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

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

and

Here, n = 50

s = 3600

We take , α = 0.05

The test statistics t is given by

t = 2.15

Now the ">" sign in the 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 . Hence the mean salary that the accounting graduates are offered from the university is more than the average salary of 48,734 dollar.

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step-by-step explanation:

step-by-step explanation:

here's what happens:

1) the graph of f   = x^2 is shifted 2 units to the right.

2) the resulting graph is reflected in the x-axis, so that the parabola now opens down.

3) this final result is transformed up by 5 units.