Provide a 95% confidence interval estimate of the difference between the proportion

Question

BusinessWeek reported that there seems to be a difference by age group in how well people like life in Russia (BusinessWeek, March 10, 2008). The following sample data are consistent with theBusinessWeek findings and show the responses by age group to the question: "Do you like life in Russia?"

 

Russian Age Group
	17-26	40 and over
Sample Size	300	260
Responded Yes	192	117

 

The question is : Provide a 95% confidence interval estimate of the difference between the proportion of young Russians aged 17 to 26 and older Russians aged 40 and over who like life in Russia

Answer 1

Let p1 be the first proportion=192/300=0.64 and p2 the second=117/260=0.45. Since we don't have population stats, we need to work out the standard error (SE) of the difference between p1 and p2, with n1 and n2 being the sample sizes:

SE=sqrt(p1(1-p1)/n1+p2(1-p2)/n2)=sqrt(0.64*0.36/300+0.45*0.55/260)=0.041472 and p1-p2=0.19.

From the confidence level of 95%, we get a=1-0.95=0.05 and the critical probability is 1-a/2=0.975. From this we can read off the z-score for the critical value: 1.96 and compute the margin of error: 1.96*0.041472=0.081285.

Now we can apply the margin of error to p1-p2: 0.19+0.08, confidence of 95% that the difference lies between 0.11 and 0.27.