a. t cannot be computed because at least one of the groups is empty. SPSS

#1
Hi there,
I've been unable to run an independent t stats on the following data and I couldn't fix the problem.
Both variables are set to Numeric and the First(Section) is Nominal while QuizGrade is Scale
Section QuizGrade
1.00 100.00
1.00 33.00
1.00 92.00
1.00 33.00
1.00 98.00
1.00 55.00
1.00 100.00
1.00 99.00
1.00 57.00
1.00 72.00
1.00 50.00
1.00 100.00
1.00 100.00
1.00 92.00
1.00 85.00
1.00 76.00
1.00 52.00
1.00 52.00
1.00 89.00
1.00 95.00
2.00 95.00
2.00 100.00
2.00 88.00
2.00 85.00
2.00 97.00
2.00 80.00
2.00 75.00
2.00 99.00
2.00 95.00
2.00 100.00
2.00 66.00
2.00 82.00
2.00 85.00
2.00 85.00
2.00 90.00
2.00 92.00
2.00 90.00
2.00 87.00
2.00 95.00
2.00 99.00
 
#2
If this is an spss question then I don't know the answer. I can't see anything wrong with the data.

But it seems to be statistically significant.

Code:
# this is an R program

d <- read.table(header = TRUE, text = "
Section QuizGrade
1.00 100.00
1.00 33.00
1.00 92.00
1.00 33.00
1.00 98.00
1.00 55.00
1.00 100.00
1.00 99.00
1.00 57.00
1.00 72.00
1.00 50.00
1.00 100.00
1.00 100.00
1.00 92.00
1.00 85.00
1.00 76.00
1.00 52.00
1.00 52.00
1.00 89.00
1.00 95.00
2.00 95.00
2.00 100.00
2.00 88.00
2.00 85.00
2.00 97.00
2.00 80.00
2.00 75.00
2.00 99.00
2.00 95.00
2.00 100.00
2.00 66.00
2.00 82.00
2.00 85.00
2.00 85.00
2.00 90.00
2.00 92.00
2.00 90.00
2.00 87.00
2.00 95.00
2.00 99.00 ")


Code:
> table(d$Section)
 1  2
20 20

> t.test(d$QuizGrade  ~ d$Section)

Welch Two Sample t-test
data:  d$QuizGrade by d$Section
t = -2.2401, df = 24.293, p-value = 0.03449
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -24.489824  -1.010176

sample estimates:
mean in group 1 mean in group 2
          76.50           89.25





> sd(d$QuizGrade[d$Section==1])
[1] 23.81839

> sd(d$QuizGrade[d$Section==2])
[1] 8.978776


> library(perm)
> permTS(d$QuizGrade  ~ d$Section, alternative="two.sided"  )

Permutation Test using Asymptotic Approximation
data:  d$QuizGrade by d$Section
Z = -2.1329, p-value = 0.03293
alternative hypothesis: true mean d$Section=1 - mean d$Section=2 is not equal to 0
sample estimates:

mean d$Section=1 - mean d$Section=2
                             -12.75