A random sample of 2,000 driver's license applications found that 17% reported their eye color as blue. What is the margin of error for the sample? Round to the nearest tenth of a percent.

1.646%

Step-by-step explanation:

Sample size (n) = 2000

Sample Proportion p = 17% = 0.17

α = 0.95

Margin of Error :

Z(1-α)/2 * √((p*(1 - p)) /n)

Z(1 - 0.95/2) * √((0.17 * (1 - 0.17) / 2000)

Zscore = 1.96

1.96 * √(0.1411/2000)

1.96 * √0.00007055

1.96 * 0.0083994

Hence,

1.96 * 0.0083994

= 0.0164628

= 0.0164628 * 100%

= 1.646%

sine

step-by-step explanation: