Question

Employees from A and company B receive annual bonuses. What information would you need to test the claim that the difference in annual bonuses is greater than $100 at the 0.5 level of significance? Write out the hypothesis and explain the testing procedure in details

Answer 1

Answer:

1. The required information are

The average annual bonuses, $\bar {x}_1$ received by employees from company A

The average annual bonuses, $\bar {x}_2$ received by employees from company B

The standard deviation, σ₁, of the average annual bonuses for employees from company A

The standard deviation, σ₂, of the average annual bonuses for employees from company A

The number of employees in company A, n₁

The number of employees in company B, n₂

2. The null hypothesis is H₀: $\bar {x}_1$ - $\bar {x}_2$ ≤ 100

The alternative hypothesis is Hₐ: $\bar {x}_1$ - $\bar {x}_2$ > 100

Step-by-step explanation:

1. The required information are

The average annual bonuses, $\bar {x}_1$ received by employees from company A

The average annual bonuses, $\bar {x}_2$ received by employees from company B

The standard deviation, σ₁, of the average annual bonuses for employees from company A

The standard deviation, σ₂, of the average annual bonuses for employees from company A

The number of employees in company A, n₁

The number of employees in company B, n₂

2. The null hypothesis is H₀: $\bar {x}_1$ - $\bar {x}_2$ ≤ 100

The alternative hypothesis is Hₐ: $\bar {x}_1$ - $\bar {x}_2$ > 100

The z value for the hypothesis testing of the difference between two means is given as follows;

$z=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{\frac{\sigma_{1}^{2} }{n_{1}}-\frac{\sigma _{2}^{2}}{n_{2}}}}$

At 0.5 level of significance, the critical $z_\alpha$ = ± 0

The rejection region is z > $z_\alpha$ and z < -$z_\alpha$

Therefore, the value of z obtained from the relation above more than or less than 0, we reject the null hypothesis, and we fail to reject the alternative hypothesis.