# 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.

• Welcome to CV. Since you’re new here, you may want to take our tour, which has information for new users. I don't understand how VAR3 and group 2 are related. But if these are the only values you have, there is no variation in Group 2. In other words, the values do not change from one observation to another. it is a constant (0). May 22, 2022 at 10:12
• I edited the question. VAR3 is the name given by the SPSS to group 2. What if I want to compare group 2 with group 1? May 22, 2022 at 10:33
• What should I answer if they ask me what distribution group 2 follows? Should I say it does not follow any distribution because the values are zero (no values)? May 22, 2022 at 10:35
• 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 May 22, 2022 at 10:42
• Please check this @T.E.G. May 22, 2022 at 11:41

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