You run a small trucking company that transports goods from Bristol, TN to Winslow, AZ. This trip should take 32 hours and 16 minutes (1936 minutes) if done in accordance with safety regulations. However, you believe that your employees have not been following these rules, because they are completing the trip too quickly. You decide to test your hypothesis. You record the time it takes for your employees to make the trip and record the following results

X =1736 minutes s = 635; n = 46

A) State your null and alternative hypotheses
B) Carry out the test with α=005, and state your conclusion
C) Carry out the test with α=0.01, and state your conclusion
D) Does your conclusion change with the change in a?

And since the p value is lower than the significance level 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly lower than 1736

c) [tex] p_v = P(t_{45} <-2.136) = 0.0191[/tex]

And since the p value is higher than the significance level 0.05 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly lower than 1736

d) For this case the answer is yes since when we change the significance level from 0.05 to 0.01 we see that the final decision changes.

Step-by-step explanation:

Part a

We are trying to proof the following system of hypothesis:

Null hypothesis: [tex]\mu \geq 1936[/tex]

Alternative hypothesis: [tex]\mu <1936[/tex]

Part b

We have the following data given:

[tex]\bar X =1736[/tex] minute s = 635; n = 46

And the statistic for this case is given by:

[tex] t = \frac{\bar X- \mu}{\frac{s}{\sqrt{n}}}[/tex]

And replacing we got:

[tex] t = \frac{1736-1936}{\frac{635}{\sqrt{46}}}= -2.136[/tex]

The degrees of freedom are:

[tex] df = n-1= 46-1=45[/tex]

The p value for this case since is a left tailed test is given by:

[tex] p_v = P(t_{45} <-2.136) = 0.0191[/tex]

And since the p value is lower than the significance level 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly lower than 1736

Part c

We have the same statistic t = -2.136

[tex] p_v = P(t_{45} <-2.136) = 0.0191[/tex]

And since the p value is higher than the significance level 0.05 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly lower than 1736

Part d

For this case the answer is yes since when we change the significance level from 0.05 to 0.01 we see that the final decision changes.