Question

In: Statistics and Probability

Fix the roads~The Township Board of Meridian Township wants to know how much public support there...

Fix the roads~The Township Board of Meridian Township wants to know how much public support there is for raising property taxes to fix and maintain the roads in the township. They randomly surveyed 120 residents and calculated a 90% confidence interval for the actual proportion of Meridian Township residents who are in favor of the tax increase to be (0.526, 0.674). Which of the following is a 99% confidence interval using the same survey data?

(.512, .688)

(.48, .715)

(.536, .664)

Expert Solution

Solution :

n = 120

Point estimate = = (Lower confidence interval + Upper confidence interval ) / 2

Point estimate = = (0.526 + 0.674) / 2

Point estimate = = 0.60

1 - = 1 - 0.60 = 0.40

At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z0.005 = 2.576

Margin of error = E = Z / 2 * $\sqrt$ (( * (1 - )) / n)

= 2.576 ( $\sqrt$ ((0.60 * 0.40) / 120 )

= 0.1152

A 99% confidence interval for population proportion p is ,

± E

= 0.60 ± 0.1152

= ( 0.48, 0.715 )

orchestra answered 1 year ago

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