Question

The U.S. Department of Transportation would like to estimate the proportion of drivers that are uninsured in the state of Mississippi. A pilot sample of 60 drivers from Mississippi found that 18 were uninsured. Determine the total sample size needed to construct a 99% confidence interval for the proportion of uninsured drivers in Mississippi with a margin of error equal to 5%.

Use the z-score with three decimal places, Remember the rounding policy for sample size

Answer #1

Solution :-

Given that,

= x / n = 18 /60 = 0.30

1 - = 1 -0.30 =0.70

margin of error = E = 0.05

At 99% confidence level

= 1-0.99% =1-0.99 =0.01

/2
=0.01/ 2= 0.005

Z/2
= Z0.005 = 2.576

Z/2 = 2.576

sample size = n = (Z / 2 / E )2 * * (1 - )

= (2.576 / 0.05)2 *0.30 *0.70

= 557

sample size = n = 557

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