Bob Nale is the owner of Nale’s Quick Fill. Bob would like to estimate the mean
ID: 3133131 • Letter: B
Question
Bob Nale is the owner of Nale’s Quick Fill. Bob would like to estimate the mean number of gallons of gasoline sold to his customers. Assume the number of gallons sold follows the normal distribution with a population standard deviation of 2.50 gallons. From his records, he selects a random sample of 55 sales and finds the mean number of gallons sold is 5.40.
What is the point estimate of the population mean? (Round your answer to 2 decimal places.)
Determine a 90% confidence interval for the population mean. (Use z Distribution Table.) (Round your answers to 2 decimal places.)
rev: 11_20_2014_QC_58943
a.What is the point estimate of the population mean? (Round your answer to 2 decimal places.)
Explanation / Answer
A)
It is the sample mean
point estimate = 5.40 [ANSWER]
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b)
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.05
X = sample mean = 5.4
z(alpha/2) = critical z for the confidence interval = 1.645
s = sample standard deviation = 2.5
n = sample size = 55
Thus,
Margin of Error E = 0.554529387
Lower bound = 4.845470613
Upper bound = 5.954529387
Thus, the confidence interval is
( 4.845470613 , 5.954529387 ) [ANSWER]
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For part B, just in case you use juts z = 1.64 for 90% confidence, here is the alternative:
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.05
X = sample mean = 5.4
z(alpha/2) = critical z for the confidence interval = 1.64
s = sample standard deviation = 2.5
n = sample size = 55
Thus,
Margin of Error E = 0.552843887
Lower bound = 4.847156113
Upper bound = 5.952843887
Thus, the confidence interval is
( 4.847156113 , 5.952843887 ) [ANSWER]