The national mean annual salary for a school administrator is $90,00 a year (The Cincinnati Enquirer, April 7, 2012). A school official took a sample of 25 school administrators in the state of Ohio to learn about the salaries in that state to see if they differed from the national average.

A) Formulate hypotheses that can be used to determine whether the population mean annual administrator salary in Ohio differs from the national mean of $90,000.

B) The sample data for 25 Ohio administrators is contained below. What is the p-value for your hypothesis test in part A?

C) At alpha = 0.05, can your null hypothesis be rejected? What is your conclusion?

c) If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean for the salary differs from 9000 at 5% of significance.

Step-by-step explanation:

1) Data given and notation

77600 ,76000 ,90700 ,97200 ,90700 ,101800 ,78700 ,81300 ,84200 ,97600 ,

77500 ,75700 ,89400 ,84300 ,78700 ,84600 ,87700 ,103400 ,83800 ,101300

94700 ,69200 ,95400 ,61500 ,68800

We can calculate the sample mean and deviation with the following formulas:

[tex]\bar X =\frac{\sum_{i=1}^n X_i}{n}[/tex]

[tex]s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]

The values obtained are:

[tex]\bar X=85272[/tex] represent the mean annual salary for the sample

[tex]s=11039.23[/tex] represent the sample standard deviation for the sample

[tex]n=25[/tex] sample size

[tex]\mu_o =90000[/tex] represent the value that we want to test

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

t would represent the statistic (variable of interest)

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

Part a: State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean salary differs from 90000, the system of hypothesis would be:

Null hypothesis:[tex]\mu = 90000[/tex]

Alternative hypothesis:[tex]\mu \neq 90000[/tex]

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

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

t-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".

Part b: Calculate the statistic

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

[tex]t=\frac{85272-90000}{\frac{11039.23}{\sqrt{25}}}=-2.141[/tex]

P-value

The first step is calculate the degrees of freedom, on this case:

[tex]df=n-1=25-1=24[/tex]

Since is a two sided test the p value would be:

[tex]p_v =2*P(t_{(24)}<-2.141)=0.043[/tex]

Part c: Conclusion

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean for the salary differs from 9000 at 5% of significance.