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HYPOTHESIS
TESTING
Chi-Square Test for Independence:
NADEEM UDDIN
ASSOCIATE PROFESSOR
OF STATISTICS
Chi-Square Test for Independence:
This lesson explains how to conduct a chi-
square test for independence. The test is
applied when you have two categorical
variables from a single population. It is
used to determine whether there is a
significant association between the two
variables.
Procedure Test Concerning for independence
1-Hypothesis
H0: The two attributes are not associated.
H1: The two attributes are associated.
2-Level of significance ο‚΅= 1% or 5% or any given
value.
3-Test of statistic 𝝌 𝟐 =
π’βˆ’π’† 𝟐
𝒆
Where o = observed value
e = estimated value
4-Critical region πœ’2
ο‚΅, π‘Ÿβˆ’1 (π‘βˆ’1)
5-Computation
6-Conclusion
If the calculated value of test statistic
falls in the area of rejection, we reject
the null hypothesis otherwise accept it.
Weather
Total
Good Bad
Result
Win 12 4 16
Draw 5 8 13
Lose 7 14 21
Total 24 26 50
Example-11:
The members of a sports team are interested in
whether the weather has an effect on their results.
They play 50 matches, with the following results:
Test the claim at 1% significance level that
the result independent of weather.
Solution:
1-Hypothesis
H0: Result independent of
weather.H1: Result dependent of
weather.
2- Level of significance ο‚΅= 0.01
3- Test of statistic πœ’2
=
π‘œβˆ’π‘’ 2
𝑒
4. Critical Region
πœ’2
ο‚΅, π‘Ÿβˆ’1 (π‘βˆ’1)
πœ’2
0.01, 3βˆ’1 (2βˆ’1)
πœ’2
ο‚΅,2 = 9.21 9.21
5. Computation
Good Bad Total
Win 12
24 Γ— 16
50
= 7.68
4
26 Γ— 16
50
= 8.32 16
Dra
w
5
13 Γ— 24
50
= 6.24 8
26 Γ— 13
50
= 6.76 13
Loos
e
7
24 Γ— 21
50
= 10.08
14
26 Γ— 21
50
= 10.92
21
Total 24 26 50
O e
12 7.68 2.43
5 6.24 0.246
7 10.08 0.941
4 8.32 2.243
8 6.76 0.227
14 10.92 0.868
 
2
o e
e
ο€­
2
6.956cal ο€½
6. Conclusion:
Accept H0.
And conclude that the team’s
result are independent of the weather.

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Hypothesis testing chi square test for independence

  • 1. HYPOTHESIS TESTING Chi-Square Test for Independence: NADEEM UDDIN ASSOCIATE PROFESSOR OF STATISTICS
  • 2. Chi-Square Test for Independence: This lesson explains how to conduct a chi- square test for independence. The test is applied when you have two categorical variables from a single population. It is used to determine whether there is a significant association between the two variables.
  • 3. Procedure Test Concerning for independence 1-Hypothesis H0: The two attributes are not associated. H1: The two attributes are associated. 2-Level of significance ο‚΅= 1% or 5% or any given value. 3-Test of statistic 𝝌 𝟐 = π’βˆ’π’† 𝟐 𝒆 Where o = observed value e = estimated value
  • 4. 4-Critical region πœ’2 ο‚΅, π‘Ÿβˆ’1 (π‘βˆ’1) 5-Computation 6-Conclusion If the calculated value of test statistic falls in the area of rejection, we reject the null hypothesis otherwise accept it.
  • 5. Weather Total Good Bad Result Win 12 4 16 Draw 5 8 13 Lose 7 14 21 Total 24 26 50 Example-11: The members of a sports team are interested in whether the weather has an effect on their results. They play 50 matches, with the following results: Test the claim at 1% significance level that the result independent of weather.
  • 6. Solution: 1-Hypothesis H0: Result independent of weather.H1: Result dependent of weather. 2- Level of significance ο‚΅= 0.01 3- Test of statistic πœ’2 = π‘œβˆ’π‘’ 2 𝑒
  • 7. 4. Critical Region πœ’2 ο‚΅, π‘Ÿβˆ’1 (π‘βˆ’1) πœ’2 0.01, 3βˆ’1 (2βˆ’1) πœ’2 ο‚΅,2 = 9.21 9.21
  • 8. 5. Computation Good Bad Total Win 12 24 Γ— 16 50 = 7.68 4 26 Γ— 16 50 = 8.32 16 Dra w 5 13 Γ— 24 50 = 6.24 8 26 Γ— 13 50 = 6.76 13 Loos e 7 24 Γ— 21 50 = 10.08 14 26 Γ— 21 50 = 10.92 21 Total 24 26 50
  • 9. O e 12 7.68 2.43 5 6.24 0.246 7 10.08 0.941 4 8.32 2.243 8 6.76 0.227 14 10.92 0.868   2 o e e ο€­ 2 6.956cal ο€½ 6. Conclusion: Accept H0. And conclude that the team’s result are independent of the weather.