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
[tex]$H_0 : \mu = 48,734$[/tex]
and [tex]$H_1 : \mu > 48,734$[/tex]
Here, n = 50
[tex]$\bar x = 49,830$[/tex]
s = 3600
We take , α = 0.05
The test statistics t is given by
[tex]$t=\frac{\bar x - \mu}{\frac{s}{\sqrt n}}$[/tex]
[tex]$t=\frac{49,830 - 48,734}{\frac{3600}{\sqrt 50}}$[/tex]
t = 2.15
Now the ">" sign in the [tex]$H_1$[/tex] 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 [tex]$H_0$[/tex]. Hence the mean salary that the accounting graduates are offered from the university is more than the average salary of 48,734 dollar.
