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Hypothesis Testing

Presented by

Dr. Bikash Barman

Assistant Professor

Department of Geography

Malda Women’s College

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Type-I Error & Type-II Error

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Types of Hypothesis Testing:�1. Parametric Test: T-test, Z-Test and ANOVA�2. Non-Parametric Test: Chi square test, Mann-Whitney U test, Kruskal-Walis test

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Steps for Hypothesis Testing

  1. Formulate null and alternative hypothesis
  2. Specify the level of significance ( alpha level)
  3. Choose the appropriate test statistics
  4. Calculate the value of test statistics
  5. Make decision with calculated value and tabulated value

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When t-test and When z-test?

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Types of T-Test

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Steps for t-test

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Degrees of Freedom

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Formula for One Sample T-Test

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Two Sample t-test

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Decision

*If calculated t value is greater than the tabulated t value then the null hypothesis (H0)can be reject and alternative hypothesis (H1) can be accepted which means that there is a significant difference between the experimental group mean and control group mean.

Calculated t> Tabulated t= Reject H0

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Paired Sample t-test: Question

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Z-Test

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One Sample Z-test

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Critical Values for Z-Test

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TWO SAMPLE Z-TEST

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Chi-Square Test

*Test of significance based on frequencies not the parameters like mean and standard deviation

*First used by Karl Pearson in 1900

*Useful measure of comparing experimentally obtained result with those expected theoretically and based on the hypothesis

*It is denoted by χ2

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Uses of Chi-Square Test

  1. Chi-square test as a test of independence (Whether two or more attributes or variables are associated or not)
  2. Chi-square test as a test of goodness of fit (Whether or not an observed frequency distribution differs from a expected frequency distribution)

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Steps for calculation of Chi-Square Test

1. Setting up the hypothesis (H0 and H1)

2. Prepare a cross tabulation observation on two variables

3. Calculate the observed and expected values

fe= (Row Total × Column Total)/N

4. Calculate Chi-Square value

5. Calculate the Degree of freedom= (Columns-1) (Rows-1)

6. Find out the tabulated value from the Critical values of Chi-Square on alpha level 0.05 (Significance level)

7. Made the decision (If calculated chi value is greater than tabulated value than null hypothesis can be rejected and alternative hypothesis can be accepted)

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Example of �Test of Independence

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Survey �onMarital Status among the Muslim Women of Maldah District

Hypothesis Setting:

H0= There is no relationship between educational attainment and marital status

H1= There is a significant relationship between educational attainment and marital status

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Step-1: Table of Observed Values

Step-2: Table of Expected Value

fe= (Row Total × Column Total)/N

Qualification/Marital Status

Middle School

High School

Bachelor's

Master's

Ph.D.

Total

Never Married

18

36

21

9

6

90

Married

12

36

45

36

21

150

Divorced

6

9

9

3

3

30

Widowed

3

9

9

6

3

30

Total

39

90

84

54

33

300

Qualification/Marital Status

Middle School

High School

Bachelor's

Master's

Ph.D.

Never Married

90*39/300=11.7

27

25.2

16.2

9.9

Married

19.5

45

42

27

16.5

Divorced

3.9

9

8.4

5.4

3.3

Widowed

3.9

9

8.4

5.4

3.3

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Step-3: Calculation of Chi value

Observed Values (O)

Expected Values (E)

(O-E)

(O-E)2

(O-E)2/E

18

11.7

6.3

39.69

3.39

36

27

9

81

3.00

21

25.2

-4.2

17.64

0.70

9

16.2

-7.2

51.84

3.20

6

9.9

-3.9

15.21

1.54

12

19.5

-7.5

56.25

2.88

36

45

-9

81

1.80

45

42

3

9

0.21

36

27

9

81

3.00

21

16.5

4.5

20.25

1.23

6

3.9

2.1

4.41

1.13

9

9

0

0

0.00

9

8.4

0.6

0.36

0.04

3

5.4

-2.4

5.76

1.07

3

3.3

-0.3

0.09

0.03

3

3.9

-0.9

0.81

0.21

9

9

0

0

0.00

9

8.4

0.6

0.36

0.04

6

5.4

0.6

0.36

0.07

3

3.3

-0.3

0.09

0.03

 

 

 

 

23.57

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Step-4: Calculate the degree of freedom:DF= (R-1)(C-1)Where, R= Total number of Rows�C= Total number of columns��DF= (4-1) ×(5-1)�= 3 × 4�= 12

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Step-5: Find out the Critical values of chi square on significance level 0.05 = 21.026

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Step-6: Decision Rule

Calculated value= 23.57

Tabulated value= 21.03

Reject H0 if the calculated value is greater than tabulated value and accept the H1

Step-7: Conclusion

There is a significant relationship between educational attainment and marital status among the Muslim Women of Maldah district.

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Example of �Test of Goodness of Fit

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Q. A dice was thrown 60 times with the following results:

Are the data consistent with the hypothesis that the dice is unbiased?

Ans: Step-1: Setting up Hypothesis

H0= The dice is unbiased

H1= The dice is biased

Then the probability of each face is 1/6

And the expected frequency is 1/6×60= 10 for each

Face

1

2

3

4

5

6

Total

Frequency

6

10

8

13

11

12

60

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Calculated Chi-Square= 3.4

Observed Frequency (O)

Expected Frequency (E)

(O-E)

(O-E)2

(O-E)2/E

6

10

-4

16

1.6

10

10

0

0

0

8

10

-2

4

0.4

13

10

3

9

0.9

11

10

1

1

0.1

12

10

2

4

0.4

Step-2: Calculation of Chi-Square

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Degrees of Freedom= (N-1)�= (6-1)= 5

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Thank You