coefficient of variance's significance
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As far as I know, the coefficient of variance (CV) is used for measuring consistency of any variable. But should one always depend on CV for taking decisions, especially when means the are different?
For instance, there are 2 companies: A and B. Company A has a mean profit of $1000 and CV is 0.816%. Company B has a mean profit of $7666.67, but CV is 26.8%.
Which company should one invest in?
variance coefficient-of-variation
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add a comment |
$begingroup$
As far as I know, the coefficient of variance (CV) is used for measuring consistency of any variable. But should one always depend on CV for taking decisions, especially when means the are different?
For instance, there are 2 companies: A and B. Company A has a mean profit of $1000 and CV is 0.816%. Company B has a mean profit of $7666.67, but CV is 26.8%.
Which company should one invest in?
variance coefficient-of-variation
$endgroup$
add a comment |
$begingroup$
As far as I know, the coefficient of variance (CV) is used for measuring consistency of any variable. But should one always depend on CV for taking decisions, especially when means the are different?
For instance, there are 2 companies: A and B. Company A has a mean profit of $1000 and CV is 0.816%. Company B has a mean profit of $7666.67, but CV is 26.8%.
Which company should one invest in?
variance coefficient-of-variation
$endgroup$
As far as I know, the coefficient of variance (CV) is used for measuring consistency of any variable. But should one always depend on CV for taking decisions, especially when means the are different?
For instance, there are 2 companies: A and B. Company A has a mean profit of $1000 and CV is 0.816%. Company B has a mean profit of $7666.67, but CV is 26.8%.
Which company should one invest in?
variance coefficient-of-variation
variance coefficient-of-variation
edited Dec 15 '18 at 16:50
Karolis Koncevičius
1,92721425
1,92721425
asked Dec 15 '18 at 12:40
nafisnafis
111
111
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2 Answers
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votes
$begingroup$
CV is a measure of the spread of a distribution, adjusted for the mean of the variable - it is defined as the standard deviation divided by the mean. So, it is only useful in situations where the means are different - if the means were the same, you could just use the standard deviation.
However, CV becomes useless in some situations - e.g. when some of the values are negative.
As to your specific question, this is far too little information to decide which company to invest in. And, since profit can be negative, the CV may be nonsensical. Suppose, for example, that company C has profit over the last three years of $1,000, $0 and -$1,000 (a loss of $1,000). Then the CV is undefined because the mean is 0. But change the first profit to $1,001 and the CV is now 3001.5 (or 300,150%). Or make the the loss in year 3 one dollar more and the CV is negative.
$endgroup$
add a comment |
$begingroup$
In addition to the very informative answer that Peter provided above, you should also take into serious consideration all of the descriptive statistics derived from your sample. Especially when it comes to optimum investment option selecting, skewness of your data plays crucial role in deciding which one would be the most promising in terms of profitability.
For instance, suppose that you have this sample: $1000,$1500,$1300,$1400,$1350,$1550,$1250,$1100,$10000,$1150,$1280. This sample implies CV=38,15% and mean profit=$2080
Then we have another sample:
$2000,$2160,$1960,$2200,-$4000,$8160,-$10000,$14160,-$15000,$19160,$2080 wich implies CV=141% and mean profit=2080
At first glance,whereas both of the samples have the same mean, we would opt for the first option as the optimum investment since it implies the smallest CV, but if you examine more thoroughly both of the data, you will find out that in the first sample we have an extreme value ($10000) wich significantly affects the distribution of the data (positive skewness),fact that renders them unreliable.
As far as the second sample is concerned, we can distinguish from the graph (and from the descriptive statistics of course) that the data is normally distributed around the mean, fact that makes them more consistent to rely on and take decissions based on them, even though it has larger volatility.
Ιn conclusion, the real challenge of a researcher is whether he/she should exclude or not the extreme values of the sample and how he/she justify such action.
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
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active
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$begingroup$
CV is a measure of the spread of a distribution, adjusted for the mean of the variable - it is defined as the standard deviation divided by the mean. So, it is only useful in situations where the means are different - if the means were the same, you could just use the standard deviation.
