Answer:
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.
