How to convert percentage values from 7 point scale to 5 point scale? Suppose in one year I had a survey with 7-point scale and the values are like
Scale7, 1: 28%
Scale7, 2: 12%
Scale7, 3: 10%
Scale7, 4: 15%
Scale7, 5: 15%
Scale7, 6: 10%
Scale7, 7: 10%

where % are the percentage of responses for each scale from the sample size.
Now for next year if I have survey on the same question but with 5-point scale like
Scale5, 1: 25%
Scale5, 2: 25%
Scale5, 3: 10%
Scale5, 4: 20%
Scale5, 5: 20%

where % are the percentage of responses for each scale from the sample size.
Then if I have to convert the 7 point scale responses to 5 point scale how do I do that?
 A: As already signalled, you can't do much -- without extra information or assumptions.
Here, without any strong claims, is a method from Mosteller, F. and Tukey, J.W. 1977. Data Analysis and Regression. Reading, MA: Addison-Wesley.
The idea is that your data on frequencies of grades define cumulative probabilities -- that is the only easy point -- and you can relate those cumulative probabilities to some reference distribution and then average over each interval to get a mean score as a numerical equivalent for each grade.
If we choose a logistic distribution as reference, then the calculation is easy, as equations feature only simple expressions in exponentials or logarithms.  So for example grade 2 in year 1 covers the interval from cumulative probability 0.28 to 0.40.
Here are some token results:
  +------------------------------------------+
  | grade   percent   year   score1   score2 |
  |------------------------------------------|
  |     1        28      1   -2.118        . |
  |     2        12      1   -0.667        . |
  |     3        10      1   -0.201        . |
  |     4        15      1    0.305        . |
  |     5        15      1    0.980        . |
  |     6        10      1    1.753        . |
  |     7        10      1    3.251        . |
  |------------------------------------------|
  |     1        25      2        .   -2.249 |
  |     2        25      2        .   -0.523 |
  |     3        10      2        .    0.201 |
  |     4        20      2        .    0.863 |
  |     5        20      2        .    2.502 |
  +------------------------------------------+

The reservations are manifold and start with

*

*There is no handle here on whether the underlying distributions are essentially the same or different, for which surveying people in both years may help to provide information.


*Why logistic? Why not normal? Or any other distribution? The method would be a poor choice if attitudes were often polarised and followed a U-shaped distribution -- or if they followed an asymmetric distribution. But that should be evident from an exploratory analysis.


*Pulling numeric scores out of a reference distribution like rabbits out of a hat does not make them more suitable as responses for (say) ordered logit modelling. But if these score variables were predictors, then results using them could be compared with the results of treating the scores as factor variables (i.e. as a set of indicator variables).
So, the idea is to let the pattern of frequencies indicate the scores: for example, if an extreme grade is rare, that implies a rather high or low numeric score.
If in doubt, try different procedures.
More mathematical detail if needed: We are relating cumulative probabilities to a logistic distribution with location parameter $0$ and scale parameter $1$. So for variable $x$ its density is
$$ \exp(-x)\ /\ [ 1 + \exp(-x)]^2$$ and its quantile function  for cumulative probability $p$ is $\text{logit} 
\ p$ or
$\ln\ [p/(1-p)]$. We want the mean between cumulative probability $A$ and $B$ or
$${1 \over B - A} \int_A^B \ln\ [p/(1 - p)]\ dp $$
which falls out easily (Mosteller and Tukey, 1977, p.245) as
$${1 \over B - A} \{ [B \ln B + (1 - B) \ln\ (1-B)]\ - \ [A \ln A + (1 - A) \ln\ (1 -A)]\}$$
which is quite programmable, a detail being to ensure that $0 \ln 0$ is returned as $0$, not some missing or undefined value. The calculus details are not spelled out by Mosteller and Tukey, but can be expanded upon.
A: To "translate" the scales, you would need to be able to make an assumption that the distributions of the underlying phenomenon did not change across time, the only difference was the measurement units that were used. Notice that this is a pretty strong assumption. If you can make it, you could use equating.
There are simpler approaches, but with tabular data like this, you could easily use equipercentile equating. If there are two distributions $F_X$ and $F_Y$, you could "translate" $X$'s by plugging the $F_X$ cumulative probabilities to the inverse of $Y$'s distribution $F_Y^{-1}$:
$$ \operatorname{Equi}_Y(x_i) = F^{-1}_Y \left[ F_X(x_i) \right] $$
or the same using a graphic

So what you need to do is to calculate the cumulative probabilities and align the values by matching their cumulative probabilities.




$y$
$Pr(Y\le y)$
$x$
$Pr(X\le x)$




1
28%
1
25%


2
40%




3
50%
2
50%




3
60%


4
65%




5
80%
4
80%


6
90%




7
100%
5
100%



A: Subject behavior between Likert-7 and Likert-5 is clearly different, but poorly understood.
Suppose the following are true.

*

*Unless bored or uninterested, subjects tend to avoid the middle value as a wasted vote.


*Subjects may regard 1 on a 7-point scale as more of a rebuke than 1 on a 5-point scale.


*Unless very highly pleased and motivated, subjects are unwilling to vote 7 on a 7-point scale.
Then, the only fair conversion is to make both Likert-7 and Likert-5 into No/Yes scales: 1,2,3 or 1,2 become No; 5,6,7 or 4,5 become Yes. (Ignore middle values.)
This would mean subjects in the first year gave $40/85 = 47\%$ Yes votes
and subjects in the second year gave $4/9 = 44\%$ Yes votes --- perhaps an unimportant difference, even if there were
enough subjects for it to be 'significant'.
For a one-sided test, a thousand subjects might not suffice for significance:
prop.test(c(.44,.47)*1000, c(1000,1000), alt="less", cor=F)

        2-sample test for equality of proportions 
        without continuity correction

data:  c(0.44, 0.47) * 1000 out of c(1000, 1000)
X-squared = 1.8147, df = 1, p-value = 0.08897
alternative hypothesis: less
95 percent confidence interval:
 -1.000000000  0.006614161
sample estimates:
 prop 1 prop 2 
   0.44   0.47 

