Chi Squared Test, tests for variance which is the difference between observed and expected numbers

Structure

O = Observed Number

E= Expected number

The numbers of degree of freedom = Number of Categories – 1

If the value is greater than the table value then null hypothesis is rejected

e.g. A gardener wanted to compare the number of worms found in his compost heap with the number in the middle of the field

results:

Compost Heap

Lawn

150

50

Null Hypothesis: There is no significant difference between the distribution of the worms in the compost heap compared to the lawn

Compost Heap

Lawn

Total

O

150

50

200

E

100

100

200

(O-E)

50

-50

0

(O-E)2

2500

-2500

0

(O-E)2

E

25

25

50

*according to the null hypothesis, there are equal quantities of the worms in both the compost and the law, therefore our “expected” (E) is the total of the Observed results divided by the 2 groups (150 + 50 = 200 200 / 2 = 100

*Remember that it is a 5% error margin, therefore use

0.05

table

*Remember that it is a 5% error margin, therefore use 0.05

Degrees of Freedom: 2 – 1 = 1

(2 Variables: Compost Heap and Lawn)

Critical Value = 3.84

The critical value is 3.84 which is lower than the , so the value is rejected therefore the null hypothesis is also rejected. There is less than a 5% probability that the difference is due to chance, so the difference is significant.

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