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
₁ = 5.2
Standard deviation
₁ = 1.1
Company B
Sample size n₂ = 21 workers
Mean
₂ = 4.6
Standard deviation
₂ = 4.6
The null hypothesis and the alternative hypothesis can be computed as follows:
![H_o : \mu _1 = \mu_2](https://tex.z-dn.net/?f=H_o%20%3A%20%5Cmu%20_1%20%3D%20%5Cmu_2)
![H_1 : \mu _1 > \mu_2](https://tex.z-dn.net/?f=H_1%20%3A%20%5Cmu%20_1%20%3E%20%5Cmu_2)
The value of the test statistics can be determined by using the formula:
![t = \dfrac{\overline {x_1}- \overline {x_2}}{\sqrt{\sigma p^2( \dfrac{1}{n_1}+\dfrac{1}{n_2})}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cdfrac%7B%5Coverline%20%7Bx_1%7D-%20%5Coverline%20%7Bx_2%7D%7D%7B%5Csqrt%7B%5Csigma%20p%5E2%28%20%5Cdfrac%7B1%7D%7Bn_1%7D%2B%5Cdfrac%7B1%7D%7Bn_2%7D%29%7D%7D)
where;
![\sigma p^2= \dfrac{(n_1 -1) \sigma_1^2+ (n_2-1)\sigma_2^2}{n_1+n_2-2}](https://tex.z-dn.net/?f=%5Csigma%20p%5E2%3D%20%5Cdfrac%7B%28n_1%20-1%29%20%5Csigma_1%5E2%2B%20%28n_2-1%29%5Csigma_2%5E2%7D%7Bn_1%2Bn_2-2%7D)
![\sigma p^2= \dfrac{(16 -1) (1.1)^2+ (21-1)4.6^2}{16+21-2}](https://tex.z-dn.net/?f=%5Csigma%20p%5E2%3D%20%5Cdfrac%7B%2816%20-1%29%20%281.1%29%5E2%2B%20%2821-1%294.6%5E2%7D%7B16%2B21-2%7D)
![\sigma p^2= \dfrac{(15) (1.21)+ (20)21.16}{35}](https://tex.z-dn.net/?f=%5Csigma%20p%5E2%3D%20%5Cdfrac%7B%2815%29%20%281.21%29%2B%20%2820%2921.16%7D%7B35%7D)
![\sigma p^2= \dfrac{18.15+ 423.2}{35}](https://tex.z-dn.net/?f=%5Csigma%20p%5E2%3D%20%5Cdfrac%7B18.15%2B%20423.2%7D%7B35%7D)
![\sigma p^2= \dfrac{441.35}{35}](https://tex.z-dn.net/?f=%5Csigma%20p%5E2%3D%20%5Cdfrac%7B441.35%7D%7B35%7D)
![\sigma p^2= 12.61](https://tex.z-dn.net/?f=%5Csigma%20p%5E2%3D%2012.61)
Recall:
![t = \dfrac{\overline {x_1}- \overline {x_2}}{\sqrt{\sigma p^2( \dfrac{1}{n_1}+\dfrac{1}{n_2})}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cdfrac%7B%5Coverline%20%7Bx_1%7D-%20%5Coverline%20%7Bx_2%7D%7D%7B%5Csqrt%7B%5Csigma%20p%5E2%28%20%5Cdfrac%7B1%7D%7Bn_1%7D%2B%5Cdfrac%7B1%7D%7Bn_2%7D%29%7D%7D)
![t = \dfrac{5.2- 4.6}{\sqrt{12.61( \dfrac{1}{16}+\dfrac{1}{21})}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cdfrac%7B5.2-%204.6%7D%7B%5Csqrt%7B12.61%28%20%5Cdfrac%7B1%7D%7B16%7D%2B%5Cdfrac%7B1%7D%7B21%7D%29%7D%7D)
![t = \dfrac{0.6}{\sqrt{12.61( \dfrac{37}{336})}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cdfrac%7B0.6%7D%7B%5Csqrt%7B12.61%28%20%5Cdfrac%7B37%7D%7B336%7D%29%7D%7D)
![t = \dfrac{0.6}{\sqrt{12.61(0.110119)}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cdfrac%7B0.6%7D%7B%5Csqrt%7B12.61%280.110119%29%7D%7D)
![t = \dfrac{0.6}{\sqrt{1.38860059}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cdfrac%7B0.6%7D%7B%5Csqrt%7B1.38860059%7D%7D)
![t = \dfrac{0.6}{1.178388981}](https://tex.z-dn.net/?f=t%20%3D%20%5Cdfrac%7B0.6%7D%7B1.178388981%7D)
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
=
= 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.