A report states that the mean yearly salary offer for students graduating with a degree in accounting is $48,722. Suppose that a random sample of 50 accounting graduates at a large university who received job offers resulted in a mean offer of $49,870 and a standard deviation of $3900. 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 $48,722? Test the relevant hypotheses using α = 0.05. State your conclusion.A. 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 $48,722.B. 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 $48,722.C. Reject H0. We have convincing evidence that the mean salary offer for accounting graduates of this university is higher than the national average of $48,722.D. 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 $48,722.

Option C - Reject H0. We have convincing evidence that the mean salary offer for accounting graduates of this university is higher than the national average of $48,722.

Step-by-step explanation:

First of all let's define the hypothesis;

Null hypothesis;H0; μ = $48,722

Alternative hypothesis;Ha; μ > $48,722

Now, let's find the test statistic for the z-score. Formula is;

z = (x' - μ)/(σ/√n)

We are given;

x' = 48,722

μ = 49,870

σ = 3900

n = 50

Thus;

z = (49870- 48722)/(3900/√50)

z = 2.08

So from online p-value calculator as attached, using z = 2.08 and α = 0.05 ,we have p = 0.037526

This p-value of 0.037526 is less than the significance value of 0.05,thus, we reject the claim that that the mean salary offer for accounting graduates of this university is higher than the national average of $48,722