Viewing satisfactions scores as levels of a categorical variable. There are various possible approaches. One of the simplest is to
put the counts into a $2 \times 5$ table and do a chi-squared test
for independence. Your satisfaction scores are essentially a Likert
scale with percentages proportional to numbers 1 through 5, used in
the fake example below:
Satis 1 2 3 4 5 TOTAL
Full 191 319 329 456 675 1970
Part 562 603 419 267 275 2126
Analysis in R:
f = c(191, 319, 329, 456, 675)
p = c(562, 603, 419, 267, 275)
TBL = rbind(f, p)
chisq.test(TBL)
Pearson's Chi-squared test
data: TBL
X-squared = 493.7, df = 4, p-value < 2.2e-16
With P-value so near zero, there is strong evidence of association
between Full/Part time and Satisfaction.
Expected counts $E_{ij}$ for the ten cells of the table are computed from row and column totals of the table of counts---assuming that the null hypothesis of no association between the two categorical variables to be true. You can see the expected counts as follows:
cq.out = chisq.test(TBL)
cq.out$exp
[,1] [,2] [,3] [,4] [,5]
f 362.1606 443.4424 359.7559 347.7319 456.9092
p 390.8394 478.5576 388.2441 375.2681 493.0908
Observed counts #X_{ij}$ are the corresponding (integer) counts in TBL.
Comparing observed and expected counts, you can see that part time
workers tend to have more than the expected number of counts in
the lower-numbered satisfaction categories.
The Pearson residuals
are the 'signed' square roots of the ten quantities
$\frac{(X_{ij} - E_{IJ})^2}{E_{ij}}$ can be displayed as follows:
cq.out$res
[,1] [,2] [,3] [,4] [,5]
f -8.994008 -5.909486 -1.621526 5.806014 10.202872
p 8.657745 5.688545 1.560901 -5.588942 -9.821412
Usually, Pearson residuals with absolute values greater than $3$
are taken to show cells with especially poor agreement between
observed and expected counts. For my fake data, the residuals of greatest
interest are for lowest and highest-numbered categorical levels of satisfaction scores.
Viewing satisfaction scores as actual numerical values. The chi-squared test essentially ignores any numerical properties associated
with satisfaction scores (even order), treating numbers only as labels for nominal categorical levels. Other kinds of tests, including
a 2-sample t test might be used if you want to treat satisfaction
scores as actual numbers.
In the data above, we could let $X$ (for full-time employees) have 'numerical' values
as follows: $191$ 1's, $319$ 2's, and so on. And similarly for $Y$ (for part-time employees). Whether to ascribe actual numerical meaning to Likert scores is controversial, but widely accepted as useful in the social sciences.
According to this scheme we have $X$ and $Y$ as follows:
x = rep(1:5, f); y = rep(1:5, p)
par(mfrow=c(2,1))
hist(x, br=(0:5)+.5, ylim=c(0,900), lab=T,
col="skyblue2", main="Full-Time")
hist(y, br=(0:5)+.5, ylim=c(0,900), lab=T,
col="skyblue2", main="Part-Time")
par(mfrow=c(1,1))
Then a Welch 2-sample t test on the 'numerical' values in $X$ and $Y$ shows a highly significant difference in population means.
Welch Two Sample t-test
data: x and y
t = 23.437, df = 4063.8, p-value < 2.2e-16
alternative hypothesis:
true difference in means is not equal to 0
95 percent confidence interval:
0.906221 1.071674
sample estimates:
mean of x mean of y
3.560914 2.571966
