Question

# Salary 77600 76000 90700 97200 90700 101800 78700 81300 84200 97600 77500 75700 89400 84300 78700...

Solution:

Given: sample of 25 observations.

A school official took a sample of 25 school administrators in the state of Ohio to learn about salaries in that state and to see if they differed from the national average.

Part a) Formulate hypotheses that can be used to determine whether the population mean annual administrator salaries in Ohio differ from the national mean of \$90,000.

vs

Part b) Use the sample data for the 25 Ohio administrators and the t distribution table (Table 2 in Appendix B) to estimate a p-value range for your hypothesis test in part (a).

Find t test statistic:

where

Thus we need to make following table:

 x Salary (x - xbar)^2 77600 58859584 76000 85969984 90700 29463184 97200 142277184 90700 29463184 101800 273174784 78700 43191184 81300 15776784 84200 1149184 97600 151979584 77500 60403984 75700 91623184 89400 17040384 84300 944784 78700 43191184 84600 451584 87700 5895184 103400 328624384 83800 2166784 101300 256896784 94700 88887184 69200 258309184 95400 102576384 61500 565107984 68800 271326784

Thus t test statistic :

For p-value , find df = n -1 = 25 - 1 = 24

Now look in t table for df = 24 row and find an interval in which t = 2.141 fall.

and then find corresponding two tail area.

t = 2.141 fall in between 2.064 and 2.492 and corresponding two tail area is in between 0.02 and 0.05

thus range of p-value = 0.02 < p-value < 0.05

Part c. At = .05, can your null hypothesis be rejected? What is your conclusion?

p-value is less than 0.05, We reject H0.

We conclude that the mean annual administrator salary in Ohio is different from the national mean annual salary.

Part d. Repeat the preceding hypothesis test using the critical value approach

find t critical value for df = 24 and two tail area = 0.05 and find t critical value.

t critical value = 2.064 = 2.06

Since this is two tailed test, t critical values are ( -2.06 , 2.06)

Since t test statistic value = -2.141 < -2.06 , we reject H0.

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