Sign of Cohen's d is unaffected by reversing order of factor levels in R












0















I'm using Cohen's d (implemented using cohen.d() from the effsize package) as a measure of effect size in my dependent variable between two levels of a factor.



My code looks like this: cohen.d(d, f) where d is a vector of numeric values and f is a factor with two levels: "A" and "B".



Based on my understanding, the sign of Cohen's d is dependent on the order of means (i.e. factor levels) entered into the formula. However, my cohen.d() command returns a negative value (and negative CIs), even if I reverse the order of levels in f.



Here is a reproducible example:



library('effsize')
# Load in Chickweight data
a=ChickWeight

# Cohens d requires two levels in factor f, so take the first two available in Diet
a=a[a$Diet==c(1,2),]
a$Diet=a$Diet[ , drop=T]

# Compute cohen's d with default order of Diet
d1 = a$weight
f1 = a$Diet
cohen1 = cohen.d(d1,f1)

# Re-order levels of Diet
a$Diet = relevel(a$Diet, ref=2)

# Re-compute cohen's d
d2 = a$weight
f2 = a$Diet
cohen2 = cohen.d(d2,f2)

# Compare values
cohen1
cohen2


Can anyone explain why this is the case, and/or if I'm doing something wrong?



Thanks in advance for any advice!










share|improve this question



























    0















    I'm using Cohen's d (implemented using cohen.d() from the effsize package) as a measure of effect size in my dependent variable between two levels of a factor.



    My code looks like this: cohen.d(d, f) where d is a vector of numeric values and f is a factor with two levels: "A" and "B".



    Based on my understanding, the sign of Cohen's d is dependent on the order of means (i.e. factor levels) entered into the formula. However, my cohen.d() command returns a negative value (and negative CIs), even if I reverse the order of levels in f.



    Here is a reproducible example:



    library('effsize')
    # Load in Chickweight data
    a=ChickWeight

    # Cohens d requires two levels in factor f, so take the first two available in Diet
    a=a[a$Diet==c(1,2),]
    a$Diet=a$Diet[ , drop=T]

    # Compute cohen's d with default order of Diet
    d1 = a$weight
    f1 = a$Diet
    cohen1 = cohen.d(d1,f1)

    # Re-order levels of Diet
    a$Diet = relevel(a$Diet, ref=2)

    # Re-compute cohen's d
    d2 = a$weight
    f2 = a$Diet
    cohen2 = cohen.d(d2,f2)

    # Compare values
    cohen1
    cohen2


    Can anyone explain why this is the case, and/or if I'm doing something wrong?



    Thanks in advance for any advice!










    share|improve this question

























      0












      0








      0








      I'm using Cohen's d (implemented using cohen.d() from the effsize package) as a measure of effect size in my dependent variable between two levels of a factor.



      My code looks like this: cohen.d(d, f) where d is a vector of numeric values and f is a factor with two levels: "A" and "B".



      Based on my understanding, the sign of Cohen's d is dependent on the order of means (i.e. factor levels) entered into the formula. However, my cohen.d() command returns a negative value (and negative CIs), even if I reverse the order of levels in f.



      Here is a reproducible example:



      library('effsize')
      # Load in Chickweight data
      a=ChickWeight

      # Cohens d requires two levels in factor f, so take the first two available in Diet
      a=a[a$Diet==c(1,2),]
      a$Diet=a$Diet[ , drop=T]

      # Compute cohen's d with default order of Diet
      d1 = a$weight
      f1 = a$Diet
      cohen1 = cohen.d(d1,f1)

      # Re-order levels of Diet
      a$Diet = relevel(a$Diet, ref=2)

      # Re-compute cohen's d
      d2 = a$weight
      f2 = a$Diet
      cohen2 = cohen.d(d2,f2)

      # Compare values
      cohen1
      cohen2


      Can anyone explain why this is the case, and/or if I'm doing something wrong?



      Thanks in advance for any advice!










      share|improve this question














      I'm using Cohen's d (implemented using cohen.d() from the effsize package) as a measure of effect size in my dependent variable between two levels of a factor.



      My code looks like this: cohen.d(d, f) where d is a vector of numeric values and f is a factor with two levels: "A" and "B".