However, CV becomes useless in some situations - e.g. when some of the values are negative.
As to your specific question, this is far too little information to decide which company to invest in. And, since profit can be negative, the CV may be nonsensical. Suppose, for example, that company C has profit over the last three years of $1,000, $0 and -$1,000 (a loss of $1,000). Then the CV is undefined because the mean is 0. But change the first profit to $1,001 and the CV is now 3001.5 (or 300,150%). Or make the the loss in year 3 one dollar more and the CV is negative.
$endgroup$
add a comment |
$begingroup$
CV is a measure of the spread of a distribution, adjusted for the mean of the variable - it is defined as the standard deviation divided by the mean. So, it is only useful in situations where the means are different - if the means were the same, you could just use the standard deviation.
However, CV becomes useless in some situations - e.g. when some of the values are negative.
As to your specific question, this is far too little information to decide which company to invest in. And, since profit can be negative, the CV may be nonsensical. Suppose, for example, that company C has profit over the last three years of $1,000, $0 and -$1,000 (a loss of $1,000). Then the CV is undefined because the mean is 0. But change the first profit to $1,001 and the CV is now 3001.5 (or 300,150%). Or make the the loss in year 3 one dollar more and the CV is negative.
$endgroup$
add a comment |
$begingroup$
CV is a measure of the spread of a distribution, adjusted for the mean of the variable - it is defined as the standard deviation divided by the mean. So, it is only useful in situations where the means are different - if the means were the same, you could just use the standard deviation.
However, CV becomes useless in some situations - e.g. when some of the values are negative.
As to your specific question, this is far too little information to decide which company to invest in. And, since profit can be negative, the CV may be nonsensical. Suppose, for example, that company C has profit over the last three years of $1,000, $0 and -$1,000 (a loss of $1,000). Then the CV is undefined because the mean is 0. But change the first profit to $1,001 and the CV is now 3001.5 (or 300,150%). Or make the the loss in year 3 one dollar more and the CV is negative.
$endgroup$
CV is a measure of the spread of a distribution, adjusted for the mean of the variable - it is defined as the standard deviation divided by the mean. So, it is only useful in situations where the means are different - if the means were the same, you could just use the standard deviation.
However, CV becomes useless in some situations - e.g. when some of the values are negative.
As to your specific question, this is far too little information to decide which company to invest in. And, since profit can be negative, the CV may be nonsensical. Suppose, for example, that company C has profit over the last three years of $1,000, $0 and -$1,000 (a loss of $1,000). Then the CV is undefined because the mean is 0. But change the first profit to $1,001 and the CV is now 3001.5 (or 300,150%). Or make the the loss in year 3 one dollar more and the CV is negative.
answered Dec 15 '18 at 14:07
Peter Flom♦Peter Flom
74.7k11107204
74.7k11107204
add a comment |
add a comment |
$begingroup$
In addition to the very informative answer that Peter provided above, you should also take into serious consideration all of the descriptive statistics derived from your sample. Especially when it comes to optimum investment option selecting, skewness of your data plays crucial role in deciding which one would be the most promising in terms of profitability.
For instance, suppose that you have this sample: $1000,$1500,$1300,$1400,$1350,$1550,$1250,$1100,$10000,$1150,$1280. This sample implies CV=38,15% and mean profit=$2080
Then we have another sample:
$2000,$2160,$1960,$2200,-$4000,$8160,-$10000,$14160,-$15000,$19160,$2080 wich implies CV=141% and mean profit=2080
At first glance,whereas both of the samples have the same mean, we would opt for the first option as the optimum investment since it implies the smallest CV, but if you examine more thoroughly both of the data, you will find out that in the first sample we have an extreme value ($10000) wich significantly affects the distribution of the data (positive skewness),fact that renders them unreliable.
As far as the second sample is concerned, we can distinguish from the graph (and from the descriptive statistics of course) that the data is normally distributed around the mean, fact that makes them more consistent to rely on and take decissions based on them, even though it has larger volatility.
