Question

The Bureau of Labor Statistics reported that the average yearly income of dentists in the year 2011 was $110,000. A sample of 81 dentists, which was taken in 2012, showed an average yearly income of $120,000. Assume the standard deviation of the population of dentists’ incomes in 2012 is $36,000.

a. We want to test to determine if there has been a significant increase in the average yearly income of dentists. Provide the null and the alternative hypotheses

b. Compute the test statistic.

c. Test the hypotheses at 95% confidence.

Answer 1

Part a)

To Test :-

H0 :- µ = 110000
H1 :- µ > 110000

Part b)

Test Statistic :-
Z = ( X - µ ) / ( σ / √(n))
Z = ( 120000 - 110000 ) / ( 36000 / √( 81 ))
Z = 2.50

Part c)

Test Criteria :-
Reject null hypothesis if Z > Z(α)
Critical value Z(α) = Z(0.05) = 1.645
Z > Z(α) = 2.5 > 1.645
Result :- Reject null hypothesis

Decision based on P value
Reject null hypothesis if P value < α = 0.05 level of significance
P value = P ( Z < 2.5 )
P value = 0.0062
Since 0.0062 < 0.05 ,hence we reject null hypothesis
Result :- Reject null hypothesis

There is sufficient evidence to support the claim that  there has been increase in the average yearly income of dentists.