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 don't 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. I think it would be about 93 people. Is that right?

ridge041h

ridge041h

Answered question

2022-09-10

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 don't 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.
I think it would be about 93 people. Is that right?

Answer & Explanation

letovanjelm

letovanjelm

Beginner2022-09-11Added 13 answers

Let n be the appropriate sample size. Then the variance of the sample mean is
p ( 1 p ) n ,
where p is the true population proportion. We will use our preliminary estimate p 0.6 to estimate the variance of the sample mean. The reason we can do that is that the standard deviation of the sample mean is fairly insensitive to the actual value of p.
Using the normal approximation, we find that the probability that the sample mean differs by more than 0.05 from the true p is approximately
Pr ( | Z | > 0.05 ( 0.6 ) ( 0.4 ) / n ) ,
where Z is standard normal. So we want
0.05 ( 0.6 ) ( 0.4 ) / n 2.575.
Now we can find n. It is quite a bit bigger than 93.

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