I have a group that contains zero data, and I want to know whether it is considered normally or not normally distributed.? I have a group (group2) that contains zero data, and I want to know whether it is considered normal or not normally distributed.
I used the SPSS software, and it showed this result.

I also tried R, and it showed this message:
Error in Shapiro.test(group2) : all 'x' values are identical
If I want to compare the following groups:
Group 1: 0.7, 0.0, 0.3, 0.5, 0.0, 0.0, 0.4, 0.3, 0.4, 0.0, 0.0, 0.0, 0.0, 0.2
Group 2: 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0
Group 1 is not normally distributed according to the Shapiro Wilk test,
group 2 is the one with the undefined Shapiro test results
What test should I use to compare the following, mean, median, variance, and difference between groups?
What should I answer if they ask me what distribution group 2 follows?
EDIT:
Just for clarification, Samples in groups 1 and 2 were prepared in the same manner, and a specific material was added to group 2 and not added to group 1 ( this material resulted in zero values). So in order to evaluate the effect of the added material, I have to compare these groups. to prove the significant difference and the good effect of the added material.

Comment in Answer format--to allow showing output from R:
Clearly, there is a difference between Groups 1 and 2. You could take the position (1) that the 0's in Group 2 are
real data, but simply too near to 0 to detect or (2) that Group 2 values are truly all exactly 0.
According to (1), one could use a Welch two-sample t test, allowing it to reduce the degrees of freedom (from $n_1 + n_2 = 2 = 26$ for a pooled 2-sample t test) nearly to the minimum possible $13.$ The Welch test finds the difference between means
for the two groups highly significant with P-value very near $0.$ This may be pushing the Welch test beyond
anticipated heteroscedasticity.
t.test(x2, rep(0,14))

        Welch Two Sample t-test

data:  x2 and rep(0, 14)
t = 8.8231, df = 19, p-value = 3.799e-08
alternative hypothesis: 
 true difference in means is not equal to 0
95 percent confidence interval:
 10.41348 16.89059
sample estimates:
mean of x mean of y 
 13.65204   0.00000

According to (2), one could do a one-sample t test to see
if the mean of Group 1 is significantly different from $0.$ It is--with a P-value below 1%.
t.test(x1, mu=0)

        One Sample t-test

data:  x1
t = 3.1798, df = 13, p-value = 0.007244
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
 0.0641191 0.3358809
sample estimates:
mean of x 
      0.2

Because we cannot know whether Group 1 has all $0$s
for its values because there is nothin to measure or whether
the values are simply beyond the range your technology can
detect, this may be abusing the one-sample t test.
For non-statisticians, I would not mention formal tests.
I would say that for Group 2 the instrumentation is not
able to detect anything but 0, but that Group 1 obviously
has $14$ positive values and therefore differs from Group 2.
If you feel that some kind of formal test is required, you
might consider Fisher's Exact Test, with 14 values at $0$ in Group 2 and
14 positive values in Group 1.
TBL = rbind(c(14,0),c(0,14));  TBL
     [,1] [,2]
[1,]   14    0
[2,]    0   14
fisher.test(TBL)$p.val
[1] 4.985467e-08
2/choose(28,14)
[1] 4.985467e-08

