Question

12. The observed frequencies and calculated expected frequencles in parenthesis) of 200 students on their preference for a university health plan is shown Plan A 40(36) 20(24) Plan B 50(54) 40(36) No Preference 0(30) 20(20) Female Male Expected frequencies are in parenthesis. You are to do test of independence with the null hypothesis, Ho: The plans A, B, and no preference are independence of gender (male and female). a) Find the degree of freedom. 2 b) Find the critical value at the d-0S level. c) Compute the test statistic, X2-3 d) Do you reject Ho at-05? e) What conclusion can you make?

Answer 1

a)

degrees of freedom (df) = (r-1)*(c-1) = (2-1)*(3-1) = 1*2 = 2

b)

Alpha = 0.05,

Critical Value = 5.99

c)

Test Statistic:

Chi Square = ∑(O_ij – E_ij)²/E_ij = (40-36)²/36 + ……………….. + (20 – 20)²/20 = 1.85

d)

Decision Rule:

If Chi Square> Chi Square _Critical reject the null hypothesis

Result

Since, Chi Square< Chi Square _Critical fail to reject the null hypothesis.

e)

At alpha = 0.05, we conclude that the plan choice and Gender are not associated.