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.

Question

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Answer 1

Question

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.

a.

Expert Solution

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,

$x¯=∑xin$

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

$x¯=(4.71+4.86+4.14+1.29+2.29+4.43+...+4.71+4.71+2.14+3.57)27=100.2927=3.7144$

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,

$Location of M=n+12=282=14$

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:

$x¯=(4.71+4.86+4.14+2.29+4.43+...+4.71+4.71+2.14+3.57)25=97.5725=3.9028$

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

Median without smallest scores of pain group:

$Location of M=n+12=262=13$

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 $1.5×IQR$ rule identify the low bonding scores as suspected outliers or not.

b.

Expert Solution

Answer to Problem 2.51E

The $1.5×IQR$ 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 $Q1$ :

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

$location of Q1=n+12=142=7$

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

The first quartile $Q1$ is, 3.43.

Third quartile $Q3$ :

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

$location of Q3=n+12=142=7$

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

The third quartile $Q3$ is, 4.43.

Interquartile range (IQR):

$IQR=Q3−Q1=4.43−3.43=1$

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

The $1.5×IQR$ rule for outliers:

An observation is suspected as an outlier if it is more than $Q3+(1.5×IQR)$ and less than $Q1−(1.5×IQR)$ .

$Q3+(1.5×IQR)=4.43+(1.5×1)=4.43+1.5=5.93$

$Q1−(1.5×IQR)=3.43−(1.5×1)=3.43−1.5=1.93$

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 $1.5×IQR$ 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.

c.

Expert Solution

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.

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Answer 2

Question

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.

a.

Expert Solution

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,

$x¯=∑xin$

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

$x¯=(4.71+4.86+4.14+1.29+2.29+4.43+...+4.71+4.71+2.14+3.57)27=100.2927=3.7144$

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,

$Location of M=n+12=282=14$

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:

$x¯=(4.71+4.86+4.14+2.29+4.43+...+4.71+4.71+2.14+3.57)25=97.5725=3.9028$

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

Median without smallest scores of pain group:

$Location of M=n+12=262=13$

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 $1.5×IQR$ rule identify the low bonding scores as suspected outliers or not.

b.

Expert Solution

Answer to Problem 2.51E

The $1.5×IQR$ 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 $Q1$ :

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

$location of Q1=n+12=142=7$

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

The first quartile $Q1$ is, 3.43.

Third quartile $Q3$ :

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

$location of Q3=n+12=142=7$

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

The third quartile $Q3$ is, 4.43.

Interquartile range (IQR):

$IQR=Q3−Q1=4.43−3.43=1$

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

The $1.5×IQR$ rule for outliers:

An observation is suspected as an outlier if it is more than $Q3+(1.5×IQR)$ and less than $Q1−(1.5×IQR)$ .

$Q3+(1.5×IQR)=4.43+(1.5×1)=4.43+1.5=5.93$

$Q1−(1.5×IQR)=3.43−(1.5×1)=3.43−1.5=1.93$

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 $1.5×IQR$ 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.

c.

Expert Solution

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.

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Answer 3

Question

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.

a.

Expert Solution

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,

$x¯=∑xin$

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

$x¯=(4.71+4.86+4.14+1.29+2.29+4.43+...+4.71+4.71+2.14+3.57)27=100.2927=3.7144$

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,

$Location of M=n+12=282=14$

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:

$x¯=(4.71+4.86+4.14+2.29+4.43+...+4.71+4.71+2.14+3.57)25=97.5725=3.9028$

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

Median without smallest scores of pain group:

$Location of M=n+12=262=13$

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 $1.5×IQR$ rule identify the low bonding scores as suspected outliers or not.

b.

Expert Solution

Answer to Problem 2.51E

The $1.5×IQR$ 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 $Q1$ :

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

$location of Q1=n+12=142=7$

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

The first quartile $Q1$ is, 3.43.

Third quartile $Q3$ :

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

$location of Q3=n+12=142=7$

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

The third quartile $Q3$ is, 4.43.

Interquartile range (IQR):

$IQR=Q3−Q1=4.43−3.43=1$

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

The $1.5×IQR$ rule for outliers:

An observation is suspected as an outlier if it is more than $Q3+(1.5×IQR)$ and less than $Q1−(1.5×IQR)$ .

$Q3+(1.5×IQR)=4.43+(1.5×1)=4.43+1.5=5.93$

$Q1−(1.5×IQR)=3.43−(1.5×1)=3.43−1.5=1.93$

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 $1.5×IQR$ 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.

c.

Expert Solution

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.

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Answer 4

Question

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.

a.

Expert Solution

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,

$x¯=∑xin$

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

$x¯=(4.71+4.86+4.14+1.29+2.29+4.43+...+4.71+4.71+2.14+3.57)27=100.2927=3.7144$

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,

$Location of M=n+12=282=14$

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:

$x¯=(4.71+4.86+4.14+2.29+4.43+...+4.71+4.71+2.14+3.57)25=97.5725=3.9028$

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

Median without smallest scores of pain group:

$Location of M=n+12=262=13$

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 $1.5×IQR$ rule identify the low bonding scores as suspected outliers or not.

b.

Expert Solution

Answer to Problem 2.51E

The $1.5×IQR$ 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 $Q1$ :

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

$location of Q1=n+12=142=7$

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

The first quartile $Q1$ is, 3.43.

Third quartile $Q3$ :

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

$location of Q3=n+12=142=7$

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

The third quartile $Q3$ is, 4.43.

Interquartile range (IQR):

$IQR=Q3−Q1=4.43−3.43=1$

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

The $1.5×IQR$ rule for outliers:

An observation is suspected as an outlier if it is more than $Q3+(1.5×IQR)$ and less than $Q1−(1.5×IQR)$ .

