Out of 200 people surveyed, 70 of them were found to be in favor of changing...

Out of 200 people surveyed, 70 of them were found to be in favor of changing the color of the five dollar bill from green to purple. Give a 99% confidence interval for the proportion of people in favor of changing the color of the five dollar bill.

Solutions

Expert Solution

olution :

Given that,

n = 200

x = 70

Point estimate = sample proportion = = x / n = 70/200=0.35

1 - = 1-0.35 =0.65

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 * ( $\sqrt$ ((( * (1 - )) / n)

= 2.576* ( $\sqrt$ ((0.35*0.65) /200 )

E = 0.087

A 99% confidence interval for proportion p is ,

- E < p < + E

0.35-0.087 < p < 0.35+0.087

0.263< p < 0.437

orchestra answered 1 year ago

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