Question

Recent evidence has surfaced regarding the effect of multiple CAT scans on the increase in risk for blood related cancers in children aged 5-18. Suppose after selecting 267 cases of childhood cancers, researchers found that 139 had received one or more CAT scans. Of the 382 matched sibling controls without blood cancers, only 89 reported having exposure to one or more CAT scan.

A) Calculate the risk difference between the groups for developing blood cancer based on exposure to one or more CAT scans versus never having received a CAT scan.

B) Construct a confidence interval about your point estimate for the difference in risk.

C) Carry out a formal statistical test to determine if there is a significant difference in risk between the groups.

Choose the most appropriate option

		option 1    A) risk difference=0.881 B) CI=(0.12, 0.63) C) ) Ho: p1=p2; Ha: p1≠p2 Z stat=9.7 Pvalue: <0.0001
		Option 2              A) risk difference=0.288 B) CI=(0.21, 0.36) C) ) Ho: p1=p2; Ha: p1≠p2 Z stat=7.7 Pvalue: <0.0001
		Option 3    A) risk difference=0.128 B) CI=(0.06, 0.16) C) ) Ho: p1=p2; Ha: p1≠p2 Z stat=5.6 Pvalue: <0.01

Answer 1

	A		B
x1                =	139	x2                =	89
p̂1=x1/n1 =	0.5206	p̂2=x2/n2 =	0.2330
n1                       =	267	n2                       =	382
estimated difference in proportion   =p̂1-p̂2   =			0.2876

B)

for 95 % CI value of z=		1.960
margin of error E=z*std error =		0.0734
lower bound=(p̂1-p̂2)-E=		0.2142
Upper bound=(p̂1-p̂2)+E=		0.3610
from above 95% confidence interval for difference in population proportion =(0.21,0.36)

C(

std error Se=√(p̂1*(1-p̂1)*(1/n1+1/n2) =		0.0381
test stat z=(p̂1-p̂2)/Se =		7.70

Pvalue: <0.0001

option 2 is corect

A) risk difference=0.288

B) CI=(0.21, 0.36)

C) ) Ho: p1=p2; Ha: p1≠p2

Z stat=7.7

Pvalue: <0.0001