Quote: Preliminary data analyses indicate that you can reasonably apply the z-interval procedure. Find a 90% confidence interval for the mean number of tongue flicks per 20 minutes for all juvenile common lizards. Assume a population standard deviation of 190.0. Note: The sum of the data is 8891. The mean is 523
Confidence interval is?:
first off
if the mean is 523, and the sum is 8891 then N = 8891/523 = 17
so we have to adjust the population standard deviation appropriately.
sigma' = sigma / sqrt(17) = 190 / sqrt(17) = 46.1
now for a 90% confidence interval we have tails of 5% each, i.e. (100-90)/2 = 5
so we hit the table and find the z-value for 0.05
we find it is z = -1.64485
so our confidence interval is [523 - 1.64485 * 46.1, 523 + 1.64485 * 46.1] = [447.2, 598.8]
+6251 
