Question

In a study, 80 men are tested and 7 of them have red/green color blindness. Construct the 99% confidence interval for the percentage of all men with red/green color blindness.Also find the margin of error. Round your answers to 3 decimal places.

Answer 1

solution

Given that,

Point estimate = sample proportion = = x / n = 7/80=0.088

1 -   = 1- 0.088 =0.912

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* (((0.088*0.912) /80 )

E = 0.082

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.088 - 0.082 < p < 0.088 +0.082

0.006< p < 0.170