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,

n = 60

x = 18

Point estimate = sample proportion = = x / n =18 / 60 = 0.3

1 - = 1 - 0.3 = 0.7

margin of error = E = 0.05

At 99% confidence level

= 1 - 99%

= 1 - 0.99 =0.01

/2
= 0.005

Z/2
= Z0.005 = 2.576

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

= (2.576 / 0.05)2 * 0.3 * 0.7

= 557.41

sample size = n = 558

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