Question

A report states that the mean yearly salary offer for students graduating with a degree in accounting is $48,734. Suppose that a random sample of 50 accounting graduates at a large university who received job offers resulted in a mean offer of $49,830 and a standard deviation of $3600. 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 $48,734

Answer 1

Solution :

To claim to be tested is whether "the mean salary is higher than 48,734".

i.e. μ > 48,734

Therefore the null and the alternative hypothesis are

$H_0 : \mu = 48,734$

and $H_1 : \mu > 48,734$

Here, n = 50

$\bar x = 49,830$

s = 3600

We take , α = 0.05

The test statistics t is given by

$t=\frac{\bar x - \mu}{\frac{s}{\sqrt n}}$

$t=\frac{49,830 - 48,734}{\frac{3600}{\sqrt 50}}$

t = 2.15

Now the ">" sign in the $H_1$ sign indicates that the right tailed test

Now degree of freedom, df = n - 1

                                              = 50 - 1

                                              = 49

Therefore, the p value = 0.02

The observed p value is less than α = 0.05, therefore we reject $H_0$. Hence the mean salary that the accounting graduates are offered from the university is more than the average salary of 48,734 dollar.