Question

The U.S. Daily Industry want to estimate the mean yearly mild consumption. A sample of 16 people reveals the mean yearly consumption to be 60 gallons with a standard deviation of 20 gallons.

a. What is the value of the population mean? What is the best estimate of this value?

b. Explain why we need to use t distribution.

c. For a 90 percent confidence interval, what is the value of t?

d. Develop the 90 percent confidence interval for the population mean.

e. Would it be reasonable to conclude that the population mean is 63 gallons?

Answer 1

a.

Population mean is mean yearly milk consumption in USA. Best estimate of this value is sample mean which is 60 gallons.

b.

Since, we do not know the population standard deviation of mean yearly consumption, we need to use t distribution.

c.

Degree of freedom = n-1 = 16 - 1 = 15

For df = 15 and 90 percent confidence interval, the value of t is 1.75 (From t distribution calculator)

d.

Standard error of mean = Sample standard deviation / = 20 / = 5 gallons

Margin of error = Standard error * t = 5 * 1.75 = 8.75

90 percent confidence interval for the population mean is Mean Margin of error

(60 - 8.75, 60 + 8.75)

(51.25, 68.75)

e.

Since the value 63 lies in the 90 percent confidence interval, it would be reasonable to conclude that the population mean is 63 gallons.