Question

Erythromycin is a drug that has been proposed to possibly lower the risk of premature delivery. A related area of interest is its association with the incidence of side effects during pregnancy. Assume that 30% of all pregnant women complain of nausea between the 24th and 28th week of pregnancy. Furthermore, suppose that of 178 women who are taking erythromycin regularly during this period, 63 complain of nausea. Find the p-value for testing the hypothesis that incidence rate of nausea for the erythromycin group is greater than for a typical pregnant woman.

Answer 1

Solution :

Given that,

= 0.30

1 - = 0.70

n = 178

x = 63

Point estimate = sample proportion = = x / n = 0.354

This a right (One) tailed test.

The null and alternative hypothesis is,

Ho: p = 0.30

Ha: p 0.30

Test statistics

z = ( - ) / *(1-) / n

= ( 0.354 - 0.30) / (0.30*0.70) /178

= 1.57

P-value = P(Z>z)

= 1 - P(Z <z )

= 1- P(Z < 1.57)

= 1 - 0.9418

= 0.0582

The p-value is p = 0.0582