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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.