The US Dairy Industry wants to estimate the mean yearly milk

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

(a-1)

What is the value of the population mean?

(a-2)

What is the best estimate of this value?

(c)

For a 90 percent confidence interval, what is the value of t? (Round your answer to 3 decimal places.)

(d)

Develop the 90 percent confidence interval for the population mean. (Round your answers to 3 decimal places.)

Solution

Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
a.1)
Mean(x)=60
Standard deviation( sd )=20
Sample Size(n)=16
c) t = 1.753
d)

Confidence Interval = [ 60 ± t a/2 ( 20/ Sqrt ( 16) ) ]
= [ 60 - 1.7531 * (5) , 60 + 1.7531 * (5) ]
= [ 51.235,68.766]