Correct question is;
The U.S. Dairy Industry wants to estimate the mean yearly milk consumption. A sample of 21 people reveals the mean yearly consumption to be 74 gallons with a standard deviation of 16 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 the t distribution. What assumption do you need to make?
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 68 gallons?
Answer:
A) Best estimate = 74 gallons
B) because the population standard deviation is unknown. The assumption we will make is that the population follows the normal distribution.
C) t = 1.725
D) 90% confidence interval for the population mean is (67.9772, 80.0228) gallons
E) Yes
Step-by-step explanation:
We are given;
Sample mean; x' = 74
Sample population; n = 21
Yearly Standard deviation; s = 16
A) We are not given the population mean.
So the closest estimate to the population mean would be the sample mean which is 74.
B) We are not given the population standard deviation and as such we can't use normal distribution. So what is used when population standard deviation is not known is called t - distribution table. The assumption we will make is that the population follows the normal distribution.
C) At confidence interval of 90% and DF = n - 1 = 21 - 1 = 20
From t-tables, the t = 1.725
D) Formula for the confidence interval is;
x' ± t(s/√n) = 74 ± 1.725(16/√21) = 74 ± 6.0228 = 67.9772 or 80.0228
Thus 90% confidence interval for the population mean is (67.9772, 80.0228) gallons
E) 68 gallons lies within the range of the confidence interval, thus we can say that "Yes, it is reasonable"
