Question

Find the margin of error and​ 95% confidence interval for the survey result described.

According to a poll of

1847

​people, about​ one-third

​(32​%)

of Americans keep a dog for protection.

Margin of

error equals= ___%

​(Round to the nearest whole number as​ needed.)Confidence interval is from

____%

to

____​%.

Answer 1

Solution :

Given that,

n = 1847

= 0.32

1 - = 1 - 0.32 = 0.68

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 = Z / 2 * (( * (1 - )) / n)

= 1.96 * (((0.32 * 0.68) / 1847)

= 0.021

A 95% confidence interval for population proportion p is ,

- E < P < + E

0.32 - 0.021 < p < 0.32 + 0.21

0.299 < p < 0.341

30% < p < 34%

The 95% confidence interval is from 30% to 34%

Margin of error = 2%