Question

in a survey of 2480 golfers, 15% said they were left - handed. the surveys margin of error was 3%. construct a confidence interval for the proportion of left - handed golfers.

oa (0.12,0.15)
ob (0.12,0.18)
oc (0.18,0.21)
od (0.11,0.19)

Explanation:

Step1: Convert percentage to decimal

The sample proportion \(p = 0.15\), and the margin of error \(E=0.03\)

Step2: Calculate the lower and upper bounds

The lower bound of the confidence interval is \(p - E\), and the upper bound is \(p + E\)
For the lower bound: \(0.15- 0.03=0.12\)
For the upper bound: \(0.15 + 0.03=0.18\) (Wait, no, wait the options have \(0.12\) and \(0.18\) not in the options. Wait, no, re - check. Wait the margin of error formula for proportion confidence interval is \(E = z\sqrt{\frac{p(1 - p)}{n}}\), but here we are given the margin of error directly as \(3\%=0.03\). The confidence interval for proportion is \((p - E,p + E)\). Given \(p = 0.15\) (15% in decimal) and \(E = 0.03\) (3% in decimal)
The lower limit \(=0.15-0.03 = 0.12\)
The upper limit \(=0.15 + 0.03=0.18\). But looking at the options, option A is \((0.12,0.18)\) (assuming it's a typo in the problem statement where maybe the upper - bound was mis - labeled as \(0.15\) in the option description. If we follow the formula \((p - E,p + E)\) with \(p = 0.15\) and \(E=0.03\))

Answer:

A. \((0.12,0.18)\) (assuming a typo in the option's upper - bound display. If we use the formula for confidence interval of proportion \((\hat{p}-E,\hat{p} + E)\) where \(\hat{p}=0.15\) (15% of golfers said left - handed) and \(E = 0.03\) (margin of error))