In a survey of 630 males ages 1864 398 say they have gone to

In a survey of 630 males ages 18-64, 398 say they have gone to the dentist in the past year. Construct 90% and 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals.

Solution

a. AT 90% confidence intervals for the population 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
No. of gone to the dentist in the past year(x)=398
Sample Size(n)=630
Sample proportion = x/n =0.632
Confidence Interval = [ 0.632 ±Z a/2 ( Sqrt ( 0.632*0.368) /630)]
= [ 0.632 - 1.645* Sqrt(0) , 0.632 + 1.65* Sqrt(0) ]
= [ 0.6,0.664]

b. AT 95% confidence intervals for the population proportion

Confidence Interval = [ 0.632 ±Z a/2 ( Sqrt ( 0.632*0.368) /630)]
= [ 0.632 - 1.96* Sqrt(0) , 0.632 + 1.96* Sqrt(0) ]
= [ 0.594,0.67]

c.
Increase the confidence widen the interval