Chapter 10.3, Problem 85E

To determine

To explain do these data provide convincing evidence at the $\alpha =0.05$ level of a difference in true mean service time inside and at the drive-thru for this Dunkin’ Donuts restaurant.

Answer to Problem 85E

There is no convincing evidence at the $\alpha =0.05$ level of a difference in true mean service time inside and at the drive-thru for this Dunkin’ Donuts restaurant.

Explanation of Solution

It is given that:

  $\begin{array}{l}\alpha =0.05\\ n=10\end{array}$

Let us find the difference between the inside time and the drive-thru time for each visit as:

Visit	Inside time	Drive-thru time	Difference
1	62	55	7
2	63	50	13
3	325	321	4
4	105	110	-5
5	135	124	11
6	55	54	1
7	92	90	2
8	75	69	6
9	203	200	3
10	100	103	-3

The mean value of the difference is as:

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{n}\\ =\frac{7+13+4-5+11+1+2+6+3-3}{10}\\ =\frac{39}{10}\\ =3.9\end{array}$

The standard deviation of the difference is as follows:

  $\begin{array}{l}s=\sqrt{\frac{{\displaystyle \sum {(x-\overline{x})}^2}}{n-1}}\\ =\sqrt{\frac{\begin{array}{}$ {(7-3.9)}^2+{(13-3.9)}^2+{(4-3.9)}^2+{(-5-3.9)}^2+{(11-3.9)}^2+$$ {(1-3.9)}^2+{(2-3.9)}^2+{(6-3.9)}^2+{(3-3.9)}^2+{(-3-3.9)}^2$\end{array}}{10-1}}\\ =5.6460\end{array}$

Given claim is that: difference.

The claim is either null hypothesis or an alternative hypothesis.

  $\begin{array}{l}{H}_0:{\mu }_D=0\\ H_a:{\mu }_D\ne 0\end{array}$

The test statistics value will be:

  $\begin{array}{l}t=\frac{{\overline{x}}_D-{\mu }_D}{\frac{s_D}{\sqrt{n}}}\\ =\frac{3.9-0}{\frac{5.6460}{\sqrt{10}}}\\ =2.184\end{array}$

And the degree of freedom will be:

  $\begin{array}{l}df=n-1\\ =10-1\\ =9\end{array}$

As we know that the P -value is the probability of obtaining the sample results or results more extreme when the null hypothesis is true.

  $0.05=2(0.025)<P<2(0.05)=0.10$

And we know that if the P -value is less than or equal to the significance level then the null hypothesis is rejected, then,

  $P>0.05\Rightarrow \text{Fail to Reject }{H}_0$

Thus, we conclude that there is no convincing evidence at the $\alpha =0.05$ level of a difference in true mean service time inside and at the drive-thru for this Dunkin’ Donuts restaurant.