Drive-thru or go inside? Many people think it’s faster to order at the drive-thru than to order inside at fast-food restaurants. To find out, Patrick and William used a random number generator to select$10$times over a $2‐$week period to visit a local Dunkin’ Donuts restaurant. At each of these times, one boy ordered an iced coffee at the drive-thru and the other ordered an iced coffee at the counter inside. A coin flip determined who went inside and who went to the drive-thru. The table shows the times, in seconds, that it took for each boy to receive his iced coffee after he placed the order.

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

We have been given that Patrick and William used a random number generator to select $10$times over a $2‐$week period to visit a local Dunkin’ Donuts restaurant.

One of them ordered an iced coffee at the drive-thru and the other ordered an iced coffee at the counter inside.

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

Given:

$$\begin{gathered} n=Samplesize=10 \\ \alpha =Significancelevel=0.05 \end{gathered}$$

Let us determine the difference between inside time and drive-thru time

Now we will determine the mean of values of the difference:

$$\begin{gathered} \overline{x}=\frac{7+13+4-5+11+1+2+6+3-3}{10} \\ =\frac{39}{10}=3.9 \end{gathered}$$

Now we will find the standard deviation:

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

Now we will carry out a hypothesis test for the population mean difference.

Here we have:

$$\begin{gathered} Populationmeandifference={\mu }_d \\ {H}_0=Nullhyphothesis \\ {H}_a=Alternativehyphothesis \\ {H}_0:{\mu }_d=0 \\ {H}_a:{\mu }_d\ne 0 \\ Nowwewilldeterminethevalueofteststatistics: \\ t=\frac{\overline{x}-{\mu }_d}{{\displaystyle \raisebox{1ex}{$s$}\!\left/ \!\raisebox{-1ex}{$\sqrt{n}$}\right.}} \\ t=\frac{3.9-0}{{\displaystyle \raisebox{1ex}{$5.6460$}\!\left/ \!\raisebox{-1ex}{$\sqrt{10}$}\right.}} \\ \approx 2.184 \end{gathered}$$

The P-value is the probability of obtaining the value of test statistics.

Degree of freedom localid="1654194023311" $=n-1=10-9=1$

The test is a two-tailed test so we double the boundaries of the P-value.

$$\begin{gathered} 0.05=2\left(0.025\right)<P<2\left(0.05\right)=0.10 \\ Nowatt=2.184anddegreeoffreedom=9,Pvalue=0.0568 \end{gathered}$$

We have to reject the null hypothesis if the probability value is less than the significance value.

$P>0.05\Rightarrow failtoreject{H}_0$

This shows that for this Dunkin' Donuts restaurant, there is no convincing evidence of a difference in true mean service time inside and at the drive-thru.