Question

Suppose researchers interested in the effects of adolescent exposure to mold in a household is associated with asthma. Researchers collected information about a sample of children living in homes determined to have a significant amount of potential mold exposure and determined that 154 out of the 213 children had asthma. After a sample of 428 households with not a sufficient amount of mold exposure, only 108 children were found to have asthma. A) Conduct a statistical test to determine if there is a significant risk difference for asthma based on exposure to mold at the alpha level of 0.05. B) Construct a confidence interval about your point estimate for the difference in risk between the groups.

Answer 1

A)

Ho : P1 = P2

Ha : P1 is not equal to P2

N1 = 213, P1 = 154/213

N2 = 428, P2 = 108/428

First we need to check the conditions of normality that is if n1p1 and n1*(1-p1) and n2*p2 and n2*(1-p2) all are greater than equal to 5 or not

N1*p1 = 154

N1*(1-p1) = 59

N2*p2 = 108

N2*(1-p2) = 320

All the conditions are met so we can use standard normal z table to conduct the test

Test statistics z = (P1-P2)/standard error

Standard error = √{p*(1-p)}*√{(1/n1)+(1/n2)}

P = pooled proportion = [(p1*n1)+(p2*n2)]/[n1+n2]

After substitution

Test statistics z = 11.42

From z table, P(Z>11.42) = 0

As our test is two tailed

So, P-Value = 2*0 = 0

As the obtained P-Value is less than the given significance (0.05)

We reject the null hypothesis

So, we have enough evidence to support the claim that there is a difference

B)

Confidence level = 1 - alpha = 1-0.05 = 0.95 (95%)

From z table critical value for 95% confidence level is 1.96

Margin of error (MOE) = Z*Standard error

Z = 1.96

Standard error = √{p1*(1-p1)/n1} + √{p2*(1-p2)/n2}

Confidence interval is given by

(P1-P2)-MOE < (P1-P2) < (P1-P2)+MOE

after substitution required interval is

0.398 < (P1-P2) < 0.544