Question

An article in the Archives of Interval Medicine reported that in a sample of 244 men, 73 had elevated total cholesterol levels (more than 200 milligrams per deciliter). In a sample of 232 women, 44 had elevated cholesterol levels. a. Find a 99% confidence interval about the difference in the proportion of men versus women who had elevated total cholesterol levels. We are 99% confident the true difference in the proportion of men versus women who had elevated total cholesterol levels lies in the interval. b. At the .01 level of significance, can you conclude that the proportion of people with elevated cholesterol levels differs between men and women? State the hypotheses:

Answer 1

a) p1 = 73/244 = 0.2992

p2 = 44/232 = 0.1897

The pooled sample proportion(P) = (p1 * n1 + p2 * n2)/(n1 + n2)

                                                      = (0.2992 * 244 + 0.1897 * 232)/(244 + 232)

                                                      = 0.2458

SE = sqrt(P * (1 - P) * (1/n1 + 1/n2))

      = sqrt(0.2458 * (1 - 0.2458) * (1/244 + 1/232)

      = 0.0395

At 99% confidence interval the critical value is z_0.005 = 2.58

The 99% confidence interval for the true difference in proportions is

(p1 - p2) +/- z_0.005 * SE

= (0.2992 - 0.1897) +/- 2.58 * 0.0395

= 0.1095 +/- 0.1019

= 0.0076, 0.2114

b) H₀: P1 = P2

H₁: P1 $\neq$ P2

The test statistic z = (p1 - p2)/SE

                             = (0.2992 - 0.1897)/0.0395

                             = 2.77

P-value = 2 * P(Z > 2.77)

             = 2 * (1 - P(Z < 2.77))

             = 2 * (1 - 0.9972)

             = 0.0056

As the P-value is less than the significance level (0.0056 < 0.01), so we should reject H₀.

So at 0.01 significance level we can conclude that the proportions of people with elevated cholesterol levels differes between men and women.