Let me get this straight -- you want to create a confidence interval for a mean that you *know* is equal to 53,000?
Even so, your formula needs a finite population size correction factor. Albert <[EMAIL PROTECTED]> wrote in message news:<[EMAIL PROTECTED]>... > A diet food company has 120 salesman. Monthly sales of each salesman is > approximately normally distributed with a mean sales amount of $53,000 and a > standard deviation of $15,000. A random sample of 10 salesman is randomly selected. > Suppose that the sample mean sales amount turn out to be $63,000, construct a 95% > confidence interval for the true mean among the 120 salesman. > > Solution: > > I calculated the mean and standard deviation of the sample mean sales to be $53,000 > and $4384.67 respectively. > > bar X = 63000, n = 120, s.d.= 15,000 > > 95% confidence = (1-a) where a = 0.05; hence a/2 = 0.025 ,so Zvalue(.025) = 1.96 > > = bar X +/- Z(a/2) (sigma/sqroot(n)) > = 63,000 +/- (1.96)*15000/sqrt(120) > = 63,000 +/- 2,683.8405 > LCL = 60316.1595 > UCL = 65683.8405 > > Is the solution correct? . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
