Question

On a survey of 2480 golfers, 372 said they were left-handed. Construct a 92% confidence interval for the proportion of left-handed golfers from this sample.

Answer 1

Solution :

Given that,

n = 2480

x = 372

Point estimate = sample proportion = = x / n = 372 / 2480 = 0.15

1 - = 1 - 0.15 = 0.85

At 92% confidence level

= 1 - 92%

= 1 - 0.92 = 0.08

/2 = 0.04

Z_/2 = Z_0.04 = 1.751

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

= 1.751 (((0.15 * 0.85) / 2480)

= 0.01

A 92% confidence interval for population proportion p is ,

- E < p < + E

0.15 - 0.013 < p < 0.15 + 0.013

0.137 < p < 0.163

( 0.137 , 0.163 )

The 92% confidence interval for the proportion is : ( 0.137 , 0.163 )