Chapter 1.3, Problem 63E

(a)

To determine

To find: The Inter Quartile Range $\left(IQR\right)$ for the data.

Answer to Problem 63E

Solution: The Inter Quartile Range $\left(IQR\right)$ is 68

Explanation of Solution

Calculation: The Inter Quartile Range $\left(IQR\right)$ can be obtained by using Minitab. Follow the steps given below:

Step 1: Enter the provided data in a Minitab worksheet.

Step 2: Go to Stat and select basic statistics.

Step 3: Then select Display Descriptive Statistics and enter the name of the column containing the data in the field marked as variables.

Step 4: Next click on Statistics tab and tick mark the option against interquartile range and click on OK twice to get the results.

From the Minitab output the Inter Quartile Range $\left(IQR\right)$ is 68.

Interpretation: The Inter Quartile Range $\left(IQR\right)$ refers to difference between Third Quartile and First quartile and indicates spread of data set. The IQR of data provided data is 68.

(b)

To determine

To find: The outliers using $1.5\times \left(IQR\right)$ rule.

Answer to Problem 63E

Solution: The value of outliers in the provided data is 428 of London. Any value above 236.5 or below –35.5 would be considered outliers as they are upper whisker and lower whisker, respectively.

Explanation of Solution

Calculation: To obtain the value of first and third quartile, follow the steps given below in Minitab,

Step 1: Enter the data in a Minitab worksheet.

Step 2: Go to Stat and select basic statistics.

Step 3: Then select Display Descriptive Statistics. Enter the name of the column containing the provided data in the variables textbox.

Step 4: Then click on Statistics tab and tick mark the option against first quartile and third quartile and click on OK twice.

The value of $Q_1$ is 66.5 and $Q_3$ is 134.5 as shown in Minitab.

The formula for upper whisker is,

$UW=Q_3+1.5\times (IQR)$.

where $Q_1$ is first quartile, $Q_3$ is third quartile, and $IQR$ is $Q_3-Q_1$.

$\begin{array}{c}UW=Q_3+1.5\times \left(Q_3-Q_1\right)\\ UW=134.5+1.5\times \left(134.5-66.5\right)\\ UW=236.5\end{array}$

The formula for lower whisker is,

$LW=Q_1-1.5\times (IQR)$.

where $Q_1$ is first, $Q_3$ is third quartile, and $IQR$ is $Q_3-Q_1$.

$\begin{array}{c}LW=Q_1-1.5\times \left(Q_3-Q_1\right)\\ LW=66.5-1.5\times \left(134.5-66.5\right)\\ LW=-35.5\end{array}$

Upper whisker of boxplot is found to be 236.5, so any values above 236.5 would be considered outliers and from data provided, value of 428 of London is outlier. There is no value below lower whisker.

Interpretation: Outliers refers to those data points that lie either above upper whisker and below lower whisker in boxplot. There was one outlier found in this data, and it is 428 of London.

(c)

To determine

To graph: A boxplot of the provided data and describe the distribution using it.

Explanation of Solution

Graph: Plot the boxplot in Minitab by performing the following steps,

Step 1: Enter the data into a Minitab worksheet.

Step 2: Go to ‘Graph’ and click on ‘Boxplot’.

Step 3: In the dialogue box that appears select ‘Simple’ and click OK.

Step 4: Next enter the name of the column containing the data in the filed marked as ‘Graph variables’ and click on OK.

The boxplot is obtained as shown below,

Interpretation: The boxplot is generally preferred to describe dataset having unsymmetrical distribution. The boxplot shows First quartile, Median, and Third quartile. The boxplot of the data is shown above and clearly displays an outlier. The boxplot marks the outlier as a part of the whisker.

(d)

To determine

To graph: A modified boxplot and describes the distribution using it.

Explanation of Solution

Graph: Plot the modified boxplot in Minitab by performing the following steps,

Step 1: Enter the data into a Minitab worksheet.

Step 2: Go to ‘Graph’ and click on ‘Boxplot’.

Step 3: In the dialogue box that appears select ‘Simple’ and click OK.

Step 4: Next enter the name of the column containing the data in the filed marked as ‘Graph variables’ and click on OK.

The boxplot is obtained as shown below,

Interpretation: The modified boxplot is used to display data graphically when the distribution of data is unsymmetrical and skewed as it can clearly show outliers. In the modified boxplot it was found there is one data value which is upper outlier. This outlier is London. The modified boxplot does not display the outlier as a part of the whisker but marks the outlier away.

(e)

To determine

To graph: A stemplot of the provided data.

Explanation of Solution

Graph: Follow the steps given below to obtain the stemplot:

Step 1: Enter the data of sales in a Minitab worksheet.

Step 2: Go to Graph and select stem and leaves.

Step 3: Enter the name of the column containing the data in the Graph variables textbox and click OK.

The required stemplot is attached below,

Interpretation: The stemplot of data is generally drawn when size of data is small and all the data values are positive. It shows all the data values on stemplot. In the stemplot shown above there is one outliers which is London whose data values is 428. Also the data does not seem to be symmetrically distributed.

(f)

To determine

To find: The comparison of Boxplot, Modified boxplot, and stemplot and mention advantages and disadvantages of each.

Answer to Problem 63E

Solution: In boxplot, data is displayed based on five-number summary, which included Minimum, Maximum, first quartile, third quartile, and Median and displaying the outlier as part of the whisker. In Modified boxplot also data is displayed based on five-number summary, but it displays the outliers such that they are not connected to the whiskers. In stemplot, data values are arranged in stem consisting of all digits except right most and leaves contain final digit. Advantage of boxplots is that it is suitable for unsymmetrical data while advantage of stemplot is that it shows all numerical value of data on graph itself. Disadvantage of Boxplot is that it is not suitable for unsymmetrical data while disadvantage of stemplot is that it is used only for positive numbers only and if the data size is small.

Explanation of Solution

The comparison of Boxplot, Modified boxplot, and stemplot is shown below:

	Boxplot	Modified Boxplot	Stemplot
Description	It displays data based on five number summary including Minimum, Maximum, First quartile and Third Quartile and Median.	It displays data based on five number summary including Minimum, Maximum, First quartile and Third Quartile and Median. The outliers are not displayed as part of the whiskers.	In stemplot data values are arranged in stem consisting of all digits except right most digit and leaves contain final digit
Advantages	1. It displays five number summary graphically. 2. It is suitable for unsymmetrical data.	1. It displays five number summary. 2. It is suitable for unsymmetrical data which is skewed. 3. It shows outliers clearly.	1. It can display both symmetrical and unsymmetrical data graphically. 2. It can indicate outliers also 3. It displays all numerical values of data on stemplot.
Disadvantages	1. It is not suitable data set having symmetrical distribution. 2. It does not display outliers on graph.	1. It is not suitable data set having symmetrical distribution.	1. It is not suitable if data size is very large. 2. It is not used fornegative numbers.