Does it make sense to use the slope of trend line from a regression as a ratio between x and y












6












$begingroup$


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?










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  • 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
    8 hours ago










  • $begingroup$
    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
    6 hours ago


















6












$begingroup$


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?










share|cite|improve this question







New contributor




user234979 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 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
    8 hours ago










  • $begingroup$
    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
    6 hours ago
















6












6








6





$begingroup$


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?










share|cite|improve this question







New contributor




user234979 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




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






share|cite|improve this question







New contributor




user234979 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




user234979 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






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asked 8 hours ago









user234979user234979

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311




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New contributor





user234979 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






user234979 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 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
    8 hours ago










  • $begingroup$
    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
    6 hours ago
















  • 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
    8 hours ago










  • $begingroup$
    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
    6 hours ago










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
8 hours ago




$begingroup$
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
8 hours ago












$begingroup$
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
6 hours ago






$begingroup$
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
6 hours ago












2 Answers
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active

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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 $xrightarrow infty$.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    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.






    share|cite|improve this answer









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      2 Answers
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      2 Answers
      2






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      active

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      4












      $begingroup$

      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 $xrightarrow infty$.






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        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 $xrightarrow infty$.






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          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 $xrightarrow infty$.






          share|cite|improve this answer









          $endgroup$



          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 $xrightarrow infty$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 8 hours ago









          gunesgunes

          3,8901112




          3,8901112

























              3












              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  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.






                  share|cite|improve this answer









                  $endgroup$



                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 8 hours ago









                  usεr11852usεr11852

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