Question

There are two (2) hypothesis testing inquiries.

Hint: one inquiry is either a left-tail test or a right-tail test and one inquiry is a two-tail test.

For each inquiry show all work, including:

1. The null (H0) and alternative (H1) hypothesis.

2. The numerical critical value (ZCrit).

3. The calculation and calculated value (ZCalc).

Recall: z = (Xbar – mu) / (sigma / square-root of the sample size)

4. Determination: Reject H0 or Do not reject H0.

5. A brief explanation as to how you determined whether to Reject H0 not.

1. Student Expenditures: The average expenditure per student (based on average daily attendance) for a certain school year was $10,338 with a population standard deviation of $1560. A survey for the next school year of 150 randomly selected students resulted in a sample mean of $10,797. At the α = 0.10 level of significance, do these results indicate that the average expenditure has changed? (50 points)

For full or partial credit, show all work (every step in your calculation).

2. Salaries of Government Employees: The mean salary of federal government employees on the General Schedule is $58,000. The average salary of 30 state employees who do similar work is $57,600 with σ = $1,500. At the 0.05 level of significance, can it be concluded that state employees earn on average less than federal employees? (50 points)

For full or partial credit, show all work (every step in your calculation).

Answer 1

1)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 10338
Alternative Hypothesis, Ha: μ ≠ 10338

Rejection Region
This is two tailed test, for α = 0.1
Critical value of z are -1.645 and 1.645.
Hence reject H0 if z < -1.645 or z > 1.645

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (10797 - 10338)/(1560/sqrt(150))
z = 3.60

Reject H0

As the value of test statistic, z is within critical value range, reject the null hypothesis

2)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 58000
Alternative Hypothesis, Ha: μ < 58000

Rejection Region
This is left tailed test, for α = 0.05
Critical value of z is -1.645.
Hence reject H0 if z < -1.645

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (57600 - 58000)/(1500/sqrt(30))
z = -1.46

fail to reject null hypothesis.

As the value of test statistic, z is outside critical value range, fail to reject the null hypothesis