The U.S. Dairy Industry wants to estimate the mean yearly milk consumption. A sa
ID: 3380933 • Letter: T
Question
The U.S. Dairy Industry wants to estimate the mean yearly milk consumption. A sample of 27 people reveals the mean yearly consumption to be 75 gallons with a standard deviation of 25 gallons. Assume that the population distribution is normal. (Use z Distribution Table.)
For a 98% confidence interval, what is the value of t? (Round your answer to 3 decimal places.)
Develop the 98% confidence interval for the population mean. (Round your answers to 3 decimal places.)
a-1. What is the value of the population mean? 25 Unknown 75Explanation / Answer
a-1. UNKNOWN [ANSWER]
Only the sample mean is given here.
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a-2.
75 gal. [ANSWER]
The best estimate of the population mean is the sample mean.
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c.
As df = n - 1 = 26, then at 98% confidence, by table/technology,
t = 2.479 [ANSWER]
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d)
Note that
Margin of Error E = t(alpha/2) * s / sqrt(n)
Lower Bound = X - t(alpha/2) * s / sqrt(n)
Upper Bound = X + t(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.01
X = sample mean = 75
t(alpha/2) = critical t for the confidence interval = 2.478629824
s = sample standard deviation = 25
n = sample size = 27
df = n - 1 = 26
Thus,
Margin of Error E = 11.9253133
Lower bound = 63.0746867
Upper bound = 86.9253133
Thus, the confidence interval is
( 63.0746867 , 86.9253133 ) [ANSWER]
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E)
NO, as 54 is not inside this interval. [ANSWER]
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