In: Statistics and Probability
MARITAL STATUS |
||||||
MARRIED |
DIVORCED |
WIDOWED |
SEPARATED |
NEVER MARRIED |
TOTAL |
|
VERY HAPPY |
600 |
93 |
63 |
19 |
144 |
919 |
PRETTY HAPPY |
720 |
304 |
142 |
51 |
459 |
1676 |
NOT TOO HAPPY |
93 |
88 |
51 |
31 |
127 |
390 |
TOTAL |
1412 |
485 |
256 |
101 |
730 |
2984 |
Let us assume the significance level to be 5%.
Chi-Square Independence test - Results |
(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: H0: The two variables - Marital Status and Happiness are independent Ha: The two variables - Marital Status and Happiness are dependent This corresponds to a Chi-Square test of independence. (2) Degrees of Freedom The number of degrees of freedom is df = (3 - 1) * (5 - 1) = 8 (3) Critical value and Rejection Region Based on the information provided, the significance level is α=0.05, the number of degrees of freedom is df = (3 - 1) * (5 - 1) = 8, so the critical value is 15.5073. Then the rejection region for this test becomes R={χ2:χ2>15.5073}. (4)Test Statistics The Chi-Squared statistic is computed as follows: (5)P-value The corresponding p-value for the test is p=Pr(χ2>236.4048)=0 (6)The decision about the null hypothesis Since it is observed that χ2=236.4048>χ2_crit=15.5073, it is then concluded that the null hypothesis is rejected. (7)Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the two variables - Marital Status and Happiness are dependent, at the 0.05 significance level. Conditions: a. The sampling method is simple random sampling. b. The data in the cells should be counts/frequencies c. The levels (or categories) of the variables are mutually exclusive. |
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