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A poll on a certain policy reports 62% approval margin with an error of 4%. What...

A poll on a certain policy reports 62% approval margin with an error of 4%. What is the confidence interval for this poll?

Math   Statistics-And-Probability   
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Answer #1

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%)

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