Hypothesis Testing
Presented by
Dr. Bikash Barman
Assistant Professor
Department of Geography
Malda Women’s College
Type-I Error & Type-II Error
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
Steps for Hypothesis Testing
When t-test and When z-test?
Types of T-Test
Steps for t-test
Degrees of Freedom
Formula for One Sample T-Test
Two Sample t-test
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
Paired Sample t-test: Question
Z-Test
One Sample Z-test
Critical Values for Z-Test
TWO SAMPLE Z-TEST
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
Uses of Chi-Square Test
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)
Example of �Test of Independence
Survey �on �Marital 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
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 |
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 |
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��
Step-5: Find out the Critical values of chi square on significance level 0.05 = 21.026
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.
Example of �Test of Goodness of Fit
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 |
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
Degrees of Freedom= (N-1)�= (6-1)= 5
Thank You