1. Company blue-collar workers
The data for this assignment are listed in the Excel file Company blue-collar workers.
In the midst of labor-management negotiations, the president of a company argues that the company’s blue-collar workers, who are paid an average of 30000 dollars per year, are well-paid because the mean annual income of all blue-collar workers in the country is less than 30000 dollars. That figure is disputed by the union, which does not believe that the mean blue-collar income is less than 30000 dollars. To test the company president’s belief, an arbitrator draws a random sample of 350 blue-collar workers from across the country and asks each to report his or her annual income. The arbitrator assumes that the blue-collar incomes are normally distributed with a standard deviation of 8000 dollars.
a. Compute the mean blue-collar annual income of the sample.
b. What would you take as null hypothesis H_0 and alternative hypothesis H_1?
c. What is the test statistic and can it be inferred at the 5% significance level that the company president is correct? Explain your answer.
d. What test statistic would you use if the population standard deviation of the blue-collar incomes is not known? Explain the consequences.


2. Variable gas production
The data for this assignment are listed in the Excel file Variable gas production.
With gasoline prices increasing, drivers are more concerned with their cars’ gasoline consumption. For the past 5 years, a driver has tracked the gas mileage of his car and found that the standard deviation from fill-up to fill-up was \sigma=4.796 miles per gallon. Now that his car is 5 years old, he would like to know whether the variability of gas mileage has changed. He recorded the gas mileage from his last eight fill-ups; these are listed in the data files.
a. Formulate the null hypothesis H_0 and the alternative hypothesis H_1.
b. Which test statistic should be used?
c. Infer at a 10% significance level whether the variability has changed. Explain your answer.

a. H_0: \sigma^2=23; H_1: \sigma^2\neq23 (4.976^2\approx23).
b. The test statistic is \displaystyle{\chi^2=\frac{(n-1)s^2}{\sigma^2}}.
c. This is a two-sided test and thus the rejection region is: \chi^2<\chi^2_{1-\alpha/2, n-1}=\chi^2_{0.95;7}=14.1.
From the data in the data file we compute s^2=16.50 (by Excel). Thus, the test statistic is \chi^2=\frac{(8-1)16.50}{23}=5.02. This number is not in the rejection region, so there is not enough evidence to infer that the population variance has changed.

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