Question

1. Suppose that you are interested in determining whether the advice given by a physician during a routine physical examination is effective in encouraging patients to 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, 23 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 based on the data provided in the question.

b. Use Chi-square test to determine whether there is an association between receiving talks and quitting smoking. Choose α = 0.05.

Answer 1

a.

	Quit Smoking	Not Quit Smoking	Total
Talk	23	150-23 = 127	150
No Talk	8	90 - 8 = 82	90
Total	31	209	240

b.

H0:  There is no association between receiving talks and quitting smoking.

Ha:  There is an association between receiving talks and quitting smoking.

Expected frequencies are calculated as,

E = (Tr * Tc) / T

where Tr and Tc are marginal totals for row r and column c respectively.

T is grand total.

	Quit Smoking	Not Quit Smoking	Total
Talk	(150 * 31)/240 = 19.375	(150 * 209)/240 = 130.625	150
No Talk	(90 * 31)/240 = 11.625	(90 * 209)/240 = 78.375	90
Total	31	209	240

Chi Square test statistic,

= 2.077

Degree of freedom = (r - 1) * (c - 1) = (2 - 1) * (2 - 1) = 1

Critical value of Chi Square test statistic at α = 0.05 and df = 1 is 3.84

Since the observed Chi Square test statistic is less than the critical value, we fail to reject null hypothesis H0 and conclude that there is no significant evidence from the data that there is an association between receiving talks and quitting smoking.