Made-to-order delivery times.Refer to the data on delivery times for a made-to-order product, Exercise 2.34 (p. 87). The delivery times (in days) for a sample of 25 orders are repeated in the accompanying table. (Times marked by an asterisk are associated with customers who subsequently placed additional orders with the company.) Identify any unusual observations (outliers) in the data set, and then use the results to comment on the claim that repeat customers tend to have shorter delivery times than one-time customers.

50* 64* 56* 43* 64* 82* 65* 49* 32* 63* 44* 71 54* 51* 102 49* 73* 50* 39* 86 33* 95 59* 51* 68

We will use the z-score method to find the outliers. For that, we will first arrange the data in ascending order.

(32, 33, 39, 43, 44, 49, 49, 50, 50, 51, 51, 54, 56, 59, 63, 64, 64, 65, 68, 71, 73, 82, 86, 95, 102)

In order to calculate the z-score we require mean and the standard deviation. So next we will calculate that.

$\begin{array}{rcl}\mathrm{Mean}& =& \frac{\mathrm{Sum}\mathrm{of}\mathrm{all}\mathrm{observations}}{\mathrm{No}.\mathrm{of}\mathrm{observations}}\\ & =& \frac{1493}{25}\\ & =& 59.72\end{array}$

$$\begin{gathered} Variance=\frac{\sum \left(x-\overline{x}\right)}{n-1} \\ =\frac{{\left(\text{3}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{3}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{39}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{43}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{44}-\text{59.72}\right)}^{\text{2}} \\ \text{+}{\left(\text{49}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{49}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{50}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{50}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{51}-5\text{9.7}2\right)}^{\text{2}} \\ \text{+ (51}-5{\text{9.72)}}^{\text{2}}\text{+}{\left(\text{54}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{56}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{59}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{63}-\text{59.72}\right)}^{\text{2}} \\ \text{+}{\left(\text{64}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{64}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{65}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{68}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{71}-\text{59.72}\right)}^{\text{2}} \\ \text{+}{\left(\text{73}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{73}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{82}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{86}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{95}-\text{59.72}\right)}^{\text{2}}\text{+}{\left(\text{102}-\text{59.72}\right)}^{\text{2}}}{\text{25 - 1}} \\ \text{=}\frac{7723.04}{24} \\ =321.79 \end{gathered}$$

$\begin{array}{rcl}\mathrm{Standard}\mathrm{Deviation}& =& \sqrt{\mathrm{Variance}}\\ & =& \sqrt{321.79}\\ & =& 17.94\end{array}$

Scores whose z score lies beyond -3 and 3 are outliers

$\begin{array}{rcl}z& =& \frac{x-\mu }{\delta }\\ & =& \frac{32-59.72}{17.94}\\ & =& -1.54\end{array}$

Since the z score for 32 is -1.54 it is not an outlier.

$\begin{array}{rcl}z& =& \frac{x-\mu }{\delta }\\ & =& \frac{95-59.72}{17.94}\\ & =& 1.966\end{array}$

As the z score of 95 is 1.966 it is not an outlier.

$\begin{array}{rcl}z& =& \frac{x-\mu }{\delta }\\ & =& \frac{102-59.72}{17.94}\\ & =& 2.35\end{array}$

Since the z score of 102 is 2.35 it is not an outlier but a suspect outlier.

Therefore, there are no outliers in this data set.

It is observed that the z scores of repeat customer’s lies well within 2 standard deviations from mean and the scores of onetime customers is close to or beyond 2 standard deviations from mean. Therefore, we can say that the delivery time for repeat customers is indeed less.