Recent evidence has surfaced regarding the effect of multiple CAT scans on the increase in risk...

3.7 Ratings (648 Votes)

	A		B
x₁ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =Â Â Â	139	x₂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =Â Â Â	89
pÌ‚₁=x₁/n₁ =	0.5206	pÌ‚₂=x₂/n₂ =	0.2330
n_{1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â} =	267	n_{2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â} =	382
estimated difference in proportionÂ Â =pÌ‚₁-pÌ‚₂Â Â =	0.2876

for 95 % CI value of z=	1.960
margin of error E=z*std error =	0.0734
lower bound=(pÌ‚1-pÌ‚2)-E=	0.2142
Upper bound=(pÌ‚1-pÌ‚2)+E=	0.3610
from above 95% confidence interval for difference in population proportion =(0.21,0.36)

std error Se=âˆš(pÌ‚1(1-pÌ‚1)(1/n1+1/n2) =	0.0381
test stat z=(pÌ‚1-pÌ‚2)/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: p1â‰ p2

Z stat=7.7

Pvalue: <0.0001

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