      Based on my understanding, the sign of Cohen's d is dependent on the order of means (i.e. factor levels) entered into the formula. However, my cohen.d() command returns a negative value (and negative CIs), even if I reverse the order of levels in f.



      Here is a reproducible example:



      library('effsize')
      # Load in Chickweight data
      a=ChickWeight

      # Cohens d requires two levels in factor f, so take the first two available in Diet
      a=a[a$Diet==c(1,2),]
      a$Diet=a$Diet[ , drop=T]

      # Compute cohen's d with default order of Diet
      d1 = a$weight
      f1 = a$Diet
      cohen1 = cohen.d(d1,f1)

      # Re-order levels of Diet
      a$Diet = relevel(a$Diet, ref=2)

      # Re-compute cohen's d
      d2 = a$weight
      f2 = a$Diet
      cohen2 = cohen.d(d2,f2)

      # Compare values
      cohen1
      cohen2


      Can anyone explain why this is the case, and/or if I'm doing something wrong?



      Thanks in advance for any advice!







      r






      share|improve this question













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      asked Nov 12 '18 at 14:07









      LyamLyam

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          I'm not entirely sure what the reasoning behind the issue in your example is (maybe someone else can comment here), but if you look at the examples under ?cohen.d, there are a few different methods for calculating it:



          treatment = rnorm(100,mean=10)
          control = rnorm(100,mean=12)
          d = (c(treatment,control))
          f = rep(c("Treatment","Control"),each=100)
          ## compute Cohen's d
          ## treatment and control
          cohen.d(treatment,control)
          ## data and factor
          cohen.d(d,f)
          ## formula interface
          cohen.d(d ~ f)


          If you use the first example of cohen.d(treatment, control) and reverse that to cohen.d(control, treatment) you get the following:



          cohen.d(treatment, control)
          Cohen's d

          d estimate: -1.871982 (large)
          95 percent confidence interval:
          inf sup
          -2.206416 -1.537547

          cohen.d(control, treatment)
          Cohen's d

          d estimate: 1.871982 (large)
          95 percent confidence interval:
          inf sup
          1.537547 2.206416


          So using the two-vector method from the examples with your data, we can do:



          a1 <- a[a$Diet == 1,"weight"]
          a2 <- a[a$Diet == 2,"weight"]
          cohen3a <- cohen.d(a1, a2)
          cohen3b <- cohen.d(a2, a1)


          I noticed that f in the ?cohen.d examples is not a factor, but a character vector. I tried playing around with the cohen.d(d, f) method, but didn't find a solution. Would like to see if someone else has anything regarding that.






          share|improve this answer























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            0














            I'm not entirely sure what the reasoning behind the issue in your example is (maybe someone else can comment here), but if you look at the examples under ?cohen.d, there are a few different methods for calculating it:



            treatment = rnorm(100,mean=10)
            control = rnorm(100,mean=12)
            d = (c(treatment,control))
            f = rep(c("Treatment","Control"),each=100)
            ## compute Cohen's d
            ## treatment and control
            cohen.d(treatment,control)
            ## data and factor
            cohen.d(d,f)
            ## formula interface
            cohen.d(d ~ f)


            If you use the first example of cohen.d(treatment, control) and reverse that to cohen.d(control, treatment) you get the following:



            cohen.d(treatment, control)
            Cohen's d

            d estimate: -1.871982 (large)
            95 percent confidence interval:
            inf sup
            -2.206416 -1.537547

            cohen.d(control, treatment)
            Cohen's d

            d estimate: 1.871982 (large)
            95 percent confidence interval:
            inf sup
            1.537547 2.206416


            So using the two-vector method from the examples with your data, we can do:



            a1 <- a[a$Diet == 1,"weight"]
            a2 <- a[a$Diet == 2,"weight"]
            cohen3a <- cohen.d(a1, a2)
            cohen3b <- cohen.d(a2, a1)


            I noticed that f in the ?cohen.d examples is not a factor, but a character vector. I tried playing around with the cohen.d(d, f) method, but didn't find a solution. Would like to see if someone else has anything regarding that.






            share|improve this answer




























              0














              I'm not entirely sure what the reasoning behind the issue in your example is (maybe someone else can comment here), but if you look at the examples under ?cohen.d, there are a few different methods for calculating it:



              treatment = rnorm(100,mean=10)
              control = rnorm(100,mean=12)
              d = (c(treatment,control))
              f = rep(c("Treatment","Control"),each=100)
              ## compute Cohen's d
              ## treatment and control
              cohen.d(treatment,control)
              ## data and factor
              cohen.d(d,f)
              ## formula interface
              cohen.d(d ~ f)


