A survey of 47 members of a health club reported that 68% prefer walking or running...
A survey of 47 members of a health club reported that 68% prefer walking or running on a treadmill to any other type of aerobic exercise. Construct and interpret a 99% confidence interval for the percentage of health club members who prefer walking/running on a treadmill.
Homework Answers
Given
Sample size n = 47
Sample proportion p^ = 0.68
The 99% confidence interval for the percentage of health club members who prefer walking/running on a treadmill.
p^ - Za/2*sqrt(p^*(1-p^)/n) < p < p^ + Za/2*sqrt(p^*(1-p^)/n)
For 
=
0.01
Za/2 = Z0.05 = 2.58
0.68 -2.58*sqrt(0.68*0.32/47) < p < 0.68 + 2.58*sqrt(0.68*0.32/47)
0.68 - 0.18 < p < 0.68 + 0.18
0.50 < p < 0.86
Interpretation of confidence interval
We are 99% confident that the percentage of health club members who prefer walking/running on a treadmill is lies between 50% to 86%
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