A survey of 2450 golfers showed that 281 of them are left-handed. Construct a 98% confidence interval for the proportion of golfers that are left-handed.
A)(0.100, 0.130)
B)(0.203, 0.293)
C)(0.369, 0.451)
D)(0.683, 0.712)

Question

Quizplus · Accepted Answer

The proportion of left-handed golfers is $ p = \frac{281}{2450} $. To construct a 98% confidence interval for the proportion, we use the formula: $$ \text{CI} = p \pm Z_{\alpha/2} \sqrt{\frac{p(1-p)}{n}} $$ where $ p $ is the sample proportion, $ n $ is the sample size, and $ Z_{\alpha/2} $ is the critical value for a 98% confidence level (approximately 2.33). Plugging in the numbers: $$ p = \frac{281}{2450} \approx 0.115 $$ $$ \text{CI} = 0.115 \pm 2.33 \sqrt{\frac{0.115(1-0.115)}{2450}} $$ This calculation results in a confidence interval approximately between 0.100 and 0.130, which matches option A.

A Survey of 2450 Golfers Showed That 281 of Them