Chi square- a non parametric test

1. NAME: NAYANIKA SENGUPTA
AFFILIATION: BANARAS HINDU
UNIVERSITY
APPLICATION NO. :2ddc6ae1e59611e9a85ed95f3e0d8b9f

2. CHI SQUARE

3. PARAMETRIC AND NON
PARAMETRIC STATISTICAL
TESTS
Parametric tests are the ones which specifies certain
conditions about the parameters of the population from
which the sample is taken.
Ex- t test, F ratio
Non parametric tests are the ones which do not specify
any conditions about the parameters of the population
from which the sample has been taken.
Ex Chi Square, Median Test, Kruskal Wallis etc.

4. CHI SQUARE
• IT INVOLVES NO ASSUMPTION REGARDING THE NORMALCY OF
DISTRIBUTION OR HOMOGENEITY OF VARIANCES.
• CHI SQUARE TESTS ARE USED FOR CATEGORICAL DATA.
• IT CAN BE USED TO ESTIMATE HOW CLOSELY THE DISTRIBUTION
OF A CATEGORICAL VARIABLE MATCHES AN EXPECTED
DISTRIBUTION ( THE GOODNESS-OF-FIT TEST)
• IT IS ALSO USED TO ESTIMATE WHETHER TWO CATEGORICAL
VARIABLES ARE INDEPENDENT OF ONE ANOTHER ( THE TEST OF
INDEPENDENCE)

5. GOODNESS-OF-FIT TEST
The χ² test formula for goodness of fit is :
χ² =
o − ⅇ 2
ⅇ
Where,
O = observed frequency
e = expected frequency

6. 2 CONDITIONS WHEN NULL HYPOTHESIS IS
REJECTED.
When the probability value ( P value) is lesser than the
specified significance level.
When the χ² value is greater than the critical value of the
specified significance level.

7. GOODNESS-OF-FIT TEST
In the game rock-paper-scissors, Kenny expects to win, tie and lose
with equal frequency. He plays rock-paper-scissors often, but he
suspected his own game were not following that pattern, so he took a
random sample of 24 games and recorded the outcomes. Here are his
results:
He wants to use these results to carry out a χ² goodness-of-fit test to
determine if the distribution of his outcomes disagrees with an even
distribution.
Outcome Win Loss Tie
Games 4 13 7

8. HO: ALL OF THE OUTCOMES ARE OF EQUAL PROBABILITY
HA: ALL OF THE OUTCOMES ARE NOT OF EQUAL PROBABILITY
Χ² =
4−8 2
8
+
13−8 2
8
+
7−8 2
8
Χ² = 5.25
df = 2
Probability value = 0.05 < 𝑃 𝑣𝑎𝑙𝑢𝑒 < 0 ⋅ 10
Fails to reject the null hypothesis.
Outcome Win Loss Tie
Games 4 13 7
Expected 8 8 8

9. TAKING INTO CONSIDERATION
THE SIGNIFICANT DIFFERENCE
BETWEEN THE EXPECTED AND
OBSERVED OUTCOME.
Ho – There is no significant difference between the
expected and observed outcome.
Ha - There is significant difference between the
expected and observed outcome.
α - 0.05
χ² - 5.25 df = 2
At significance level of 0.05 with df of 2, the expected outcome is not
significantly different from the observed outcome. Therefore, the null
hypothesis holds.

10. TEST OF INDEPENDENCE
THE CHI-SQUARE TEST OF INDEPENDENCE IS USED TO DETERMINE IF THERE IS A
SIGNIFICANT RELATIONSHIP BETWEEN TWO NOMINAL (CATEGORICAL) VARIABLES.
Right foot
longer
Left foot
longer
Both feet
same
Total
Right hand
longer
11 3 8 22
Expected 5.5 5.5 11
Left hand
longer
2 9 14 25
Expected 6.25 6.25 12.5
Both hands
same
12 13 28 53
Expected 13.25 13.25 26.5
Total 25 25 50 100

11. • Ho = There is no association between hand and foot length.
• Ha = There is association between hand and foot length.
χ² =
11 − 5.5 2
5 ⋅ 5
+
3 − 5 ⋅ 5 2
5.5
+
8 − 11 2
11
+
2 − 6 ⋅ 25 2
6 ⋅ 25
+
9 − 6 ⋅ 25 2
6 ⋅ 25
+
14 − 12 ⋅ 5 2
12 ⋅ 5
+
12 − 13 ⋅ 25 2
13 ⋅ 25
+
13 − 13 ⋅ 25 2
13 ⋅ 25
+
28 − 26 ⋅ 5 2
26 ⋅ 5
= 11.942
df = 4
α - 0.05
There is significant difference between
expected and observed values of foot and hand length.
Thus, the null hypothesis is rejected and concluding that there is association
between hand and foot length.

12. ACKNOWLEDGEMENT
• Academic Writing Course (SWAYAM)