Question

# The U.S. Department of Transportation would like to estimate the proportion of drivers that are uninsured...

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))

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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