Question

The state highway department is studying traffic patterns on one of the busiest highways in the state. As part of the study, the department needs to estimate the average number of vehicles that pass an intersection each day. A random sample of 64 days gives us a sample mean of 14,205 cars and a sample standard deviation of 1,010 cars. Compute the 99% confidence interval for the mean ? number of cars passing the intersection.

Note : the Answer should be computerized

Answer 1

The standard deviation for this sample is 1010. But this is the standard deviation for the sample, For the greatest precsion we should multiply this by sqrt(64/63) to get a slightly larger estimate for the population standard deviation of 1018.

Without this correction if we use a sample of N days we get

1010/sqrt(N) as the standard deviation for our sample means and multiplying this by 2.575 which is the number of standard deviations needed to give us a 0.005 tail in each direction on a 2 sided test we get 2.575 times 1010/sqrt(N) = 300.

Solving for sqrt(N) we get 2.575 times 1010/300 = 8.669166667 and squaring this we get 75.15445. If this is the approach you are expected to take we round up.

75 is closer but it gives less than 99% confidence so you need to go up to 76.

Doing the same calculation with the corrected variance of 1018 we 76.34973

Which would have to round up to 77. I conclude that this is not the approach you are expected to take. Your answer is 76