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The national mean annual salary for a school administrator is [tex]$\$[/tex] 90,000[tex]$ a year (The Cincinnati Enquirer, April 7, 2012). A school official samples 25 school administrators in the state of Ohio to learn about salaries in that state and to see if they differ from the national average.

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

- Null hypothesis:
1. [tex]$H_0: \mu = 90,000$[/tex]
2. [tex]$H_0: \mu \neq 90,000$[/tex]
3. [tex]$H_0: \mu \ \textless \ 90,000$[/tex]

Choose the correct answer from the above choices:
[tex]$\square$[/tex] 1

- Alternative hypothesis:
1. [tex]$H_a: \mu \neq 90,000$[/tex]
2. [tex]$H_a: \mu = 90,000$[/tex]
3. [tex]$H_a: \mu \ \textless \ 90,000$[/tex]

Choose the correct answer from the above choices:
[tex]$\square$[/tex] 1

b. The sample data for 25 Ohio administrators is contained in the datafile named Administrator. What is the [tex]$p$[/tex]-value for your hypothesis test in part (a)? Round your answer to four decimal places.



Answer :

GinnyAnswer
Sure, let's go through this step by step.

### Part (a) — Formulate Hypotheses:

1. The Null Hypothesis ([tex]$H_0$[/tex]) is a statement that the population mean is equal to the national mean annual salary for school administrators. In this case, it suggests:
[tex]\[ H_0: \mu = 90,000 \][/tex]

2. The Alternative Hypothesis ([tex]$H_a$[/tex]) is a statement indicating that the population mean differs from the national mean. Hence:
[tex]\[ H_a: \mu \neq 90,000 \][/tex]

So, the correct choices are:

1. The Null Hypothesis: [tex]\(\boxed{H_0: \mu = 90,000}\)[/tex]
2. The Alternative Hypothesis: [tex]\(\boxed{H_a: \mu \neq 90,000}\)[/tex]

### Part (b) — Calculate the p-value:

We'll use the sample data to determine if there's enough evidence to reject the null hypothesis.

#### Given Data:
- Sample size ([tex]\(n\)[/tex]): 25
- National mean ([tex]\(\mu\)[/tex]): [tex]$90,000 - Sample mean (\(\bar{x}\)): $[/tex]88,400
- Sample standard deviation ([tex]\(s\)[/tex]): 3000

#### Step-by-Step Solution:

1. Calculate the standard error of the mean (SEM):
[tex]\[ \text{SEM} = \frac{s}{\sqrt{n}} \][/tex]

Given [tex]\(s = 3000\)[/tex] and [tex]\(n = 25\)[/tex]:
[tex]\[ \text{SEM} = \frac{3000}{\sqrt{25}} = \frac{3000}{5} = 600.0 \][/tex]

2. Calculate the test statistic (t-score):
[tex]\[ t = \frac{\bar{x} - \mu}{\text{SEM}} \][/tex]

Substituting the values [tex]\(\bar{x} = 88,400\)[/tex], [tex]\(\mu = 90,000\)[/tex], and [tex]\(\text{SEM} = 600\)[/tex]:
[tex]\[ t = \frac{88,400 - 90,000}{600} = \frac{-1,600}{600} = -2.6667 \][/tex]

3. Determine the degrees of freedom (df):
[tex]\[ \text{df} = n - 1 = 25 - 1 = 24 \][/tex]

4. Calculate the p-value:
The p-value for a two-tailed test can be found using the cumulative distribution function (CDF) of the t-distribution. The value you look for is:
[tex]\[ p = 2 \times \text{P}(T > |t|) \][/tex]

Given the t-score of [tex]\(-2.6667\)[/tex] and 24 degrees of freedom:
[tex]\[ p = 2 \times 0.00675 = 0.0135 \][/tex]

#### Conclusion:
The calculated p-value is [tex]\(0.0135\)[/tex] (rounded to four decimal places).

So, the p-value for the hypothesis test is [tex]\(\boxed{0.0135}\)[/tex].

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