Question

. The Bureau of Labor Statistics reported that the average yearly income of dentists in the year 2009 was $106,000. Dr. Smith thinks there has been a significant increase in the average yearly income of dentists. He has a dataset of a sample of 81 dentists, which was taken in 2010, showed an average yearly income of $112,800. Assume the standard deviation of the population of dentists in 2010 is $36,000. Please answer the following questions:

(a)	Develop appropriate null and alternative hypotheses such that rejection of H0 will support Dr. Smith’s argument.
(b)	Compute the test statistic.
(c)	Determine the p-value; and at 95% confidence, test the hypotheses.

Answer 1

(a) We have the population mean as , and the sample mean as . The argument given by Dr. Smith is that there has been a significant increase in the average yearly income, ie . Hence, the null hypothesis would be and the alternate hypothesis would be . Rejection of null would mean that alternate hypothesis of is correct, which is the argument of Dr. Smith.

Note that this is would be a one sample one-tailed t-test.

(b) The test statistic would be as or . For the given values, we have or or or .

(c) The p-value of the given t-statistic would be or , for x have t-distribution with df=n-1=81-1=80.

At 95% confidence or 5% significance, we have the critical t as . Since , we may reject the null hypothesis and conclude that the sample mean is greater than the population mean. This basically means that Dr. Smith's assertion is correct at 5% significance level, and the income is significantly (significant at 5%) increased.

This can also be done by considering the p-value. We have the p-value as 0.0465, and since this p-value is less than the significance level (of 5% or 0.05), we may reject the null and conclude the same.

Note that we have used the one-tailed t-test to achieve the results. If this was a two tailed t test, then the p-value would have been and the critical t would have been . As can be seen, either way the test would fail to reject the null at 5% significance. But since the assertion is whether the sample mean of Dr. Smith is GREATER than the population mean provided by BLS, we must use one-tailed test. The two-tailed test would be done if the assertion was that the sample mean is different from the population mean.