Question

A random sample of 225 people shows that 25 are left-handed. Form a 99% confidence interval for the true proportion of left-handers.

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 25 / 225 = 0.111

1 - = 1 - 0.111 = 0.889

Z/2 = Z_0.005 = 2.576

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

= 2.576 (((0.111 * 0.889 ) / 225)

= 0.054

A 99% confidence interval for population proportion p is ,

± E  

= 0.111 ± 0.054

= ( 0.057, 0.165 )