2
$\begingroup$

I am new at this and could use some help with a conversion.

Survey participants were given the answer choices on a ranking scale of 1-9 for each question. We then averaged the results for each question to understand what the average answer was for each question. For example, for question B, the average answer was 8.1 out of 9.

How would I convert the average to what the number would be on a 1-5 scale? So using the example above, 8.1, what formula would I have to apply to that number so that the average would be a number on the 1-5 scale? i.e. "the average answer was X.X out of 5.

Thanks for your help. The answer may be very simple, but at this point, I think I have over-thought this problem.

$\endgroup$
4
  • 2
    $\begingroup$ Subtract 1 from the value and range, and work with 7.1/8. That is the same as 3.55/4. Add 1 back in and you have 4.55. There is a strong assumption in there about how people would have answered had the scale been different, which you make and are responsible for and which the arithmetic doesn't guarantee. $\endgroup$
    – Nick Cox
    Commented Oct 3, 2014 at 17:28
  • $\begingroup$ That looks like an answer, @Nick :-). $\endgroup$
    – whuber
    Commented Oct 3, 2014 at 17:31
  • $\begingroup$ @whuber Your wish is my command. $\endgroup$
    – Nick Cox
    Commented Oct 3, 2014 at 17:39
  • $\begingroup$ Thanks Nick Cox and whuber. I appreciate the help. You folks rock! $\endgroup$
    – fugie
    Commented Oct 4, 2014 at 2:28

2 Answers 2

4
$\begingroup$

Let's do this by arithmetic:

Subtract $1$ from the value and range, and work with $7.1/8$. That is the same as $3.55/4$.

Add $1$ back in and you have $4.55$.

There is a strong assumption in there about how people would have answered had the scale been different, which you make and are responsible for and which the arithmetic doesn't guarantee.

If you want the algebra, then

$(\text{new} - 1)/4 = (\text{old} - 1)/8$

so shuffling it all around

$\text{new} = 0.5\ (1 + \text{old})$

which checks at $0.5\ (1 + 8.1) = 4.55$.

Geometrically, think about the conversion as given by a line connecting $(1,1)$ and $(9,5)$.

It often helps to check with extreme cases. Clearly the average can't go below $1$ in both cases, or above $5$ or $9$, depending on the upper limit. So, any conversion must respect those extremes.

$\endgroup$
8
  • 1
    $\begingroup$ Would it also work just to set $\frac{8.1}{9} = \frac{x}{5}$ and solve? $\endgroup$
    – 114
    Commented Oct 3, 2014 at 18:16
  • 1
    $\begingroup$ No; you can try it to see. That is $40.5/9 = 4.5$. Geometrically, that is assuming that you are using the line from $(0,0)$ to $(9,5)$, which doesn't have the same slope. $\endgroup$
    – Nick Cox
    Commented Oct 3, 2014 at 18:24
  • 1
    $\begingroup$ +1 The geometric approach is convincing and helpful. Another way to think about it is to realize that going (in a linear manner) from a 1 to 9 scale (which has a range of $9-1=8$) to a 1 to 5 scale (with its range of $5-1=4$) must exactly halve the differences among the response levels: $1/2 = (5-1)/(9-1)$. Thus, if the $1$'s are intended to correspond, then $2$ must correspond to $(2-1)/2 = 1/2$ more than $1$, or $1.5$; $3$ must correspond to $(3-1)/2=1$ more than $1$, or $2$; and so on, making $9$ (which is $8$ more than $1$) correspond to $8/2=4$ more than $1$, or $5$, as intended. $\endgroup$
    – whuber
    Commented Oct 3, 2014 at 18:34
  • 1
    $\begingroup$ Agreed. A simple counter-example to @114's suggestion is that a value of $1$, the minimum possible, would map to $5/9$, which is outside the possible limits. $\endgroup$
    – Nick Cox
    Commented Oct 3, 2014 at 19:10
  • $\begingroup$ This equates the endpoints of both scales ($1_{(9)} \to 1_{(5)}$ and $9_{(9)}\to 5_{(5)}$) - and then interpolating linearly, which is one reasonable mapping. But the meaning of $1_{(5)}$ is not necessarily identical to $1_{(9)}$. One might in some situations argue that $1_{(5)}$ is a category that encompasses both $1_{(9)}$ and part of $2_{(9)}$ (which are themselves categories that might encompass several values on still larger scales) and that the "typical" value of a collection of $1_{(5)}$ scores should map to something between $1_{(9)}$ and $2_{(9)}$. $\endgroup$
    – Glen_b
    Commented May 18, 2016 at 12:06
1
$\begingroup$

The best way is to use a linear transformation. Basically you use 2. First convert the 9 point scale to a 0-1 interval and then convert it to a 5 point interval. This isn't dissimilar to the method used by random number generators, they often start by generating a random number from 0-1 and then transforming it to the desired scale.

So in your case it looks something like this: Let X = score from 9 point scale (I'll use the example of 8.1 below) y = transformed score to 5 point scale

Step 1: Recode to 0 - 1 scale Y_1 = (X-min_old scale)/(max_old scale -min_old scale) Y_1 = (8.1-1)/(9 - 1) = 0.8875

Step 2: transformed to 5 point scale Y = Y_1 (max_new scale - min_new scale) + min_new scale Y = 0.8875(5-1)+1 = 4.55

Couple things to keep in mind about this: 1) I would convert your underlying scores first and then find its average, that way you can get its SD too and do any statistical analysis you want on the recoded scores 2) It's much harder going the other way e.g. 5 to 9 point since you are missing some datum e.g. even after recoding you only have 5 scores not 9. This can make direct comparisons a bit "patchy"

$\endgroup$
1
  • 1
    $\begingroup$ Please avoid texting vulgarities such as "U" for "you". $\endgroup$
    – Nick Cox
    Commented May 18, 2016 at 8:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.