A poll on a certain policy reports 62% approval margin with an error of 4%. What is the confidence interval for this poll?
Solution :
Point estimate = sample proportion = = 62% = 0.62
1 - = 1 - 0.62 = 0.38
At 95% confidence level the z is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
Z/2 = Z0.025 = 1.96
Margin of error = E = 4% = 0.04
A 95% confidence interval for population proportion p is ,
- E < p < + E
0.62 - 0.04 < p < 0.62 + 0.04
0.58 < p < 0.66
At 95% the confidence interval for this poll is : (0.58 , 0.66) or (58% , 66%)
Get Answers For Free
Most questions answered within 1 hours.