Question

Calculate the test statistic.

Chapter 12. Chi-Square Test Saved Help Save & Exit Su Check my w Given the following contingency table, conduct a test for Independence at the 5% significance level. (You may find It useful to reference the appropriate table: chi-square table or F table) Variable A points Variable B 36 35 31 51 eBook a. Choose the null and alternative hypotheses. Hint Ho: The two variables are independent.; HA: The two variables are dependent. O Ho: The two variables are dependent.; HA: The two variables are independent. Print b. Calculate the value of the test Statist(Round intermediate calculations to at least 4 decimal places and final answer to 3 decimal places.) References Test statistic c. Find the p-value. 0.05 s p-value < 0.10 O 0.025 s p-value < 0.05

Answer 1

The table for observed frequency is ,

	Variable A
Variable B	1	2	Total
1	36	31	$(A_{1})=67$
2	35	51	$(A_{2})=86$
Total	$(B_{1})=71$	$(B_{2})=82$	N=153

The table for expected frequency is ,

$e_{ij}=\frac{(A_{i})(B_{j})}{N}$ ; i=1,2 and j=1,2

	Variable A
Variable B	1	2
1	31.0915	35.9085
2	39.9085	46.0915

The table for $\frac{O_{ij}^2}{e_{ij}}$ is ,

	Variable A
Variable B	1	2
1	41.68341	26.76247
2	30.69522	56.43123

Therefore , the test statistic is ,

$\chi^2=\sum_{i}\sum_{j}\frac{O_{ij}^2}{e_{ij}}-N=155.5723-153=2.5723$