Homework SourseThe manager of a paint supply store wants to estimate the ac
The manager of a paint supply store wants to estimate the actual amount of paint contained in 1-gallon cans purchased from a nationally known manufacturer. The manufacturer’s specifications state that the standard deviation of the amount of paint is equal to 0.03 gallon. A random sample of 36 cans is selected and the sample mean amount of paint per 1-gallon can is 0.998 gallon.
Construct a 95% confidence interval estimate for the population mean amount of paint included in a 1-gallon can.
Step 1: figure out a point estimate x-bar:
Step 2: Is (population standard deviation) known? If yes, then find sample sizenand /n
Step 3: find the confidence level (1-) and the critical value z(1-/2).
Step 4: construct a confidence interval for population mean (the actual amount of paint contained in 1-gallon cans purchased). Please interpret the confidence interval.
Solution
1.
Xbar = 0.998 gal [ANSWER]
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2.
sigma = 0.03
n = 36
signa/sqrt(n) = 0.03/sqrt(36) = 0.005 [ANSWER]
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3.
1 - alpha = 0.95 [ANSWER]
Here,
alpha/2 = (1 - confidence level)/2 = 0.025
z(alpha/2) = critical z for the confidence interval = 1.959963985 [ANSWER]
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4.
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 = 0.998
z(alpha/2) = critical z for the confidence interval = 1.959963985
s = sample standard deviation = 0.03
n = sample size = 36
Thus,
Margin of Error E = 0.00979982
Lower bound = 0.98820018
Upper bound = 1.00779982
Thus, the confidence interval is
( 0.98820018 , 1.00779982 ) [ANSWER]
Thus,we are 95% confident that the true mean amount of paint per gallon is from 0.9882 gal to 1.0078 gal. [ANSWER]
