Question

The results of a recent study regarding smoking and three types of illness are shown in the following table. (60 points)

Illness	Non-Smokers	Smokers	Totals
Emphysema	20	60	80
Heart problem	70	80	150
Cancer	30	40	70
Totals	120	180	300

We are interested in determining whether or not illness is independent of smoking.

1. State the null and alternative hypotheses to be tested. (10 points)

b) Show the contingency table of the expected frequencies. (20 points)

Illness	Non-Smokers	Smokers
Emphysema
Heart problem
Cancer

c) Compute the test statistic (chi-square). The null hypothesis is to be tested at 95% confidence. Determine the critical value for this test. What do you conclude? (30 points)

Answer 1

12:47 6 Yol Bill 36% 11 LTE1 | cola chi sa fx < A B с D ד G 1 2 3 4 a) Ho : illness is NOT independent of smoking. Ha : Illness is independent of smoking. Observed Values illness Non-Smokers Smokers Row Totals Emphysema 20 60 80 Heart problem 70 80 150 Cancer 30 40 70 Column Totals 120 180 300 5 6 7 8 10 11 12 Row Total These are expected values assuming null hypothesis to be true (row total)*(column total)/(grand total) 13 Calculate Expected Frequencies b) Expected Values Non-Smokers Smokers Emphysema 32.00 48.00 Heart problem 60.00 90.00 Cancer 28.00 42.00 Column Total 120.00 180.00 80.00 14 150.00 70.00 15 16 300.00 17 19 Calculation: 20 Emphysema Heart problem Cancer Non-Smokers Smokers 4.5000 3.0000 1.6667 1.1111 0.1429 0.0952 Obtained by using chi-sq formula : (Oi-Ei)^2/Ei 21 22 23 24 25 c) Chi-Sq Test Statistic P-Value 27 Solution: 10.52 Obtained by summing above values 0.0052 Using Formula Echisq.Test Test Statistic 10.52 DF 2 using formula (r-1)*(c-1) Critical value 5.992 28 29 30 31 32 Since the P- Value < 0.05, we Reject the null hypothesis at (a = 0.05) and support the claim that the illness is independent of smoking. 33 34 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 EJ III O <