Question

Use a .05 significance level and the table below to test the claim that gender and left-handedness are independent.

Left-Handed Not Left-handed

Male	83	17
Female	184	16

Step 1: Claim: H0: Ha:

Step 2: Significance level: alpha=

Step 3: Test:

Values for required entry on calculator:

P-value:

Step 4:

Decision and Reason:

Conclusion (Complete sentence, in context)

Answer 1

Step 1: Claim : the Gender and left-handedness are independent.

Null Hypothesis H0: The gender and left-handedness are independent.

Alternative Hypothesis H1: The gender and  left-handedness are not independent.

Step 2: Significance Level

Step 3: Under H0, the expected frequencies are

E(83) = 267*100/300 = 89

E(17) = 100-89 = 11

E(184) = 267-89 = 178

E(16) = 200-178 = 22

The Calculation table for Chi square test for independence is given below:

	O	E	(O-E)^2/E
	83	89	0.404
	17	11	3.273
	184	178	0.202
	16	22	1.636
Total	300	300	5.516

Degrees of freedom = (r-1)*(c-1) = 1

The P-Value is 0.0188

Step 4:

Decision and Reason : Reject H0 because p value is less than significance level

Conclusion : Hence at 5% Significance level, we donot have enough evidenec to support that he claim that gender and left-handedness are independent.