Company a samples 16 workers, and their average time with the company is 5.2 years with a standard deviation of 1.1. Company b samples 21 workers and their average time with the company is 4.6 years with a standard deviation 4.6 years
The populations are normally distributed. Determine the:
Hypothesis in symbolic form?
Determine the value of the test statistic?
Find the critical value or value?
determine if you should reject null hypothesis or fail to reject?
write a conclusion addressing the original claim?
Answer:
Step-by-step explanation:
GIven that :
Company A
Sample size n₁ = 16 workers
Mean [tex]\mu[/tex]₁ = 5.2
Standard deviation [tex]\sigma[/tex]₁ = 1.1
Company B
Sample size n₂ = 21 workers
Mean [tex]\mu[/tex]₂ = 4.6
Standard deviation [tex]\mu[/tex]₂ = 4.6
The null hypothesis and the alternative hypothesis can be computed as follows:
[tex]H_o : \mu _1 = \mu_2[/tex]
[tex]H_1 : \mu _1 > \mu_2[/tex]
The value of the test statistics can be determined by using the formula:
[tex]t = \dfrac{\overline {x_1}- \overline {x_2}}{\sqrt{\sigma p^2( \dfrac{1}{n_1}+\dfrac{1}{n_2})}}[/tex]
where;
[tex]\sigma p^2= \dfrac{(n_1 -1) \sigma_1^2+ (n_2-1)\sigma_2^2}{n_1+n_2-2}[/tex]
[tex]\sigma p^2= \dfrac{(16 -1) (1.1)^2+ (21-1)4.6^2}{16+21-2}[/tex]
[tex]\sigma p^2= \dfrac{(15) (1.21)+ (20)21.16}{35}[/tex]
[tex]\sigma p^2= \dfrac{18.15+ 423.2}{35}[/tex]
[tex]\sigma p^2= \dfrac{441.35}{35}[/tex]
[tex]\sigma p^2= 12.61[/tex]
Recall:
[tex]t = \dfrac{\overline {x_1}- \overline {x_2}}{\sqrt{\sigma p^2( \dfrac{1}{n_1}+\dfrac{1}{n_2})}}[/tex]
[tex]t = \dfrac{5.2- 4.6}{\sqrt{12.61( \dfrac{1}{16}+\dfrac{1}{21})}}[/tex]
[tex]t = \dfrac{0.6}{\sqrt{12.61( \dfrac{37}{336})}}[/tex]
[tex]t = \dfrac{0.6}{\sqrt{12.61(0.110119)}}[/tex]
[tex]t = \dfrac{0.6}{\sqrt{1.38860059}}[/tex]
[tex]t = \dfrac{0.6}{1.178388981}[/tex]
t = 0.50917
degree of freedom df = ( n₁ + n₂ - 2 )
degree of freedom df = (16 + 21 - 2)
degree of freedom df = 35
Using Level of significance ∝ = 0.05, From t-calculator , given that t = 0.50917 and degree of freedom df = 35
p - value = 0.3069
The critical value [tex]t_{\alpha ,d.f}[/tex] = [tex]t_{0.05 , 35}[/tex] = 1.6895
Decision Rule: Reject the null hypothesis if the test statistics is greater than the critical value.
Conclusion: We do not reject the null hypothesis because, the test statistics is lesser than the critical value, therefore we conclude that there is no sufficient information that the claim that company a retains it workers longer than more than company b.
