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 |
= | |

-valueP |
= |

1. Do not reject *H*_{0}. 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 *H*_{0}. 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 *H*_{0}. 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 *H*_{0}. 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.

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