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Novosadov [1.4K]
3 years ago
15

The manager of a paint supply store wants to estimate the actual amount of paint contained in 1​-gallon cans purchased from a na

tionally 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.
Mathematics
1 answer:
lorasvet [3.4K]3 years ago
8 0

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.  <u>No,</u> because a​ 1-gallon paint can containing exactly​ 1-gallon of paint lies <u>within</u> 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 <u>within</u> the 90​% confidence interval. The manager <u>doesn't have</u> 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

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