Bicycle Commuting Times. A city planner working on bikeways designs a questionnaire to obtain information about local bicycle commuters. One of the questions asks how long it takes the rider to pedal from home to his or her destination. A sample of local bicycle commuters yields the following times, in minutes.

a. Find a $90\%$confidence interval for the mean commuting time of all local bicycle commuters in the city. (Note: The sample mean and sample standard deviation of the data are $25.82$minutes and $7.71$minutes, respectively.)

b. Interpret your result in part (a).

c. Graphical analyses of the data indicate that the time of $48\min$ utes may be an outlier. Remove this potential outlier and repeat part (a). (Note: The sample mean and sample standard deviation of the abridged data are $24.76$ and $6.05$, respectively.)

d. Should you have used the procedure that you did in part (a)? Explain your answer.

The formula used: Confidence interval $\left(\overline{x}-t_{\frac{\alpha }{2}}\frac{s}{\sqrt{n}},\overline{x}+t_{\frac{\alpha }{2}}\frac{s}{\sqrt{n}}\right)$

The population mean commuting time is$\mu$, while the population standard deviation is $\sigma$

Sample size $n=22$

Sample mean, $\overline{x}=25.82$

Sample S.D, $s=7.71$

Confidence level $=90\%$

$$\begin{gathered} =0.90\times 100\% \\ \therefore 1-\alpha =0.90 \\ \Rightarrow \alpha =0.10 \\ \Rightarrow \frac{\alpha }{2}=0.05 \end{gathered}$$

$100(1-\alpha )\%$Confidence interval of $\mu$, using $t-$interval procedure is given by $\left(\overline{x}-t_{\frac{\alpha }{2}}\frac{s}{\sqrt{n}},\overline{x}+t_{\frac{\alpha }{2}}\frac{s}{\sqrt{n}}\right)$

Where $t_{\frac{\alpha }{2}}$is the $t-$value having area $\frac{\alpha }{2}$ to its right with $df=n-1$ $=22-1$ $=21$

$\therefore$ For $\therefore$, the tabulated value, $t_{\alpha /2}=t_{0.05}$ $=1.721$

$\therefore 90\%$ confidence interval of $\mu$

$$\begin{gathered} =\left(\overline{x}-t_{0.05}\frac{s}{\sqrt{n}},\overline{x}+t_{0.05}\frac{s}{\sqrt{n}}\right) \\ =\left(25.82-1.721\times \frac{7.71}{\sqrt{22}},25.82+1.721\times \frac{7.71}{\sqrt{22}}\right) \\ =(22.99,28.65) \end{gathered}$$

We can be $90\%$ certain that the average commuting time for all local bicycle commuters in the city, $\mu$, is between $22.99$ and $28.65$ minutes.

Assume that the data graphical analyses suggest that the time of $48$minutes is an outlier. Observation We deleted the $48-$minute period as an outlier.

As a result, the abbreviated sample has a sample size of$n=21$

$\therefore$The sample size of the abridged sample is $n=21$

$\therefore$Here $df=n-1$

$$\begin{gathered} =21-1 \\ =20 \end{gathered}$$

$\therefore$For $df=20$, the tabulated value, $t_{\frac{\alpha }{2}}=t_{0.05}$$=1.725$

The abridged sample mean, $\overline{x}=24.76$

And sample standard deviation, $s=6.05$

Now, the $90\%$Confidence interval of $\mu$

$$\begin{gathered} =\left(\overline{x}-t_{\frac{\alpha }{2}}\frac{s}{\sqrt{n}},\overline{x}+t_{\frac{\alpha }{2}}\frac{s}{\sqrt{n}}\right) \\ =\left(24.76-1.725\times \frac{6.05}{\sqrt{21}},24.76+1.725\times \frac{6.05}{\sqrt{21}}\right) \\ =(22.48,27.04) \end{gathered}$$

No, the $t-$interval procedure should not have been employed in part (a). Because the sample size is moderate and the data contains outliers,