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Magazine Chapter 1: Q57E (page 46)
A sample of 20 glass bottles of a particular type was selected, and the internal pressure strength of each bottle was determined. Consider the following partial sample information:
median = 202.2 lower fourth = 196.0
upper fourth = 216.8
Three smallest observations 125.8 188.1 193.7
Three largest observations 221.3 230.5 250.2
a. Are there any outliers in the sample? Any extreme outliers?
b. Construct a boxplot that shows outliers, and comment on any interesting features.
Short Answer
- Yes, there are outliers in the sample. Also, there are extreme outliers in the sample.
- The boxplot is:
The box-plot shows the extreme and mild outliers exists.
Step by step solution
Given information
There are 20 glass bottles, so the sample size is 20. Following are the sample information provided:
\(Median = 202.20\)
\(Lower{\rm{ }}Fourth = 196\)
\(Upper{\rm{ }}Fourth = 216.8\)
Compute the inter-quartile range
Let\(IQR\)be the required inter-quartile range. The inter-quartile range is calculated by using the following formula:
\(IQR = {Q_3} - {Q_1}\)
\({Q_1}\)and\({Q_3}\)are lower fourth and upper fourth respectively, then substitute the values of\({Q_1} = 196\)and\({Q_3} = 216.8\)in the above formula,
\(\begin{aligned}IQR &= 216.8 - 196\\ &= 20.8\end{aligned}\)
Compute lower inner fence and upper inner fence
To check whether there are any outliers in the given data, we need to calculate the lower inner fence and upper inner fence by using the formulas:
\(Lower\,fence = {Q_1} - 1.5IQR\)
\(Upper\,fence = {Q_3} + 1.5IQR\)
Substitute the values of \({Q_1} = 196\), \({Q_3} = 216.8\) and \(IQR = 20.8\)in the above formulas,
\(\begin{aligned}Lower\,fence &= 196 - 1.5\left( {20.8} \right)\\ &= 196 - 31.2\\ &= 164.8\end{aligned}\)
\(\begin{aligned}Upper\,fence &= 216.8 + 1.5\left( {20.8} \right)\\ &= 216.8 + 31.2\\ &= 248\end{aligned}\)
From lower and upper inner fence, it can be observed that there are two outliers exists in the given sample i.e. 125.8 and 250.2.
Compute lower outer fence and upper outer fence
To check whether there are any extreme outliers in the given data, we need to calculate the lower outer fence and upper outer fence by using the formulas:
\(Lower\,fence = {Q_1} - 3IQR\)
\(Upper\,fence = {Q_3} + 3IQR\)
Substitute the values of \({Q_1} = 196\), \({Q_3} = 216.8\) and \(IQR = 20.8\)in the above formulas,
\(\begin{aligned}Lower\,fence &= 196 - 3\left( {20.8} \right)\\ &= 196 - 62.4\\ &= 133.6\end{aligned}\)
\(\begin{aligned}Upper\,fence &= 216.8 + 3\left( {20.8} \right)\\ &= 216.8 + 62.4\\ &= 279.2\end{aligned}\)
From lower and upper inner fence, it can be observed that there is only one extreme outlier exists in the given sample i.e. 125.8.
Construct boxplot for the given data
Following are the steps to make boxplot by hand:
- Draw a plot line of range 120 to 270.
- Draw three vertical lines that consists of first quartile, second quartile and third quartile and make two horizontal lines to make it in rectangular form like a box.
- Draw whiskers on both sides and set the minimum and maximum value with respect to the obtained lower fence and upper fence. Since, 125.8 and 250.2 are outliers so ignore it and take minimum value as 188.1 and maximum value as 130.5.
- Mark all the outliers with each numerical value.
Draw conclusion from box-plot
From the obtained box-plot of given data, it is visually observed that there are two outliers exists in the data. One is a mild outlier i.e. 250.2 and the other called an extreme outlier i.e. 125.8
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