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 0.02 gallon. A random sample of 50 cans is​ selected, and the sample mean amount of paint per 1​-gallon can is 0.998 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.

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

___because a​ 1-gallon paint can containing exactly​ 1-gallon of paint lies _____the 95​% confidence interval.

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

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

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

A​ 1-gallon paint can containing exactly​ 1-gallon of paint lies ____the 90​% confidence interval. The manager _______a right to complain to the manufacturer.

Answer 1

Answer:

a. 95​% confidence interval estimate for the population mean amount of paint included in a​ 1-gallon can is 0.998±0.0055

b.  No, because a​ 1-gallon paint can containing exactly​ 1-gallon of paint lies within the 95​% confidence interval.

c. Yes.  The population amount of paint per can is assumed normally distributed, because confidence interval calculations assume normal distribution of the parameter.

d. 90% confidence interval is 0.998±0.0046. ​The answer in b. didn't change; 1-gallon paint can containing exactly​ 1-gallon of paint lies within the 90​% confidence interval. The manager doesn't have a right to complain to the manufacturer.

Step-by-step explanation:

Confidence Interval can be calculated using M±ME where

M is the sample mean amount of paint per 1​-gallon can (0.998 gallon)

ME is the margin of error from the mean

And margin of error (ME) can be calculated using the equation

ME=$\frac{z*s}{\sqrt{N} }$ where

• z is the corresponding statistic in the 95% confidence level (1.96)

• s is the sample standard deviation (0.02 gallon)

• N is the sample size (50)

Then ME=$\frac{1.96*0.02}{\sqrt{50} }$≈0.0055

95% confidence interval is 0.998±0.0055

90% confidence interval can be calculated similary, only z statistic is 1.64.

ME=$\frac{1.64*0.02}{\sqrt{50} }$  ≈0.0046

90% confidence interval is 0.998±0.0046