The monthly incomes from a random sample of workers in a fac

The monthly incomes from a random sample of workers in a factory are shown below. Be aware that this is the standard deviation of the population unknown case. Monthly Income (In $) 4000 5000 7000 4000 6000 6000 7000 9000 a. Compute the point estimate of population mean and the point estimate of the population standard deviation (in dollars) b. Compute the margin of error (in dollars) at 95% confidence c. Compute a 95% confidence interval for the mean of the population. Assume the population has a normal distribution. Give your answer in dollars.

Solution

a. Compute the point estimate of population mean and the point estimate of the population standard deviation (in dollars)

the point estimate of population mean= (4000+5000+...+9000)/8 = 6000

poulation standard deviation = 1690.309

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b. Compute the margin of error (in dollars) at 95% confidence

The sample size is n=8

The degree of freedom =n-1=8-1=7

Given a=0.05, t(0.025, df=7) =2.36 (from student t table)

So the the margin of error is

t*s/vn = 2.36*1690.309/sqrt(8) =1410.37

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c. Compute a 95% confidence interval for the mean of the population. Assume the population has a normal distribution.

So 95% confidence interval is

xbar +/- t*s/vn

--> 6000 +/- 1410.37

--> (4589.63, 7410.37)