Does it make sense to use the slope of trend line from a regression as a ratio between x and y
$begingroup$
The regression plots Hours (y) vs jobs (x). Let's say the equation is: y = 0.4x + 90
Is it okay to say that the time for each job is 0.4 hours?
regression
asked Jan 22 at 20:57
user234979user234979
311
$endgroup$
add a comment |
$begingroup$
The regression plots Hours (y) vs jobs (x). Let's say the equation is: y = 0.4x + 90
Is it okay to say that the time for each job is 0.4 hours?
regression
asked Jan 22 at 20:57
user234979user234979
311
$endgroup$
3
$begingroup$
The incremental time for each job is 0.4 hours. For the actual time you have to take into account the 90 hour constant
$endgroup$
– Jake Westfall
Jan 22 at 21:12
$begingroup$
The estimate of the expected marginal time for each job would be 0.4 hours (i.e. adding a job will on average add 0.4 hours). It may be that no job will take 0.4 hours, even on average, because that intercept term is large - it depends on how that 90 is allocated to jobs
$endgroup$
– Glen_b♦
Jan 22 at 23:15
add a comment |
$begingroup$
The regression plots Hours (y) vs jobs (x). Let's say the equation is: y = 0.4x + 90
Is it okay to say that the time for each job is 0.4 hours?
regression
asked Jan 22 at 20:57
user234979user234979
311
$endgroup$
The regression plots Hours (y) vs jobs (x). Let's say the equation is: y = 0.4x + 90
Is it okay to say that the time for each job is 0.4 hours?
regression
regression
asked Jan 22 at 20:57
user234979user234979
311
asked Jan 22 at 20:57
user234979user234979
311
asked Jan 22 at 20:57
user234979user234979
311
asked Jan 22 at 20:57
user234979user234979
311
asked Jan 22 at 20:57
user234979user234979
311
311
3
$begingroup$
The incremental time for each job is 0.4 hours. For the actual time you have to take into account the 90 hour constant
$endgroup$
– Jake Westfall
Jan 22 at 21:12
$begingroup$
The estimate of the expected marginal time for each job would be 0.4 hours (i.e. adding a job will on average add 0.4 hours). It may be that no job will take 0.4 hours, even on average, because that intercept term is large - it depends on how that 90 is allocated to jobs
$endgroup$
– Glen_b♦
Jan 22 at 23:15
add a comment |
3
$begingroup$
The incremental time for each job is 0.4 hours. For the actual time you have to take into account the 90 hour constant
$endgroup$
– Jake Westfall
Jan 22 at 21:12
$begingroup$
The estimate of the expected marginal time for each job would be 0.4 hours (i.e. adding a job will on average add 0.4 hours). It may be that no job will take 0.4 hours, even on average, because that intercept term is large - it depends on how that 90 is allocated to jobs
$endgroup$
– Glen_b♦
Jan 22 at 23:15
3
3
$begingroup$
The incremental time for each job is 0.4 hours. For the actual time you have to take into account the 90 hour constant
$endgroup$
– Jake Westfall
Jan 22 at 21:12
$begingroup$
The incremental time for each job is 0.4 hours. For the actual time you have to take into account the 90 hour constant
$endgroup$
– Jake Westfall
Jan 22 at 21:12
$begingroup$
The estimate of the expected marginal time for each job would be 0.4 hours (i.e. adding a job will on average add 0.4 hours). It may be that no job will take 0.4 hours, even on average, because that intercept term is large - it depends on how that 90 is allocated to jobs
$endgroup$
– Glen_b♦
Jan 22 at 23:15
$begingroup$
The estimate of the expected marginal time for each job would be 0.4 hours (i.e. adding a job will on average add 0.4 hours). It may be that no job will take 0.4 hours, even on average, because that intercept term is large - it depends on how that 90 is allocated to jobs
$endgroup$
– Glen_b♦
Jan 22 at 23:15
add a comment |
2 Answers
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$begingroup$
I presume $x$ is number of jobs and $y$ is number of hours to complete it. It's not correct to say that time for each job is 0.4 hours because you have a (pretty large) bias term. This means you have a fix cost. Performing one job takes $90.4$ hours, two jobs $90.8$ hours etc. So, you can say that each additional job takes 0.4 hours. And, time for each job asymptotically approaches $0.4$ hours, i.e. as $xrightarrow infty$.
answered Jan 22 at 21:08
gunesgunes
4,1961112
$endgroup$
add a comment |
$begingroup$
Not exactly. Assuming a simple linear model $y = beta_0 + beta_1 x + epsilon$, the parameter $beta_1$ (in this case $0.4$) represents the effect of a one-unit change in the corresponding $x$ (here jobs) covariate on the mean value of the dependent variable, $y$ (here hours), assuming that any other covariates remain constant at some value.
As the $beta_0$ (the intercept) is $90$, probably there are some other factors that require on average $90$ hours to be taken into account.
