Question

A certain daily delivery route for Hostess breads and snack cakes includes eight grocery stores and four convenience stores. The historical mean tome to complete these deliveries (to the 12 stores) and return to the distribution center is 6.5 hours. A new deliver has been assigned to this route, and a random sample of his route completion times (in hours) was obtained. The data are given below:

6.61, 6.25, 6.40, 6.57, 6.35, 5.95, 6.53, 6.29

Required:
Assume the underlying population is normal. Is there evidence to suggest that the new driver has been able to shorten the route completion time, at the level of significance 0.01?

a. 0.05 b. 0.1 < p-value < 0.2: There is no evidence to suggest tha tthe new driver has been able to shorten the mean delivery time for this route.
c. 0.1 < p-value < 0.2: There is an enough evidence to suggest thatthe new driver has been able to shorten the mean delivery time for this route.
d. 0.05 < p-value < 0.1: There is no evidence to suggest tha tthe new driver has been able to shorten the mean delivery time for this route.

Answer 1

d. 0.05 < p-value < 0.1: There is no evidence to suggest tha tthe new driver has been able to shorten the mean delivery time for this route.

Explanation:

We calculate the mean and standard deviation of the sample:

`M=(1)/(8)\sum_(i=1)^(8)(6.61+6.25+6.4+6.57+6.35+5.95+6.53+6.29)\\\\\\ M=(50.95)/(8)=6.369`

`s=\sqrt{(1)/((n-1))\sum_(i=1)^(8)(x_i-M)^2}\\\\\\s=\sqrt{(1)/(7)\cdot [(6.61-(6.369))^2+...+(6.29-(6.369))^2]}\\\\\\s=\sqrt{(0.3216875)/(7)}=√(0.046)\\\\\\s=0.214`

This is a hypothesis test for the population mean.

The claim is that the new driver has been able to shorten the route completion time (significantly less than 6.5 hours).

Then, the null and alternative hypothesis are:

`H_0: \mu=6.5\\\\H_a:\mu< 6.5`

The significance level is 0.01.

The sample has a size n=8.

The sample mean is M=6.369.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.214.

The estimated standard error of the mean is computed using the formula:

`s_M=(s)/(√(n))=(0.214)/(√(8))=0.08`

Then, we can calculate the t-statistic as:

`t=(M-\mu)/(s/√(n))=(6.369-6.5)/(0.08)=(-0.13)/(0.08)=-1.73`

The degrees of freedom for this sample size are:

`df=n-1=8-1=7`

This test is a left-tailed test, with 7 degrees of freedom and t=-1.73, so the P-value for this test is calculated as (using a t-table):

`P-value=P(t<-1.73)=0.063`

As the P-value (0.063) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the new driver has been able to shorten the route completion time (significantly less than 6.5 hours).