A sanitation department is interested in estimating the mean amount of garbage p
ID: 3134732 • Letter: A
Question
A sanitation department is interested in estimating the mean amount of garbage per bin for all bins in the city. In a random sample of 40 bins, the sample mean amount was 51.5 pounds and the sample standard deviation was 3.6 pounds. Construct 95% and 99% confidence intervals for the mean amount of garbage per bin for all bins in the city.
a) What is the lower limit of the 95% interval? Give your answer to three decimal places.
b) What is the upper limit of the 95% interval? Give your answer to three decimal places.
c) What is the lower limit of the 99% interval? Give your answer to three decimal places.
d) What is the upper limit of the 99% interval? Give your answer to three decimal places.
Explanation / Answer
a-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 = 0.025
X = sample mean = 51.5
z(alpha/2) = critical z for the confidence interval = 1.959963985
s = sample standard deviation = 3.6
n = sample size = 40
Thus,
Margin of Error E = 1.115631058
Lower bound = 50.38436894 [ANSWER, A]
Upper bound = 52.61563106 [ANSWER, B]
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c-d)
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 = 0.005
X = sample mean = 51.5
z(alpha/2) = critical z for the confidence interval = 2.575829304
s = sample standard deviation = 3.6
n = sample size = 40
Thus,
Margin of Error E = 1.466187743
Lower bound = 50.03381226 [ANSWER, C]
Upper bound = 52.96618774 [ANSWER, D]
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