You wish to estimate,with 99% confidence, the proportion of Canadian drivers who want the speed limit raised to 130 kph. Your estimate must be accurate to within 5%. How many drivers must you survey,if your initial estimate of the proportion is 0.60? I know that 99% is 2.575 but i dont know how to set up the problem. I don't think that 130 kph even has anything to do with the problem. I think i am over thinking this question.

Hrefnui9 2022-09-14 Answered
You wish to estimate,with 99% confidence, the proportion of Canadian drivers who want the speed limit raised to 130 kph. Your estimate must be accurate to within 5%. How many drivers must you survey,if your initial estimate of the proportion is 0.60?
I know that 99% is 2.575 but i dont know how to set up the problem. I don't think that 130 kph even has anything to do with the problem. I think i am over thinking this question.
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Answers (1)

trabadero2l
Answered 2022-09-15 Author has 15 answers
The estimated interval for p is
[ p ^ z ( 1 α 2 ) p ^ ( 1 p ^ ) n ; p ^ + z ( 1 α 2 ) p ^ ( 1 p ^ ) n ]
Thus z ( 1 α 2 ) p ^ ( 1 p ^ ) n has to be 0.025 ( = 5 % 2 )
The given values are: p ^ = 0.6 ; α = 1 0.99 = 0.01
Therefore the (in-)equation is
z 0.995 0.6 0.4 n 0.025
2.575 0.24 n 0.025
The only remaining work is to solve the inequality for n.
Solving for n
0.24 n 0.025 2.575
0.24 n ( 0.025 2.575 ) 2
n 0.24 ( 2.575 0.025 ) 2
The inequality sign turns around, if you take the reziprocals on both sides.
n ( 2.575 0.025 ) 2 0.24
n 2546.16
Thus you have to survey 2,547 drivers.
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