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# Statistics

If 70 percent or more of the old bricks in a brickyard meet certain standards, it will be profitable to hire persons to sort the bricks. According to past experience with this brickyard, on average 70% of all the old bricks meet these standards. A random sample of 100 old bricks is taken from all the (many, many) bricks in the brickyard and the bricks are examined to see if they meet the standards.
Let X = the number of bricks in the sample that meet the standards
Let (p-hat = the sample proportion of bricks meeting the standards

a) Explain how the sample proportion (p-hat) is distributed in this sample
(p-hat is a _____ RV with ______because....)
b) Find the probability that the sample proportion, p-hat is between 0.61 and 0.79.
c) If 60% of bricks are found to conform to the standards, is it likely that the true proportion of old bricks meeting the standards is at least 70%? Explain briefly.

In contract negotiations, a company claims that a new incentive scheme has resulted in average weekly earnings of at least \$400 for all customer services workers. The union is skeptical. A union representative, recalling his college statistics course, decides to test the company's claim. She takes a random sample of 81 workers and finds that these workers have average weekly earnings of \$385 with a sample standard deviation of \$45. The representative uses H0: μ = \$400 and HA < \$400 and a significance level of 0.025.

a) Why didn't the union representative choose μ ≠ \$400?
b) What is the appropriate estimator for this test? Explain exactly how the estimator is distributed for this hypothesis test. (____is a ____ RV with ____ because)
c) Write out the decision rule for this test.
d) Using a significance level of 0.025, should the null hypothesis be rejected?
e) Is the company's claim correct?

#### Solution Summary

There are two statistics problems here regarding estimators, using the null hypothesis, probabilities, and sample proportions.

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