SOLUTION: Suppose we are interested in a population of 20 industrial units of the same size, all of which are experiencing excessive labour turnover problems. The past records show that the

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Question 1203417: Suppose we are interested in a population of 20 industrial units of the same size, all of which are experiencing excessive labour turnover problems. The past records show that the mean of the distribution of annual turnover is 320 employees, with a standard deviation of 75 employees. A sample of 5 of these industrial units is taken at random which gives a mean of annual turnover as 300 employees. Is the sample mean consistent with the population mean? Test at 5% level

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
population mean is 320 employees with population standard deviation of 75.
population size is 20 industrial units.
sample size is 5 of these 20 industrial units.
sample mean is 300
z-score is indicated since standard deviation is taken from population.

standard error = standard deviation of population divided by square root of sample size = 75 / sqrt(50) = 33.54102.

z = (x-m)/s = (300-320)/33.54102 = -.59628

z is the z-score
x is the sample mean
m is the population mean
s is the standard error

critical z-score for two tailed confidence interval of 95% is z = plus or minus 1.95996.
test z-score of -.59628 is within these limits, indicating that the sample mean is consistent with the population mean.