Why is it that I need the finite population size correction factor? and how?
On 11 Jul 2003 07:15:38 -0700, [EMAIL PROTECTED] (Jason Owen) wrote: >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/ . =================================================================