Ιn conclusion, the real challenge of a researcher is whether he/she should exclude or not the extreme values of the sample and how he/she justify such action.
$endgroup$
add a comment |
$begingroup$
In addition to the very informative answer that Peter provided above, you should also take into serious consideration all of the descriptive statistics derived from your sample. Especially when it comes to optimum investment option selecting, skewness of your data plays crucial role in deciding which one would be the most promising in terms of profitability.
For instance, suppose that you have this sample: $1000,$1500,$1300,$1400,$1350,$1550,$1250,$1100,$10000,$1150,$1280. This sample implies CV=38,15% and mean profit=$2080
Then we have another sample:
$2000,$2160,$1960,$2200,-$4000,$8160,-$10000,$14160,-$15000,$19160,$2080 wich implies CV=141% and mean profit=2080
At first glance,whereas both of the samples have the same mean, we would opt for the first option as the optimum investment since it implies the smallest CV, but if you examine more thoroughly both of the data, you will find out that in the first sample we have an extreme value ($10000) wich significantly affects the distribution of the data (positive skewness),fact that renders them unreliable.
As far as the second sample is concerned, we can distinguish from the graph (and from the descriptive statistics of course) that the data is normally distributed around the mean, fact that makes them more consistent to rely on and take decissions based on them, even though it has larger volatility.
Ιn conclusion, the real challenge of a researcher is whether he/she should exclude or not the extreme values of the sample and how he/she justify such action.
$endgroup$
add a comment |
$begingroup$
In addition to the very informative answer that Peter provided above, you should also take into serious consideration all of the descriptive statistics derived from your sample. Especially when it comes to optimum investment option selecting, skewness of your data plays crucial role in deciding which one would be the most promising in terms of profitability.
For instance, suppose that you have this sample: $1000,$1500,$1300,$1400,$1350,$1550,$1250,$1100,$10000,$1150,$1280. This sample implies CV=38,15% and mean profit=$2080
Then we have another sample:
$2000,$2160,$1960,$2200,-$4000,$8160,-$10000,$14160,-$15000,$19160,$2080 wich implies CV=141% and mean profit=2080
At first glance,whereas both of the samples have the same mean, we would opt for the first option as the optimum investment since it implies the smallest CV, but if you examine more thoroughly both of the data, you will find out that in the first sample we have an extreme value ($10000) wich significantly affects the distribution of the data (positive skewness),fact that renders them unreliable.
As far as the second sample is concerned, we can distinguish from the graph (and from the descriptive statistics of course) that the data is normally distributed around the mean, fact that makes them more consistent to rely on and take decissions based on them, even though it has larger volatility.
Ιn conclusion, the real challenge of a researcher is whether he/she should exclude or not the extreme values of the sample and how he/she justify such action.
$endgroup$
In addition to the very informative answer that Peter provided above, you should also take into serious consideration all of the descriptive statistics derived from your sample. Especially when it comes to optimum investment option selecting, skewness of your data plays crucial role in deciding which one would be the most promising in terms of profitability.
For instance, suppose that you have this sample: $1000,$1500,$1300,$1400,$1350,$1550,$1250,$1100,$10000,$1150,$1280. This sample implies CV=38,15% and mean profit=$2080
Then we have another sample:
$2000,$2160,$1960,$2200,-$4000,$8160,-$10000,$14160,-$15000,$19160,$2080 wich implies CV=141% and mean profit=2080
At first glance,whereas both of the samples have the same mean, we would opt for the first option as the optimum investment since it implies the smallest CV, but if you examine more thoroughly both of the data, you will find out that in the first sample we have an extreme value ($10000) wich significantly affects the distribution of the data (positive skewness),fact that renders them unreliable.
As far as the second sample is concerned, we can distinguish from the graph (and from the descriptive statistics of course) that the data is normally distributed around the mean, fact that makes them more consistent to rely on and take decissions based on them, even though it has larger volatility.
Ιn conclusion, the real challenge of a researcher is whether he/she should exclude or not the extreme values of the sample and how he/she justify such action.
answered Dec 15 '18 at 23:02
LogicseekerLogicseeker
184
184
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