The national mean annual salary for a school administrator is $90,000 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 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 in the DATAfile named Administrator. What is the p-value for your hypothesis test in part (a)

Step-by-step explanation:

The question is incomplete. The missing part is given below:

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?

D) Repeat the preceding hypothesis test using the critical value approach?

Salary Data

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

Solution:

Mean = total sum of salaries/number of school administrators

Mean = (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)/25 = 85272

Standard deviation = √(summation(x - mean)²/n

n = 25

√Summation(x - mean)²/n = √[(77600 - 85272)² + (76000 - 85272)² + (90700 - 85272)² + (97200 - 85272)² + (90700 - 85272)² + (101800 - 85272)² + (78700 - 85272)² + (81300 - 85272)² + (84200 - 85272)² + (97600 - 85272)² + (77500 - 85272)² + (75700 - 85272)² + (89400 - 85272)² + (84300 - 85272)² + (78700 - 85272)² + (84600 - 85272)² + (87700 - 85272)² + (103400 - 85272)² + (83800 - 85272)² + (101300 - 85272)² + (94700 - 85272)² + (69200 - 85272)² + (95400 - 85272)² + (61500 - 85272)² + (68800 - 85272)²]/25 = 11039.23

A) We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0 : µ = 90000

For the alternative hypothesis,

H1 : µ ≠ 90000

This is a 2 tailed test.

B) Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 25

Degrees of freedom, df = n - 1 = 25 - 1 = 24

t = (x - µ)/(s/√n)

Where

x = sample mean = 85272

µ = population mean = 90000

s = samples standard deviation = 11039.23

t = (85272 - 90000)/(11039.23/√25) = - 2.14

We would determine the p value using the t test calculator. It becomes

p = 0.043

C) Since alpha, 0.05 > the p value, 0.043, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that the population mean annual administrator salary in Ohio differs from the national mean of $90,000.

D) Since α = 0.05, the critical value is determined from the normal distribution table.

For the left, α/2 = 0.05/2 = 0.025

The z score for an area to the left of 0.005 is - 1.96

For the right, α/2 = 1 - 0.025 = 0.975

The z score for an area to the right of 0.975 is 1.96

In order to reject the null hypothesis, the test statistic must be smaller than - 1.96 or greater than 1.96

Since - 2.14 < - 1.96 and 2.14 > 1.96, we would reject the null hypothesis.