Chi Square

Classifying yourself as studious or not.

Yes No Total58 42 100 Are they significantly different?

12 18 30

46 24 70

58 42 100

Yes No Total

Rea

d ah

ead Yes

No

Total

Does reading ahead make a difference? Independence?

Studious

One variable

Choice of PSYA01 Section

L01 L02 L03 L30 Total25 40 15 36 116

Is this more than a chance difference?

22

( )O E

E

O = the observed frequency in a categoryE = the expected frequency in that category

We may expect different categories to have the same frequency if chance alone is at work.

( ) ( ) ( ) ( )2 5 2 9

2 9

4 0 2 9

2 9

1 5 2 9

2 9

3 6 2 9

2 9

2 2 2 2

= .55 + 4.17 + 6.79 + 1.69

= 13.17 Is this significant? Go to the table. df = k - 1

Two Variable

Are the two variables independent of each other?

Contingency Table

37 16 53

47 62 109

84 78 162

Nat. Sci. Soc. Sci Totals

Male

Female

Totals

Career Choice

Marginal Totals

The key is determining the expected frequencies of the four observed frequencies (the 4 colored cells).

contingency is another word for “possibility”

So this is a “table of possibilities”

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

84 78 162

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)

27.48 25.52

56.52 52.48

Expected Frequencies Now we can use…..

22

( )O E

E

Expected Frequencies and Alternative Calculations

ER C

Nij

i j R = the row totalC = the column total

E 11

5 3

1 6 22 7 4 8 (8 4 )

. E 1 2

5 3 7 8

1 6 22 5 5 2 ( )

. E 2 2

1 0 9 7 8

1 6 25 2 4 8 ( )

.E 2 1

1 0 9

1 6 25 6 5 2 (8 4 )

.

( . )

.

( . )

.

( . )

.

( . )

.

3 7 2 7 4 8

2 7 4 8

1 6 2 5 5 2

2 5 5 2

4 7 5 6 5 2

5 6 5 2

6 2 5 2 4 8

5 2 4 8

2 2 2 2

= 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 rowsC = 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.