Question

In: Statistics and Probability

In 2003 a survey of 35,000 adults showed that 26% were smokers. In 2010 a survey...

In 2003 a survey of 35,000 adults showed that 26% were smokers. In 2010 a survey of 45,000 adults showed that 21% were smokers. Construct a 95% confidence interval for the true difference between proportions.

a) (4.41% , 5.59%)

b) (3.33% , 6.67%)

c) (4.82% , 5.18%)

d) (3.83% , 6.17%)

Solutions

Expert Solution

Total number of sample 1 (n1) = 35000
Total number of sample 2 (n2) = 45000

Confidence interval(in %) = 95

Required Confidence interval = (0.05 - 1.96(0.003), 0.05 + 1.96(0.003))
Required Confidence interval = (0.05 - 0.0059, 0.05 + 0.0059)
Required Confidence interval = (0.0441, 0.0559)
Required Confidence interval = (4.41% , 5.59%)
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