Question

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

Answer 1

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 $\mu$₁ = 5.2

Standard deviation $\sigma$₁ = 1.1

Company B

Sample size n₂ = 21 workers

Mean $\mu$₂ = 4.6

Standard deviation $\mu$₂ = 4.6

The null hypothesis and the alternative hypothesis can be computed as follows:

$H_o : \mu _1 = \mu_2$

$H_1 : \mu _1 > \mu_2$

The value of the test statistics can be determined by using the formula:

$t = \dfrac{\overline {x_1}- \overline {x_2}}{\sqrt{\sigma p^2( \dfrac{1}{n_1}+\dfrac{1}{n_2})}}$

where;

$\sigma p^2= \dfrac{(n_1 -1) \sigma_1^2+ (n_2-1)\sigma_2^2}{n_1+n_2-2}$

$\sigma p^2= \dfrac{(16 -1) (1.1)^2+ (21-1)4.6^2}{16+21-2}$

$\sigma p^2= \dfrac{(15) (1.21)+ (20)21.16}{35}$

$\sigma p^2= \dfrac{18.15+ 423.2}{35}$

$\sigma p^2= \dfrac{441.35}{35}$

$\sigma p^2= 12.61$

Recall:

$t = \dfrac{\overline {x_1}- \overline {x_2}}{\sqrt{\sigma p^2( \dfrac{1}{n_1}+\dfrac{1}{n_2})}}$

$t = \dfrac{5.2- 4.6}{\sqrt{12.61( \dfrac{1}{16}+\dfrac{1}{21})}}$

$t = \dfrac{0.6}{\sqrt{12.61( \dfrac{37}{336})}}$

$t = \dfrac{0.6}{\sqrt{12.61(0.110119)}}$

$t = \dfrac{0.6}{\sqrt{1.38860059}}$

$t = \dfrac{0.6}{1.178388981}$

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 $t_{\alpha ,d.f}$ = $t_{0.05 , 35}$ = 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.