Question

An allergy to peanuts is increasingly comment in Western countries. A randomized controlled trial enrolled infants with a diagnosed peanut sensitivity. Infants were randomized to either avoid peanuts or consume them regularly until they reached age 5. At the end of the study 18 out of the 51 randomized to avoidance were allergic to peanuts compared to 5 out of the 47 randomized to consuming them regularly. Based on this data, conduct the following:

4.a) Estimate the difference between the two proportions.

4. b) Use the plus four method to find a 99% confidence interval for the difference
between the two groups.

Why would it have been inappropriate to use the large sample method to create
a 99% CI?

Perform a two-sided hypothesis test for the difference between the groups.
Start by stating the null and alternative hypotheses, then calculate the test
statistic, the p-value, and conclude with your interpretation of the p-value.

Answer 1

a) the difference between the two proportions is:

18/51 - 5/47 = 0.247

b) 99% confidence interval for the difference using +4 method:

In the plus 4 method, we add two success and 4 sample size to each group:

So p1 = 20/55 = 0.364, p2 = 7/51= 0.137

Now 99% confidence interval for the difference is:

(0.364-0.137) +- 2.576*√0.364*0.636/55 + 0.137*0.863/51

= (0.019, 0.435)

It would be inappropriate to use large sample method because for proportion 2, np= 5, and np(1-p) <5 so the conditions are not met

We are testing,

H0: p1=p2 vs H1: p1 not equal p2

Under H0, test statistic: 0.247/√0.353*0.647/51 + 0.106*0.894/47 = 3.065

The p value of this two sided z test is, 2P(Z>3.065) = 0.0022

Since this is less than the significance level of 0.01, we have sufficient evidence to reject H0 and conclude that the population proportions are different.