The 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 .02 gallon. A random sample of 50 cans is selected, and the sample mean amount of paint per1-gallon can is 0.986 gallon. Complete parts (a) through (d).

a.Construct a 95% confidence interval estimate for the population mean amount of paint included in a 1-gallon can. u (Round to five decimal places as needed.)

b.On the basis of these results, do you think the manager has a right to complain to the manufacturer? Why?

Yes or No because a 1-gallon paint can containing exactly 1-gallon of paint lies

outside or within the 95% confidence interval.

c.Must you assume that the population amount of paint per can is normally distributed here? Explain.

A.Yes, because the Central Limit Theorem almost always ensures that X overbar is normally distributed when n is large. In this case, the value of n is small.

B. Yes, since nothing is known about the distribution of the population, it must be assumed that the population is normally distributed.

C.No, because the Central Limit Theorem almost always ensures that X overbar normally distributed when n is small. In this case, the value of n is small.

D. No, because the Central Limit Theorem almost always ensures that X overbar is normally distributed when n is large. In this case, the value of n is large.

d. Construct a 90% confidence interval estimate. How does this change your answer to part (b)?

(Round to five decimal places as needed.)

How does this change your answer to part (b)?

A 1-gallon paint can containing exactly 1-gallon of paint lies outside or within the 90% confidence interval. The manager still has, still does not have, or now does not have a right to complain to the manufacturer.

Solution

a)
Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=0.986
Standard deviation( sd )=0.02
Sample Size(n)=50
Confidence Interval = [ 0.986 ± Z a/2 ( 0.02/ Sqrt ( 50) ) ]
= [ 0.986 - 1.96 * (0.00283) , 0.986 + 1.96 * (0.00283) ]
= [ 0.98046,0.99154 ]

b)
No because a 1-gallon paint can containing exactly 1-gallon of paint lies
outside the 95% confidence interval.

c)
Yes, since nothing is known about the distribution of the population,
it must be assumed that the population is normally distributed.

d)
Confidence Interval = [ 0.986 ± Z a/2 ( 0.02/ Sqrt ( 50) ) ]
= [ 0.986 - 1.64 * (0.00283) , 0.986 + 1.64 * (0.00283) ]
= [ 0.98136,0.99064 ]

1-gallon paint can containing exactly 1-gallon of paint lies outside the 90% confidence interval. The manager still has, still does not have, or now does not have a right to complain to the manufacturer.