency in a category
E = the expected frequency in that category
We may expect different categories to have the same frequency if chance alone is at work.

(25  29) 2 (40  29) 2 (15  29) 2 (36  29) 2
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
29
29
29
29
= .55 + 4.17 + 6.79 + 1.69
= 13.17
Is this significant? Go to the table.
df = k - 1
Two Variable
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Are the two variables independent of each other?
Contingency Table
Career Choice
Nat. Sci.
Soc. Sci
Totals
37
16
53
Female
47
62
109
Totals
84
78
162
contingency is another
word for “possibility”
Male
So this is a
“table of
possibilities”
Marginal Totals
The key is determining the expected frequencies of the four observed frequencies (the 4 colored cells).
Two Variables – Expected Frequencies
Testing the null hypothesis that the variables are independent
We know that the probability of the joint occurrence of two independent events is the
product of their separate probabilities.
37
16
53
47
62
109
e.g., (84/162) X (53/162) = .1696 or 16.96% of
the observations are expected in the upper left
hand cell.
But, N (162) times = 27.48 (expected frequency)
84
78
162
Expected Frequencies
27.48
25.52
Now we can use…..
(O  E ) 2
 
E
2
56.52
52.48
Expected Frequencies and Alternative Calculations
Eij 
Ri C j
N
53(84)
E11 
 27.48
162


R = the row total
C = the column total
E12 
53(78)
 2552
.
162
E21 
109(84)
 56.52
162
E22 
109(78)
 52.48
162
(37  27.48) 2 (16  2552
. ) 2 (47  5652
. ) 2 (62  52.48) 2




27.48
2552
.
5652
.
52.48
= 3.30 + 3.55 + 1.60 + 1.78
= 10.18
Is the probability of this Chi-Square value (or larger) less than .05?
Degrees of Freedom for Two Variables
df = (R-1)(C-1)
R = the number of rows
C = the number of columns
With our example: df = (2-1)(2-1) = 1
Go to Chi-Square Table and you find that the critical value is 3.84.
Our Chi-Squared obtained must be larger than 3.84 for us to reject the null hypothesis.
What was the null hypothesis?

Phi Coefficient


Will establish (at the .05 alpha level) whether two variables are related.
A significant Chi-Square means we reject the null hypothesis (which
assumes that the two variable are independent. We feel we have evidence
That the two variable are related.
Gives the numerical value to the relation. The value can range from
zero to one. Zero meaning no relation at all (independence) and one
indicating a prefect relations. If you know one variable’s value you,
you can perfectly predict the value of the other variable.

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