# Comparing percentages based on likert scale by year

I'm reviewing an analysis someone else has done on some Likert scale data. They've assigned each point on the scale 1-5 (1 = bad, 2 = poor etc.), found the average score in each area, and then converted to a percentage (by multiplying by 20) to give a percentage of total score (100% being the best, 20% being the worst).

I'm okay with this, but then they're computed a significance test as if the percentages were actual percentages, like if they'd gone out and asked people "Do you own your own home? Yes/ no". They've used a method similar to the one described here:

https://www.dummies.com/education/math/statistics/how-to-compare-two-population-proportions/

I want to tell them that this is a completely invalid way of analysing the data, and they've ignored the variance in the scores by collapsing everything into a percentage. I feel they should use ordinary t-tests on the data to determine significant difference. But I'm doubting myself. Any thoughts appreciated.

This should help you understand better. I have chosen Paired test =False.

> #create random numbers between 1-5
> x = round(runif(10, 1, 5), 0)
> x
 3 3 2 4 1 1 3 4 4 4
> y = round(runif(10, 1, 5), 0)
> y
 2 2 3 4 5 2 1 3 5 4
>
> #Perform T-test
> t.test(x,y, paired = FALSE, conf.level = 0.95)

Welch Two Sample t-test

data:  x and y
t = -0.34757, df = 17.681, p-value = 0.7323
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-1.410481  1.010481
sample estimates:
mean of x mean of y
2.9       3.1

>
> #multiply X20 to scale it between 1 and 100 [Does not convert to %]
> x1 = x*20
> x1
 60 60 40 80 20 20 60 80 80 80
> y1 = y*20
> y1
  40  40  60  80 100  40  20  60 100  80
>
> #perform t-test on new data
> t.test(x1,y1, paired = FALSE, conf.level = 0.95)

Welch Two Sample t-test

data:  x1 and y1
t = -0.34757, df = 17.681, p-value = 0.7323
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-28.20962  20.20962
sample estimates:
mean of x mean of y
58        62

>
> #convert to percentage
> x2 = x/5
> x2
 0.6 0.6 0.4 0.8 0.2 0.2 0.6 0.8 0.8 0.8
> y2 = y/5
>
> #perform t-test on new data
> t.test(x2,y2, paired = FALSE, conf.level = 0.95)

Welch Two Sample t-test

data:  x2 and y2
t = -0.34757, df = 17.681, p-value = 0.7323
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.2820962  0.2020962
sample estimates:
mean of x mean of y
0.58      0.62


Irrespective of the scale you end up with the same conclusion. So, Technically it should not affect it.

• Sorry, I should have been clearer in the question. When they analyse the percentages, they use a test to compare proportions, not a paired sample t test. As described in the link – Chris Beeley Jan 30 at 14:08