Respuesta :
Answer:
1. The required information are
The average annual bonuses, [tex]\bar {x}_1[/tex] received by employees from company A
The average annual bonuses, [tex]\bar {x}_2[/tex] 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: [tex]\bar {x}_1[/tex] - [tex]\bar {x}_2[/tex] ≤ 100
The alternative hypothesis is Hₐ: [tex]\bar {x}_1[/tex] - [tex]\bar {x}_2[/tex] > 100
Step-by-step explanation:
1. The required information are
The average annual bonuses, [tex]\bar {x}_1[/tex] received by employees from company A
The average annual bonuses, [tex]\bar {x}_2[/tex] 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: [tex]\bar {x}_1[/tex] - [tex]\bar {x}_2[/tex] ≤ 100
The alternative hypothesis is Hₐ: [tex]\bar {x}_1[/tex] - [tex]\bar {x}_2[/tex] > 100
The z value for the hypothesis testing of the difference between two means is given as follows;
[tex]z=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{\frac{\sigma_{1}^{2} }{n_{1}}-\frac{\sigma _{2}^{2}}{n_{2}}}}[/tex]
At 0.5 level of significance, the critical [tex]z_\alpha[/tex] = ± 0
The rejection region is z > [tex]z_\alpha[/tex] and z < -[tex]z_\alpha[/tex]
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.
