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 17 people reveals the mean yearly consumption to be 66 gallons with a standard deviation of 26 gallons.

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

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

Solution

a-1) Mean = 66
a-2) best estimate = 66
c) value of t = 1.746
d)
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)=66
Standard deviation( sd )=26
Sample Size(n)=17
Confidence Interval = [ 66 ± t a/2 ( 26/ Sqrt ( 17) ) ]
= [ 66 - 1.746 * (6.306) , 66 + 1.746 * (6.306) ]
= [ 54.99,77.01 ]
e) no,because it is less than lower interval