Question

The Bureau of Labor Statistics reported that the average yearly income of dentists in the year 2012 was $110,000.  A sample of 81 dentists, which was taken in 2013, showed an average yearly income of $120,000.  Assume the standard deviation of the population of dentists’ incomes in 2013 is $36,000.

Test the hypotheses at 95% confidence.

Answer 1

Given: = 110000, = 120000, = 36000, n = 81, = 0.05

The Hypothesis:

H0: = 110,000

Ha: 110,000

This is a 2 tailed test

The Test Statistic: Since the population standard deviation is known, we use the z test.

The test statistic is given by the equation:

The p Value:    The p value (2 Tail) for Z= 2.5, is; p value = 0.0124

The Critical Value:   The critical value (2 Tail) at = 0.05, Zcritical= +1.96 and - 1.96

The Decision Rule:   If Zobservedis > Zcritical or if Zobserved is < -Zcritical, Then reject H0.

Also if P value is < , Then Reject H0.

The Decision:   Since Zobserved (2.5) is > Zcritical (1.96), We Reject H0.

Also since P value (0.0124) is < (0.05) , We Fail to Reject H0.

The Conclusion: There is sufficient evidence at the 95% significance level to warrant rejection of the claim that the average yearly income of dentists in the year 2012 was equal to $110,000.