              If you use the first example of cohen.d(treatment, control) and reverse that to cohen.d(control, treatment) you get the following:



              cohen.d(treatment, control)
              Cohen's d

              d estimate: -1.871982 (large)
              95 percent confidence interval:
              inf sup
              -2.206416 -1.537547

              cohen.d(control, treatment)
              Cohen's d

              d estimate: 1.871982 (large)
              95 percent confidence interval:
              inf sup
              1.537547 2.206416


              So using the two-vector method from the examples with your data, we can do:



              a1 <- a[a$Diet == 1,"weight"]
              a2 <- a[a$Diet == 2,"weight"]
              cohen3a <- cohen.d(a1, a2)
              cohen3b <- cohen.d(a2, a1)


              I noticed that f in the ?cohen.d examples is not a factor, but a character vector. I tried playing around with the cohen.d(d, f) method, but didn't find a solution. Would like to see if someone else has anything regarding that.






              share|improve this answer


























                0












                0








                0







                I'm not entirely sure what the reasoning behind the issue in your example is (maybe someone else can comment here), but if you look at the examples under ?cohen.d, there are a few different methods for calculating it:



                treatment = rnorm(100,mean=10)
                control = rnorm(100,mean=12)
                d = (c(treatment,control))
                f = rep(c("Treatment","Control"),each=100)
                ## compute Cohen's d
                ## treatment and control
                cohen.d(treatment,control)
                ## data and factor
                cohen.d(d,f)
                ## formula interface
                cohen.d(d ~ f)


                If you use the first example of cohen.d(treatment, control) and reverse that to cohen.d(control, treatment) you get the following:



                cohen.d(treatment, control)
                Cohen's d

                d estimate: -1.871982 (large)
                95 percent confidence interval:
                inf sup
                -2.206416 -1.537547

                cohen.d(control, treatment)
                Cohen's d

                d estimate: 1.871982 (large)
                95 percent confidence interval:
                inf sup
                1.537547 2.206416


                So using the two-vector method from the examples with your data, we can do:



                a1 <- a[a$Diet == 1,"weight"]
                a2 <- a[a$Diet == 2,"weight"]
                cohen3a <- cohen.d(a1, a2)
                cohen3b <- cohen.d(a2, a1)


                I noticed that f in the ?cohen.d examples is not a factor, but a character vector. I tried playing around with the cohen.d(d, f) method, but didn't find a solution. Would like to see if someone else has anything regarding that.






                share|improve this answer













                I'm not entirely sure what the reasoning behind the issue in your example is (maybe someone else can comment here), but if you look at the examples under ?cohen.d, there are a few different methods for calculating it:



                treatment = rnorm(100,mean=10)
                control = rnorm(100,mean=12)
                d = (c(treatment,control))
                f = rep(c("Treatment","Control"),each=100)
                ## compute Cohen's d
                ## treatment and control
                cohen.d(treatment,control)
                ## data and factor
                cohen.d(d,f)
                ## formula interface
                cohen.d(d ~ f)


                If you use the first example of cohen.d(treatment, control) and reverse that to cohen.d(control, treatment) you get the following:



                cohen.d(treatment, control)
                Cohen's d

                d estimate: -1.871982 (large)
                95 percent confidence interval:
                inf sup
                -2.206416 -1.537547

                cohen.d(control, treatment)
                Cohen's d

                d estimate: 1.871982 (large)
                95 percent confidence interval:
                inf sup
                1.537547 2.206416


                So using the two-vector method from the examples with your data, we can do:



                a1 <- a[a$Diet == 1,"weight"]
                a2 <- a[a$Diet == 2,"weight"]
                cohen3a <- cohen.d(a1, a2)
                cohen3b <- cohen.d(a2, a1)


                I noticed that f in the ?cohen.d examples is not a factor, but a character vector. I tried playing around with the cohen.d(d, f) method, but didn't find a solution. Would like to see if someone else has anything regarding that.







                share|improve this answer












                share|improve this answer



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                answered Nov 12 '18 at 17:00









                QwfqwfQwfqwf

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