8 Computing confidence intervals for a population proportion

8. Computing confidence intervals for a population proportion A random sample of n people is the US is asked whether they approve of Barack Obama\'s performance as President. Give an approximate 95% confidence interval for the population proportion who approve of Obama?s performance when (a) n=10, 5 say yes and 5 say no. (b) n=100, 50 say yes and 50 say no. (c) n=l00, 60 say yes and 40 say no. (d) n=100, 90 say yes and 10 say no. (e) n=1000, 600 say yes and 400 say no. (f) n=1 million, 600,000 say yes and 400,000 say no.

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)=5
Sample Size(n)=10
Sample proportion = x/n =0.5
Confidence Interval = [ 0.5 ±Z a/2 ( Sqrt ( 0.5*0.5) /10)]
= [ 0.5 - 1.96* Sqrt(0.025) , 0.5 + 1.96* Sqrt(0.025) ]
= [ 0.1901,0.8099] ~ [ 19.01% , 80.99%]

b)
Mean(x)=50
Sample Size(n)=100
Sample proportion = x/n =0.5
Confidence Interval = [ 0.5 ±Z a/2 ( Sqrt ( 0.5*0.5) /100)]
= [ 0.5 - 1.96* Sqrt(0.0025) , 0.5 + 1.96* Sqrt(0.0025) ]
= [ 0.402,0.598]

c)
Sample Size(n)=100
Sample proportion = x/n =0.6
Confidence Interval = [ 0.6 ±Z a/2 ( Sqrt ( 0.6*0.4) /100)]
= [ 0.6 - 1.96* Sqrt(0.0024) , 0.6 + 1.96* Sqrt(0.0024) ]
= [ 0.504,0.696]

d)
Mean(x)=90
Sample Size(n)=100
Sample proportion = x/n =0.9
Confidence Interval = [ 0.9 ±Z a/2 ( Sqrt ( 0.9*0.1) /100)]
= [ 0.9 - 1.96* Sqrt(0.0009) , 0.9 + 1.96* Sqrt(0.0009) ]
= [ 0.8412,0.9588]