Question

The table below shows the results of a study on smoking and three illnesses. We are interested in determining if the proportions of smokers in the three categories are different from each other.

Emphysema- Heart Problem- Cancer- Total
Smoker 145 140 300 585
Non-smoker 60 100 230 390
Total 205 240 530 975

p1 = proportions of smokers with emphysema

p2 = proportions of smokers with heart problem

p3 = proportions of smokers with cancer

a. What represents the null and alternative hypotheses in this case?

b. The expected frequency of smokers with emphysema is?

c. The chi-square value is?

d. With α = 0.1, the critical X2 value is?

e. P1 – P2 is? P2 - P3 is?

f. Using α = 0.1, the conclusion of the pairwise comparison is?

Answer 1

Answer on Question #57133 – Math – Statistics and Probability

The table below shows the results of a study on smoking and three illnesses. We are interested in determining if the proportions of smokers in the three categories are different from each other.

Emphysema - Heart Problem - Cancer - Total

p1 = proportions of smokers with emphysema

p2 = proportions of smokers with heart problem

p3 = proportions of smokers with cancer

a. What represents the null and alternative hypotheses in this case?

b. The expected frequency of smokers with emphysema is?

c. The chi-square value is?

d. With $\alpha = 0.1$, the critical X2 value is?

e. P1 – P2 is? P2 - P3 is?

f. Using $\alpha = 0.1$, the conclusion of the pairwise comparison is?

Solution

Observed data

Expected data

$$df = (2 - 1)(3 - 1) = 2$$

a. What represents the null and alternative hypotheses in this case?

Null hypothesis: the proportions of smokers in the three categories are the same.

Alternative hypothesis: the proportions of smokers in the three categories are not the same.

b. The expected frequency of smokers with emphysema is

$$\frac {205 \cdot 585}{975} = 123$$

c. The chi-square value is

$$\chi^{2} = \frac{(145 - 123)^{2}}{123} + \frac{(140 - 144)^{2}}{144} + \frac{(300 - 318)^{2}}{318} + \frac{(60 - 82)^{2}}{82} + \frac{(100 - 96)^{2}}{96} + \frac{(230 - 212)^{2}}{212} = 12.662$$

d. With $\alpha = 0.1$, the critical chi-square value is

$$\chi_{\text{crit}}^{2} = \chi^{2}(2; 0.1) = 4.605$$

e. $P1 - P2$ is? $P2 - P3$ is?

$$P_{1} - P_{2} = \frac{145}{205} - \frac{140}{240} = 0.124.$$ $$P_{2} - P_{3} = \frac{140}{240} - \frac{300}{530} = 0.017.$$

f. Using $\alpha = 0.1$, the conclusion of the pairwise comparison is?

We used the Marascuillo procedure.

For an overall level of significance of 0.1, the critical value of the chi-square distribution having $3 - 1 = 2$ degrees of freedom is $\chi^2(2; 0.1) = 4.605$.

$$r_{12} = \sqrt{\chi^{2}(2; 0.1)} \sqrt{\frac{p_{1}(1 - p_{1})}{n_{1}} + \frac{p_{2}(1 - p_{2})}{n_{2}}} = \sqrt{4.605} \sqrt{\frac{\frac{145}{205} \left(1 - \frac{145}{205}\right)}{205} + \frac{\frac{140}{240} \left(1 - \frac{140}{240}\right)}{240}} = 0.0965.$$ $$r_{23} = \sqrt{\chi^{2}(2; 0.1)} \sqrt{\frac{p_{3}(1 - p_{3})}{n_{3}} + \frac{p_{2}(1 - p_{2})}{n_{2}}} = \sqrt{4.605} \sqrt{\frac{\frac{300}{530} \left(1 - \frac{300}{530}\right)}{530} + \frac{\frac{140}{240} \left(1 - \frac{140}{240}\right)}{240}} = 0.0824.$$

The conclusion of the pairwise comparison: the proportion of smokers with emphysema is different from two others; there is not enough data to conclude that the proportion of smokers with heart problem and the proportion of smokers with cancer are different.

www.AssignmentExpert.com