Question

They screened 8069 people for the genes that cause the disease. They found 524 carriers for the disease.. Calculate an approximate 98% confidence interval proportion of this population that are carriers for the disease.

Answer 1

Solution :

Given that,

n = 8069

x = 524

Point estimate = sample proportion = = x / n = 524/8069=0.065

1 - = 1-0.065=0.935

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table    )

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 2.326 (((0.065*0.935) /8069 )

E = 0.00638

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.065-0.00638 < p <0.065+ 0.00638

0.0586< p < 0.0714

The 98% confidence interval for the population proportion p is : 0.0586, 0.0714