Question

Suppose that you are interested in determining whether the advice given by a physician during a routine physical examination is effective in encouraging patientsto stop smoking. In a study of current smokers, one group of patients was given a brief talk about the hazards of smoking and was encouraged to quit. A second group received no talk related to smoking. All patients were given a follow-up exam. In the sample of 150 patients who had received the talk, 25 reported that they had quit smoking. In the sample of 90 patients that hadn’t received a talk, 8 had ceased smoking.

a. Create a 2x2 table out of the data provided in the question.

b. Use Chi-square test to determine whether a difference exists in the proportion of smokers that had quit smoking between the two groups. Choose α = 0.05.

Answer 1

a)

below is contingency ta ble:

		Smoking
		quit	not quit	total
Talk	received	25	125	150
	not received	8	82	90
	total	33	207	240

b)

null hypothesis: there is no  difference exists in the proportion of smokers that had quit smoking between the two groups

alternate hypothesis: there is a difference exists in the proportion of smokers that had quit smoking between the two groups

degree of freedom (row-1)*(column-1)=(2-1)*(2-1)=1

for 1 df and 0.05 level rejection region   $\chi$ ² >3.841

applying chi square test:

Expected	Ei=Σrow*Σcolumn/Σtotal	quit	not quit	Total
	received	20.625	129.375	150
	not received	12.375	77.625	90
	Total	33	207	240
chi square    χ2	=(Oi-Ei)2/Ei	quit	not quit	Total
	received	0.928	0.148	1.076
	not received	1.547	0.247	1.793
	Total	2.475	0.395	2.869

as test statsiic 2.869 is not in rejection region ; we can not rject null hypothesis

we do not have evidence at 0.05 level to conclude that there is a  difference exists in the proportion of smokers that had quit smoking between the two groups.