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 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.)

Solution

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]