$Q3+(1.5×IQR)=4.43+(1.5×1)=4.43+1.5=5.93$

$Q1−(1.5×IQR)=3.43−(1.5×1)=3.43−1.5=1.93$

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 $1.5×IQR$ 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.

c.

Expert Solution

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.

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Answer 5

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.

a.

Expert Solution

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,

$x¯=∑xin$

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

$x¯=(4.71+4.86+4.14+1.29+2.29+4.43+...+4.71+4.71+2.14+3.57)27=100.2927=3.7144$

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,

$Location of M=n+12=282=14$

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:

$x¯=(4.71+4.86+4.14+2.29+4.43+...+4.71+4.71+2.14+3.57)25=97.5725=3.9028$

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

Median without smallest scores of pain group:

$Location of M=n+12=262=13$

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 $1.5×IQR$ rule identify the low bonding scores as suspected outliers or not.

b.

Expert Solution

Answer to Problem 2.51E

The $1.5×IQR$ 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 $Q1$ :

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

$location of Q1=n+12=142=7$

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

The first quartile $Q1$ is, 3.43.

Third quartile $Q3$ :

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

$location of Q3=n+12=142=7$

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

The third quartile $Q3$ is, 4.43.

Interquartile range (IQR):

$IQR=Q3−Q1=4.43−3.43=1$

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

The $1.5×IQR$ rule for outliers:

An observation is suspected as an outlier if it is more than $Q3+(1.5×IQR)$ and less than $Q1−(1.5×IQR)$ .

$Q3+(1.5×IQR)=4.43+(1.5×1)=4.43+1.5=5.93$

$Q1−(1.5×IQR)=3.43−(1.5×1)=3.43−1.5=1.93$

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 $1.5×IQR$ 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.

c.

Expert Solution

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.

Answer 6

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.

a.

Expert Solution

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,

$x¯=∑xin$

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

$x¯=(4.71+4.86+4.14+1.29+2.29+4.43+...+4.71+4.71+2.14+3.57)27=100.2927=3.7144$

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,

$Location of M=n+12=282=14$

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:

$x¯=(4.71+4.86+4.14+2.29+4.43+...+4.71+4.71+2.14+3.57)25=97.5725=3.9028$

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

Median without smallest scores of pain group:

$Location of M=n+12=262=13$

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.

Answer 7

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 8

a.

Expert Solution

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.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

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,

$x¯=∑xin$

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

$x¯=(4.71+4.86+4.14+1.29+2.29+4.43+...+4.71+4.71+2.14+3.57)27=100.2927=3.7144$

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,

$Location of M=n+12=282=14$

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:

$x¯=(4.71+4.86+4.14+2.29+4.43+...+4.71+4.71+2.14+3.57)25=97.5725=3.9028$

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

Median without smallest scores of pain group:

$Location of M=n+12=262=13$

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.

Answer 13

b.

To determine

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

b.

Expert Solution

Answer to Problem 2.51E

The $1.5×IQR$ 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 $Q1$ :

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

$location of Q1=n+12=142=7$

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

The first quartile $Q1$ is, 3.43.

Third quartile $Q3$ :

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

$location of Q3=n+12=142=7$

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

The third quartile $Q3$ is, 4.43.

Interquartile range (IQR):

$IQR=Q3−Q1=4.43−3.43=1$

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

The $1.5×IQR$ rule for outliers:

An observation is suspected as an outlier if it is more than $Q3+(1.5×IQR)$ and less than $Q1−(1.5×IQR)$ .

$Q3+(1.5×IQR)=4.43+(1.5×1)=4.43+1.5=5.93$

$Q1−(1.5×IQR)=3.43−(1.5×1)=3.43−1.5=1.93$

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

Answer 14

b.

To determine

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

Answer 15

b.

Expert Solution

Answer to Problem 2.51E

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

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

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

Calculation:

First quartile $Q1$ :

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

$location of Q1=n+12=142=7$

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

The first quartile $Q1$ is, 3.43.

Third quartile $Q3$ :

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

$location of Q3=n+12=142=7$

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

The third quartile $Q3$ is, 4.43.

Interquartile range (IQR):

$IQR=Q3−Q1=4.43−3.43=1$

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

The $1.5×IQR$ rule for outliers:

An observation is suspected as an outlier if it is more than $Q3+(1.5×IQR)$ and less than $Q1−(1.5×IQR)$ .

$Q3+(1.5×IQR)=4.43+(1.5×1)=4.43+1.5=5.93$

$Q1−(1.5×IQR)=3.43−(1.5×1)=3.43−1.5=1.93$

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

Answer 20

c.

To determine

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

c.

Expert Solution

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.

Answer 21

c.

To determine

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

Answer 22

c.

Expert Solution

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.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

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.

Answer 27

Ch. 2.1 - Prob. 1AYK Ch. 2.1 - Prob. 2AYK Ch. 2.3 - Prob. 3AYK Ch. 2.3 - Prob. 4AYK Ch. 2.5 - Prob. 6AYK Ch. 2.5 - Prob. 7AYK Ch. 2.6 - Prob. 8AYK Ch. 2.6 - Prob. 9AYK Ch. 2.8 - Prob. 10AYK Ch. 2.8 - Prob. 11AYK

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.

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

Explanation of Solution

To find: The 1.5×IQR<m:math><m:mrow><m:mn>1.5</m:mn><m:mo>×</m:mo><m:mi>I</m:mi><m:mi>Q</m:mi><m:mi>R</m:mi></m:mrow> </m:math> rule identify the low bonding scores as suspected outliers or not.

Answer to Problem 2.51E

Explanation of Solution

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

Answer to Problem 2.51E

Explanation of Solution

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Chapter 2 Solutions

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.

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