Chapter 2, Problem 2.51E

a.

To determine

To compute: The mean and the median of bonding score of the pain group with and without the two smallest scores.

To explain: which one is more effected the mean or the median and why.

Answer to Problem 2.51E

The mean of pain group with smallest Scores is 3.71.

The median of pain-group with smallest Scores is 4.00.

The mean of pain-group without smallest Scores is 3.90.

The median of pain-group without smallest Scores is 4.14.

The mean is more affected than the median because mean is affected by the extreme values.

Explanation of Solution

Given info: From exercise 2.6, the bonding scores in the pain-group.

Calculation:

Mean with smallest scores of pain-group:

The formula for mean is,

$\overline{x}=\frac{{\displaystyle \sum x_i}}{n}$

The total number of observations in the pain-group are 27. The value of mean,

$\begin{array}{c}\overline{x}=\frac{\left(\begin{array}{l}4.71+4.86+4.14+1.29+2.29+4.43+\\ \mathrm{...}+4.71+4.71+2.14+3.57\end{array}\right)}{27}\\ =\frac{100.29}{27}\\ =3.7144\end{array}$

Thus, the mean of pain group with smallest observations is, 3.71.

Median with smallest scores of pain-group:

The total number of observation is 27. The value of median is,

$\begin{array}{c}\text{Location}\text{of}M=\frac{n+1}{2}\\ =\frac{28}{2}\\ =14\end{array}$

The 14^th term in the observations of pain group is 4.00.

Thus, the median of pain-group with smallest observations is 4.00.

Mean without smallest scores of pain group:

$\begin{array}{c}\overline{x}=\frac{\left(4.71+4.86+4.14+2.29+4.43+\mathrm{...}+4.71+4.71+2.14+3.57\right)}{25}\\ =\frac{97.57}{25}\\ =3.9028\end{array}$

Thus, the mean of pain-group without smallest observations is 3.90.

Median without smallest scores of pain group:

$\begin{array}{c}\text{Location}\text{of}M=\frac{n+1}{2}\\ =\frac{26}{2}\\ =13\end{array}$

The 13^th term in the observations of pain-group is 4.14.

Thus, the median of pain-group without smallest observations is, 4.14.

Conclusion:

The mean and median of pain group with smallest observations is compared to the mean and median of the pain group without smallest observations, the effect is more on mean than the median.

The mean is affected by the extreme values because it is calculated based on all observations of the data, when the smallest observations are omitted from the distribution then mean is more effected. The median is middle value of the distribution and is not affected by the extreme values.

b.

To determine

To find: The $1.5\times IQR$ rule identify the low bonding scores as suspected outliers or not.

Answer to Problem 2.51E

The $1.5\times IQR$ rule identifies the low bonding scores 1.29, 1.43 in the pain-group as suspected outliers.

Explanation of Solution

Given info: From exercise 2.6, the bonding scores in the pain-group

Calculation:

First quartile $Q_1$:

The observations of pain group that are left to the location of median is 13.

$\begin{array}{c}\text{location}\text{of}{Q}_1=\frac{n+1}{2}\\ =\frac{14}{2}\\ =7\end{array}$

Here, 3.43 is the 7^th term in the observations of pain group that are left to location of median.

The first quartile $Q_1$ is, 3.43.

Third quartile $Q_3$:

The observations of pain group that are right to the location of median is 13.

$\begin{array}{c}\text{location}\text{of}{Q}_3=\frac{n+1}{2}\\ =\frac{14}{2}\\ =7\end{array}$

Here, 4.43 is the 7^th term in the observations of pain group that are right to median.

The third quartile $Q_3$ is, 4.43.

Interquartile range (IQR):

$\begin{array}{c}IQR=Q_3-Q_1\\ =4.43-3.43\\ =1\end{array}$

For bonding scores in pain-group, $IQR=1$.

The $1.5\times IQR$ rule for outliers:

An observation is suspected as an outlier if it is more than $Q_3+\left(1.5\times IQR\right)$ and less than $Q_1-\left(1.5\times IQR\right)$.

$\begin{array}{c}{Q}_3+\left(1.5\times IQR\right)=4.43+\left(1.5\times 1\right)\\ =4.43+1.5\\ =5.93\end{array}$

$\begin{array}{c}{Q}_1-\left(1.5\times IQR\right)=3.43-\left(1.5\times 1\right)\\ =3.43-1.5\\ =1.93\end{array}$

Justification:

Outlier:

The data points that lie outside the overall distribution is termed as outlier.

Here, in the pain-group 1.29, 1.43 are far from the remaining observations and are considered as outliers.

The $1.5\times IQR$ rule identifies the two low bonding scores as suspected outliers because they are not in between 1.93 and 5.93.

c.

To determine

To explain: how randomization of the students of two groups led to outliers.

Answer to Problem 2.51E

The randomization of students in the two groups led to the outliers because through randomization the students with little pain are misplaced in the pain-group.

Explanation of Solution

Given info: A small percentage of subjects would experience little bonding regardless whether they were in the pain-group or the no-pain group.

Justification:

Randomization:

A technique of using chance method to assign subjects to the treatments equally is termed as randomization.

Randomization is the process in which all the students are assigned randomly within two groups ‘Pain, No pain’. Due to this randomization students who are having less pain are incorrectly placed in the pain group, these misplaced students are considered as outliers in the pain group.