It is more relevant to say that each additional job requires an additional $0.4$ hours.
answered Jan 22 at 21:24
usεr11852usεr11852
18.3k14173
$endgroup$
add a comment |
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2 Answers
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$begingroup$
I presume $x$ is number of jobs and $y$ is number of hours to complete it. It's not correct to say that time for each job is 0.4 hours because you have a (pretty large) bias term. This means you have a fix cost. Performing one job takes $90.4$ hours, two jobs $90.8$ hours etc. So, you can say that each additional job takes 0.4 hours. And, time for each job asymptotically approaches $0.4$ hours, i.e. as $xrightarrow infty$.
answered Jan 22 at 21:08
gunesgunes
4,1961112
$endgroup$
add a comment |
$begingroup$
I presume $x$ is number of jobs and $y$ is number of hours to complete it. It's not correct to say that time for each job is 0.4 hours because you have a (pretty large) bias term. This means you have a fix cost. Performing one job takes $90.4$ hours, two jobs $90.8$ hours etc. So, you can say that each additional job takes 0.4 hours. And, time for each job asymptotically approaches $0.4$ hours, i.e. as $xrightarrow infty$.
answered Jan 22 at 21:08
gunesgunes
4,1961112
$endgroup$
add a comment |
$begingroup$
I presume $x$ is number of jobs and $y$ is number of hours to complete it. It's not correct to say that time for each job is 0.4 hours because you have a (pretty large) bias term. This means you have a fix cost. Performing one job takes $90.4$ hours, two jobs $90.8$ hours etc. So, you can say that each additional job takes 0.4 hours. And, time for each job asymptotically approaches $0.4$ hours, i.e. as $xrightarrow infty$.
answered Jan 22 at 21:08
gunesgunes
4,1961112
$endgroup$
I presume $x$ is number of jobs and $y$ is number of hours to complete it. It's not correct to say that time for each job is 0.4 hours because you have a (pretty large) bias term. This means you have a fix cost. Performing one job takes $90.4$ hours, two jobs $90.8$ hours etc. So, you can say that each additional job takes 0.4 hours. And, time for each job asymptotically approaches $0.4$ hours, i.e. as $xrightarrow infty$.
answered Jan 22 at 21:08
gunesgunes
4,1961112
answered Jan 22 at 21:08
gunesgunes
4,1961112
answered Jan 22 at 21:08
gunesgunes
4,1961112
answered Jan 22 at 21:08
gunesgunes
4,1961112
4,1961112
add a comment |
add a comment |
$begingroup$
Not exactly. Assuming a simple linear model $y = beta_0 + beta_1 x + epsilon$, the parameter $beta_1$ (in this case $0.4$) represents the effect of a one-unit change in the corresponding $x$ (here jobs) covariate on the mean value of the dependent variable, $y$ (here hours), assuming that any other covariates remain constant at some value.
As the $beta_0$ (the intercept) is $90$, probably there are some other factors that require on average $90$ hours to be taken into account.
It is more relevant to say that each additional job requires an additional $0.4$ hours.
answered Jan 22 at 21:24
usεr11852usεr11852
18.3k14173
$endgroup$
add a comment |
$begingroup$
Not exactly. Assuming a simple linear model $y = beta_0 + beta_1 x + epsilon$, the parameter $beta_1$ (in this case $0.4$) represents the effect of a one-unit change in the corresponding $x$ (here jobs) covariate on the mean value of the dependent variable, $y$ (here hours), assuming that any other covariates remain constant at some value.
As the $beta_0$ (the intercept) is $90$, probably there are some other factors that require on average $90$ hours to be taken into account.
It is more relevant to say that each additional job requires an additional $0.4$ hours.
answered Jan 22 at 21:24
usεr11852usεr11852
18.3k14173
$endgroup$
add a comment |
$begingroup$
Not exactly. Assuming a simple linear model $y = beta_0 + beta_1 x + epsilon$, the parameter $beta_1$ (in this case $0.4$) represents the effect of a one-unit change in the corresponding $x$ (here jobs) covariate on the mean value of the dependent variable, $y$ (here hours), assuming that any other covariates remain constant at some value.
As the $beta_0$ (the intercept) is $90$, probably there are some other factors that require on average $90$ hours to be taken into account.
It is more relevant to say that each additional job requires an additional $0.4$ hours.
answered Jan 22 at 21:24
usεr11852usεr11852
18.3k14173
$endgroup$
Not exactly. Assuming a simple linear model $y = beta_0 + beta_1 x + epsilon$, the parameter $beta_1$ (in this case $0.4$) represents the effect of a one-unit change in the corresponding $x$ (here jobs) covariate on the mean value of the dependent variable, $y$ (here hours), assuming that any other covariates remain constant at some value.
As the $beta_0$ (the intercept) is $90$, probably there are some other factors that require on average $90$ hours to be taken into account.
It is more relevant to say that each additional job requires an additional $0.4$ hours.
answered Jan 22 at 21:24
usεr11852usεr11852
18.3k14173
answered Jan 22 at 21:24
usεr11852usεr11852
18.3k14173
answered Jan 22 at 21:24
usεr11852usεr11852
18.3k14173
answered Jan 22 at 21:24
usεr11852usεr11852
18.3k14173
18.3k14173
add a comment |
add a comment |
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$begingroup$
The incremental time for each job is 0.4 hours. For the actual time you have to take into account the 90 hour constant
$endgroup$
– Jake Westfall
Jan 22 at 21:12
$begingroup$
The estimate of the expected marginal time for each job would be 0.4 hours (i.e. adding a job will on average add 0.4 hours). It may be that no job will take 0.4 hours, even on average, because that intercept term is large - it depends on how that 90 is allocated to jobs
$endgroup$
– Glen_b♦
Jan 22 at 23:15