Question

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 0.022 gallon. A random sample of 64 cans is selected, and the sample mean amount of paint per 1-gallon can is 0.988 gallon. a. Construct a 99% confidence interval estimate of the population mean amount of paint included in a 1-gallon can. b. On the basis of your results, do you think that the manager has a right to complain to the manufacturer? Why? c. Must you assume that the population amount of paint per can is normally distributed here? Explain. d. Construct a 95% confidence interval estimate. How does this change your answer to (b)?

Answer 1

a) The confidence interval is calculated as, assuming that the distribution is normal :

μ = M ± Z(s_M)

where:

M = sample mean
Z = Z statistic determined by confidence level
s_M = standard error = √(s²/n)

M = 0.988
Z = 2.58, computed using Z table shown below.
s_M = √(0.022²/64) = 0

μ = M ± Z(s_M)
μ = 0.988 ± 2.58*0
μ = 0.988 ± 0.00708

99% CI [0.98092, 0.99508].

b) Since the interval at 99% confidence level does not include 1 gallon hence we can say that yes the manager has the right to complaint to the manufacturer.

c) At 95% confidence level the interval is calculated as:

M = 0.988
Z = 1.96, computed from Z table below at 95% confidence level.
s_M = √(0.022²/64) = 0

μ = M ± Z(s_M)
μ = 0.988 ± 1.96*0
μ = 0.988 ± 0.00539

95% CI [0.98261, 0.99339].

d) Since the 955 confidence interval also doesnot include the 1 gallon value hence we can say that it does not affects the managers decision of complaining to the manufacturer,

The Z table: