Question

A medical study showed that girls under sixteen who are treated for Hodgkin’s disease face an exceptionally high risk of developing a second type of cancer later in life (New York Times, 3/21/96). The study followed 1,380 girls after their original treatment, and 88 of them, or 0.0638, developed second cancers. Only four of them, or 0.0029, would have been expected based on data for the population at large.

Test the null hypothesis that girls treated for Hodgkin’s disease are not at increased risk for a second cancer (H0: p ≤ 0.0029) against the alternative hypothesis that they are at increased risk (H0: p > 0.0029). Use a 10% level of significance.

What is your critical value?

Answer 1

Answer & Step-by-step explanation:

H0: p ≤ 0.0029

H1: p > 0.0029

The z-statistic: (x/n)-p/sqrt[(p*(1-p))/n]

x= number of occurrences

n= observations

p= expected proportion

z= (88/1380)-0.0029/sqrt[(0.0029*(1-0.0029)/1380]

z=0.0608/0.00144

z=42.04

The critical value at a 10% level of significance is : 1.28

Because the z-statistic is greater than 1.28, we must reject the null hypothesis. And so, there is sufficient evidence to claim that the proportion of girls treated for Hodgkin´s disease are at an increase rick for a second cancer.