t-test where one sample has zero variance?

Question

I'm very rusty on statistics, so go easy on me haha.

I'm trying to compare two sample sets of values to see if there is any significance difference between the two means. From memory, an unpaired t-test should be able to do this? The issue is that one set has zero variance, so do I use an equal variance test or an unequal?

If there's a better test for this then feel free to let me know :)

Group 1	Group 2
3.97	4.00
3.63	4.00
3.63	4.00
4.00	4.00
4.29	4.00
3.95	4.00
3.63	4.00
3.63	4.00
3.63	4.00
3.96	4.00
3.63
4.00
3.63
3.63
3.93
3.97
4.00
3.63
4.00
3.92
4.26
4.08
3.43
4.26
4.17
3.63

Answer 1

This answer doesn't address the issue of why the second group has no variation (and I really do suggest you get to the bottom of that).

If you're comfortable saying the yield of group 2 is definitely known to be 4.0 you can use a one-sample t-test to compare group 1 against that value.

Update

I should clarify some things, since this answer is getting more upvotes than it deserves. First, with the two-sample test you're testing the hypothesis that group 1 differs from group 2. With the one-sample test, you're testing the hypothesis that group 1 differs from a value of 4.0. These are subtly different questions.

Second, I should stress that treating the value of 4.0 as known is an assumption, and the one-sample test is only valid if this assumption is correct. All statistical tests involve some assumptions like this, for instance assuming that sampling is not biased. It's your job as the analyst to be clear about what assumptions you're making, to decide whether you think they're justified, and to bear in mind that if the assumptions turn out to be wrong then so do your conclusions.

Answer 2

It requires a bit of explaining how or why this second sample is so "precise" so to speak. Is this rounding error, or are there accidental replications? There is a lot of science going on, not explained, that would inform which option(s) would or would not work. However, it's not a dead-end. Here are 3 suggestions (non-exhaustive) in order of difficulty depending on your chops and software at hand.

1. Fit a T-test with an equal variance assumption: even if the assumption is clearly violated, you can argue that the test will be conservative. Similarly, you could jitter the response to within a rounding range - i.e. add random noise.

2. Use a resampling based test: the permutation test is a powerful and flexible means to assess the test of mean difference under the assumption the two samples (under the null) come from the same distribution - note that that distribution would be approximately normal with a "point mass" at X=4.00.

3. Use a Bayesian approach similar to option 2, it's just a convenient way to calculate the sampling distribution of the mean difference, and make inference correspondingly.

4. Use expectation maximization: OK so you rounded all the variability out of one sample - generally not a good idea. When you report the result as 4.00 - appreciably we all expect that the result must not be 3.99 or 4.01, so for God's sake, use significant figures, you! Anyway, most of the tests above involve collecting the sample into a single group and assuming that follows some kind of normal distribution - the distribution under the null. Clearly that's not the case, if a value of 4.00 is really 4 and could in fact be 3.50 up to 4.49, then you have to integrate up whatever area of the normal curve of the assumed parametric framework at that iteration to find the true likelihood contribution at that time point. This is a somewhat standard, and most correct, approach to measurement error - which is the specific problem you have here - but is sadly the most technically difficult to implement.

Answer 3

One solution is to use a one-sample t-test of the null hypothesis:

$$\text{H}_{0}\text{: }\mu_{1} = 4\text{, with H}_{\text{A}}\text{: }\mu_{1} \ne 4$$

If $n_1=26$, $\bar{x}_1=3.865$ and $s_{1} = .24428$, our test statistic for this null would be:

$$t = \frac{\bar{x}_{1}-4}{\frac{s_{1}}{\sqrt{n_{1}}}} = \frac{3.865-4}{\frac{.24428}{\sqrt{26}}} = -2.82$$

At the $\alpha=0.05$ level of significance we would reject the two-sided null hypothesis above, and conclude that we found evidence that Group 1 was sampled from a population with a mean different (less than) than 4.

Alternately, you might try adding a tiny amount of artificial noise to estimate a nearly, but not exactly, zero variance to Group 2 and conduct a two-sample t test assuming different variances (this will produce almost identical results to the one-sample test).

Answer 4

As others have already said, it would be better to address the issue with rounding to get the original data. However, if you have to work with the data you have here's another fairly simple option that may have some advantages over current answers:

Do a t-test under the assumption of equal variance where the variance is calculated using only Group 1. Basically, using the procedure for unequal sample sizes, similar variances but replacing the variance of group 2 ($s^{2}_{X_{2}}$ on the wiki page) with the variance of group 1 ($s^{2}_{X_{1}}$).

You would have to do this by hand but it's should be doable even in e.g. Excel if you don't have much experience with software like R or python.

I think the problem with pre-packaged t-test either assuming equal or unequal variance is that Group 2 is going to bias downwards the estimate of variance (assuming you don't believe Group 2 has 0 variance). Similarly, testing Group 1 against a fixed mean of 4 (like t.test(Group_1, mu=4)) ignores the variance in Group 2.

More sophisticated solutions based on re-sampling, Bayesian statistics, etc (see @AdamO answer) may be better but more difficult to implement.

---

Sextus Empiricus in comment says:

If group2 has an unexpectedly low variance, then it is arbitrary to just decide to replace it with the variance of group 1

I'd say that eventually every analysis has some arbitrary elements. The question is how you justify them and how they impact the interpretation.

My suggestion assumes that the lack of variation in group 2 is unrealistic and all data points have been reset to 4.00. In such case, I think using the variance from group 1 is more sensible than taking the values in group 2 at face value or, equivalently, testing if group 1 differs from the point value of 4.00. I think it is more reasonable to assume that different farms have more or less the same variance than assuming no variance at all. I guess you could even present results for a range of variances for group 2 (from no variance to the same variance of group 1) and make an informed decision based on that.

In the end it's up to the OP to decide what assumptions are more sensible and make sure that collaborators are aware of those decisions. I would also consider whether it is more costly a false positive (claim farms are different when they are not) or a false negative (farms are indistinguishable when instead they are).