In: Statistics and Probability
A retirement planning firm conducted a survey of 80 working Americans and found that only 40% of them had adequate retirement savings for their age group. Create the 99% confidence interval for the proportion of working Americans with adequate retirement savings.
Solution :
Given that,
n = 80
Point estimate = sample proportion = = 0.40
1 - = 1- 0.40 =0.60
At 99% confidence level the z is ,
= 1 - 99% = 1 - 0.99 = 0.01
/ 2 = 0.01 / 2 = 0.005
Z/2 = Z0.005 = 2.576 ( Using z table )
Margin of error = E Z/2 * (( * (1 - )) / n)
= 2.576*((0.40*0.60) /80 )
=0.1411
A99% confidence interval for population proportion p is ,
- E < p < + E
0.40-0.1411 < p < 0.40+0.1411
0.2589< p < 0.5411
The 909% confidence interval for the population proportion p is : 0.2589, 0.5411