Question

A manufacturer of golf equipment wishes to estimate the number of left-handed golfers. How large a sample is needed in order to be 98% confident that the sample proportion will not differ from the true proportion by more than 5%? A previous study indicates that the proportion of left-handed golfers is 8%.

Answer 1

Solution :

Given that,

$\hat p$ = 8% = 0.08

1 - $\hat p$ = 1 - 0.08 = 0.92

Margin of error = E = 5% = 0.05

At 98% confidence level the z is ,

$\alpha$ = 1 - 98% = 1 - 0.98 = 0.02

$\alpha$ / 2 = 0.02 / 2 = 0.01

Z_$\alpha$/2 = Z_0.01 = 2.326

Sample size = (  Z_$\alpha$/2 / E)² * $\hat p$ * (1 - $\hat p$)

= (2.326 / 0.05)² * 0.08 * 0.92

= 159.27 = 160

Sample size = n = 160