91影视

We have been given that Patrick and William used a random number generator to select $10$times over a $2鈥�$week period to visit a local Dunkin鈥� 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 $伪=0.05$level of a difference in the true mean service time inside and at the drive-thru for this Dunkin鈥� Donuts restaurant.

Given:

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

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

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

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

Now we will find the standard deviation:

$$\begin{gathered} s=\sqrt{\frac{{\displaystyle \underset{n=1}{\overset{10}{鈭�}}}{\left(Difference-\stackrel{炉}{x}\right)}^2}{n-1}} \\ =\sqrt{\frac{{\left(7-3.9\right)}^2+{\left(13-3.9\right)}^2+{\left(4-3.9\right)}^2+{\left(-5-3.9\right)}^2+{\left(11-3.9\right)}^2+{\left(1-3.9\right)}^2+{\left(2-3.9\right)}^2+{\left(6-3.9\right)}^2+{\left(3-3.9\right)}^2+{\left(-3-3.9\right)}^2}{10-1}} \\ 鈮�5.6460 \end{gathered}$$

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

Here we have:

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

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.

$P>0.05鈬�failtoreject{H}_0$

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.