Question

The table represents a survey of a random sample of 229 residents on whether names of convicted sexual offenders should be publicized.

	Male	Female
Yes	41	85
No	59	44

Test hypothesis of no difference between males and females on the question of whether sex offenders' names should be publicized on their release from jail. Use 0.05 significance level.

1. (Multiple Choice) What are the null and alternative hypotheses?

(a) H0: Dependence; H1: Independence

(b) H0: u1=µ2; H1: µ1?µ2

(c) H0: µ1?µ2 ; H1: u1=µ2

(d) H0: Independence; H1: dependence

2. Calculate the test statistic.

3. (Multiple Choice) Do you accept or reject the null hypothesis?

(a) accept the null hypothesis and conclude there is a relationship btwn the variables

(b) accept the null hypothesis and conclude there is no relation btwn the variables

(c) reject the null hypothesis and conclude there is a relation btwn the variables

(d) reject the null hypothesis and conclude there is no relation btwn the variables

4. What is the critical value?

5. Compute Phi-adjusted.

Answer 1

		c1	c2	total
	r1	41	85	126
	r2	59	44	103
	total	100	129	229
	Oi	41	85	59	44
	Ei	55.02183	70.97817	44.97817	58.02183
						TS
	(O--Ei)^2/Ei	3.573342	2.770033	4.371273	3.388584	14.10323
	critical value	3.84
p-value	0.000173046
phi	0.248165591
V	0.488914932

		c1	c2	total
	r1	41	85	126
	r2	59	44	103
	total	100	129	229
	Oi	41	85	59	44
	Ei	55.02183	70.97817	44.97817	58.02183
						TS
	(O--Ei)^2/Ei	3.573342	2.770033	4.371273	3.388584	14.10323
	critical value	3.84
p-value	0.000173046
phi	0.248165591
V	0.488914932

2)
using chi-square method
TS = 14.10323

3)

since TS> critical value (3.84)

hence

we reject the null hypothesis

c) is correct

4) critical value = 3.84

5)

phi-value = sqrt(14.10323/229)

= 0.24816