Question

In preparation for upcoming wage negotiations with the union, the managers for the Bevel Hardware Company want to establish the time required to assemble a kitchen cabinet. A first line supervisor believes that the job should take 40 minutes on average to complete. A random sample of 120 cabinets has an average assembly time of 42 minutes with a standard deviation of 8 minutes. Is there overwhelming evidence to contradict the first line supervisors belief at a 0.05 significance level

Answer 1

Answer:

$t=\frac{42-40}{\frac{8}{\sqrt{120}}}=2.739$  

The p value for this case would be given by:

$p_v =2*P(z>2.739)=0.0616$  

Since the p value is lower than the significance level of 0.05 we have enough evidence to conclude that the true mean for the assembly time is significantly different from 40 minutes.

Step-by-step explanation:

Information provided

$\bar X=42$ represent the sample mean for the assembly time

$s=8$ represent the sample deviation

$n=120$ sample size  

$\mu_o =40$ represent the value to verify

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value for the test

System of hypothesis

We want to conduct a hypothesis in order to see if the true mean is equal to 40 minutes or not, the system of hypothesis would be:  

Null hypothesis:$\mu =40$  

Alternative hypothesis:$\mu \neq 40$  

The statistic for this case is given by:

$z=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)  

Replacing the info given we got:

$t=\frac{42-40}{\frac{8}{\sqrt{120}}}=2.739$  

The p value for this case would be given by:

$p_v =2*P(z>2.739)=0.0616$  

Since the p value is lower than the significance level of 0.05 we have enough evidence to conclude that the true mean for the assembly time is significantly different from 40 minutes.