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 24 people reveals the mean yearly consumption to be 80 gallons with a standard deviation of 22 gallons. Assume that the population distribution is normal. (Use z Distribution Table.)

For a 99% confidence interval, what is the value of t?

Develop the 99% confidence interval for the population mean.

Solution

c)
Value of t at 99% is = 2.8073
d)
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
Mean(x)=80
Standard deviation( sd )=22
Sample Size(n)=24
Confidence Interval = [ 80 ± t a/2 ( 22/ Sqrt ( 24) ) ]
= [ 80 - 2.81 * (4.491) , 80 + 2.81 * (4.491) ]
= [ 67.381,92.619 ]