Question 1178141: A recent study examined the association between high blood pressure and increased risk of death from cardiovascular disease. There were 2676 men with low blood pressure and 3338 men with high blood pressure. In the low-blood-pressure group, 21 men died from cardiovascular disease; in the high-blood-pressure group, 55 died.
Compute the 95% confidence interval for the difference in
Proportions.
Do the study data confirm that death rates are higher among men with high blood pressure?
State hypotheses, carry out a significance test, and give your conclusions.
(Thank you) (Please)
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! p1=low BP=0.00784
p2=high BP=0.01648
the difference is 0.00863
the 95% half-interval is z* diff (means)*SE, the general form.
the SE is sqrt[((0.00784*0.99216/2676+(.01648*0.98312/3338))]=0.000002907+0.000004854=0.0028
Here, it is 1.96*0.0028=0.0055
the interval is (0.0031, 0.0141), and since the interval does not contain 0, there is a difference between the two proportions.
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the p-value will be 0.0126
It is worth doing a couple by brute force to get an idea how the test works, but these are extremely easy to put in a calculator or program.
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