The dean of a business school claims that the average starting salary of its graduates is more than 85 (in $000’s). It is known that the population standard deviation is 10 (in $000’s). Sample data on the starting salaries of 64 randomly selected recent graduates yielded a mean of 88 (in $000s).

A) What is the null hypothesis

B) What is the critical value for the rejection region if the level of significance is 5%?

C) What is the value of your test statistic

D) What is your decision

D. If we compare the p value and a significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we can reject the null hypothesis, and the actual true mean for the salary is significantly different from 85. Using the critical region founded on part B we agree with the decision obtained with the p value since 2.4 is on the critical zone, so we reject the null hypothesis.

Step-by-step explanation:

Data given and notation

Let's work without ($000s)

[tex]\bar X=88[/tex] represent the sample mean

[tex]\sigma=10[/tex] represent the standard deviation for the population

[tex]n=64[/tex] sample size

[tex]\mu_o =85[/tex] represent the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value for the test (variable of interest)

A) What is the null hypothesis

We need to conduct a hypothesis in order to determine if the mean salary for graduates its higher than 85, the system of hypothesis would be:

Null hypothesis:[tex]\mu \leq 85[/tex]

Alternative hypothesis:[tex]\mu > 85[/tex]

B) What is the critical value for the rejection region if the level of significance is 5%?

For this case we need one critical values since we are conducting a one right tailed test. We have this equality:

[tex]P(Z>a)=0.05[/tex]

And the value of a that satisfy this is a=1.64. So our critical regions is: [tex] (1.64,\infty)[/tex]

C) What is the value of your test statistic

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can replace in formula (1) the info given like this:

[tex]z=\frac{88-85}{\frac{10}{\sqrt{64}}}=2.4[/tex]

D) What is your decision

Since is a one right tailed test the p value would be:

[tex]p_v =P(Z>2.4)=0.016[/tex]

Conclusion

If we compare the p value and a significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we can reject the null hypothesis, and the actual true mean for the salary is significantly different from 85. Using the critical region founded on part B we agree with the decision obtained with the p value since 2.4 is on the critical zone, so we reject the null hypothesis.