Question

A report states that the mean yearly salary offer for students graduating with a degree in accounting is $48,716. Suppose that a random sample of 50 accounting graduates at a large university who received job offers resulted in a mean offer of $49,860 and a standard deviation of $3700. Do the sample data provide strong support for the claim that the mean salary offer for accounting graduates of this university is higher than the national average of $48,716? Test the relevant hypotheses using

α = 0.05.

(Use a statistical computer package to calculate the P-value. Round your test statistic to two decimal places and your P-value to three decimal places.)

t	=
P-value	=

1. Do not reject H₀. We have convincing evidence that the mean salary offer for accounting graduates of this university is higher than the national average of $48,716.

2. Reject H₀. We do not have convincing evidence that the mean salary offer for accounting graduates of this university is higher than the national average of $48,716.    

3. Do not reject H₀. We do not have convincing evidence that the mean salary offer for accounting graduates of this university is higher than the national average of $48,716.

4. Reject H₀. We have convincing evidence that the mean salary offer for accounting graduates of this university is higher than the national average of $48,716.

Answer 1

To Test ;-

H0 :- µ = 48716
H1 :- µ > 48716

Test Statistic :-
t = ( X̅ - µ ) / (S / √(n) )
t = ( 49860 - 48716 ) / ( 3700 / √(50) )
t = 2.19

P - value = P ( t > 2.1863 ) = 0.017

Reject null hypothesis if P value < α = 0.05 level of significance
P - value = 0.0168 < 0.05 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis

4. Reject H0. We have convincing evidence that the mean salary offer for accounting graduates of this university is hig+her than the national average of $48,716.