It seems you have already seen a similar analysis using R. Here is output from Minitab 17. Observed cell counts are consistently about the same as expected cell counts and standardized residuals are all small in absolute value. There is no reason to suspect grade distributions differ between the two years.
Rows: Columns: Worksheet columns
2017 2018 All
A 75 60 135
77.43 57.57
-0.2767 0.3209
B 176 110 286
164.05 121.95
0.9333 -1.0824
C 85 84 169
96.94 72.06
-1.2124 1.4061
F 81 56 137
78.58 58.42
0.2728 -0.3164
All 417 310 727
Cell Contents: Count
Expected count
Standardized residual
Pearson Chi-Square = 5.844, DF = 3, P-Value = 0.119
Likelihood Ratio Chi-Square = 5.827, DF = 3, P-Value = 0.120
Of course the proportional distributions of grades A, B, C, F are not
exactly the same in the two years, but the chi-squared test shows no
evidence of statistically significant differences.
In view of this, it hardly seems worthwhile to try to answer whether
grades in one year were 'better' than in the other. However, it may be
worthwhile to consider how this could be done if a difference were
suspected. One method would be to assign numerical values to the
letter grades, perhaps A=4, B=3, C=2, F=0.
If the original data on all the students are available, then it should be
easy to substitute numbers for grades. But using R, it is relatively easy to
construct numerical vectors x.17 and x.18:
x.17 = rep(c(4, 3, 2, 0), times=c(75, 176,85,81))
table(x.17)
x.17
0 2 3 4
81 85 176 75 # verify same as in contingency table
summary(x.17); sd(x.17); length(x.17)
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.000 2.000 3.000 2.393 3.000 4.000
[1] 1.329674 # sample SD 2017
[1] 417 # sample size 2018
x.18 = rep(c(4, 3, 2, 0), times=c(60, 110,30,56))
table(x.18)
x.18
0 2 3 4
56 30 110 60
summary(x.18); sd(x.18); length(x.18)
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.000 2.000 3.000 2.461 3.000 4.000
[1] 1.427475
[1] 256
The sample lower quartiles, medians, and upper quartiles are the same for
the two years. The sample means are very similar. It is not surprising
that a Welch (separate-variances) two-sample t test does not show a significant
difference in these numerical scores. [Data are not normal, but sample sizes
are moderately large, and there are no outliers or indications of extreme skewness, so it seems reasonable to use a t test, known to be robust in these circumstances. In view of the fact that the data take so few distinct values,
I would be reluctant to use a Wilcoxon rank sum test.]
t.test(x.17, x.18)
Welch Two Sample t-test
data: x.17 and x.18
t = -0.6125, df = 510.27, p-value = 0.5405
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.2846483 0.1493440
sample estimates:
mean of x mean of y
2.393285 2.460938
In summary, there is no reason to believe student performance was significantly different in
in one year than in the other.
