We know that 25 of married women have high blood pressure We

We know that 25% of married women have high blood pressure. We suspect that single women have a lower incidence of hypertension. Therefore, a sample of 100 unmarried women is tested and it is found that 20 have high blood pressure.

a) Construct a 95% confidence interval (by hand) for the actual proportion of single women with high blood pressure.

b) Construct a 95% confidence interval (using Minitab) for the actual proportion of single women with high blood pressure.

Solution

a)
Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=20
Sample Size(n)=100
Sample proportion = x/n =0.2
Confidence Interval = [ 0.2 ±Z a/2 ( Sqrt ( 0.2*0.8) /100)]
= [ 0.2 - 1.96* Sqrt(0.002) , 0.2 + 1.96* Sqrt(0.002) ]
= [ 0.122,0.278]