# Test for significant difference from zero

I have a set of actor's activity coefficients for different subjects for every day in a given time range, which looks like this:

date            actor1  actor2  actor3  actor4  actor5
15-Mar-2020     -,344   -,250   ,322    -,452   ,950
16-Mar-2020     -,260   ,135    ,094    -,508   ,305
17-Mar-2020     -,034   -,188   ,287    ,055    1,559
...


For every actor, I would like to test whether their coefficients are significantly different from zero over the course of the whole time range. At first instance, I thought of conducting a simple one-sample t-test, but this does not account for the distance between each observation and the 0 test value. Consider this small example:

date            actor1  actor2
15-Mar-2020     0       -2
16-Mar-2020     0       1
17-Mar-2020     0       0
18-Mar-2020     0       1


While both actors have an average coefficient of 0 (and thus, both don't differ siginficantly from zero), the second actor has a higher standard deviation. Is there any statistical test I could use to see whether this standard deviation is significantly different from zero, preferably using SPSS or R? Or would you suggest taking this on from a completely different angle?

If data are normal, you can use the relationship $$\frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(\nu = n-1)$$ to make inferences about population variance $$\sigma^2.$$ (If all observations in a sample are the same, then the sample variance is $$S^2=0,$$ and one wonders if the population 'distribution' may be degenerate.)

In your case, it might be more appropriate to get 95% confidence intervals for variances of actors' activity scores, than to test hypotheses about particular values. Because 'variances are very variable' 95% CIs may be annoyingly long for small samples. (For shorter, less precise, intervals, some people use 90% CIs for variances.)

A two-sided 95% confidence interval for $$\sigma^2$$, is $$\left(\frac{(n-1)S^2}{U},\, \frac{(n-1)S^2}{L}\right),$$ where $$L$$ and $$U$$ cut probability 0.025 from the lower and upper tails, respectively, of $$\mathsf{Chisq}(\nu=n-1).$$

Consider a sample of size $$n = 20$$ from $$\mathsf{Norm}(\mu=0, \sigma=2),$$ simulated in R:

x = rnorm(20, 0, 2)

summary(x); sd(x)
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
-4.6496 -1.2126 -0.2458 -0.1545  1.7583  3.2671
[1] 2.143002   # sample SD


Here is a 2-sided 95% CI $$(2.66, 9.80)$$ for $$\sigma^2.$$

19*var(x)/qchisq(c(.975,.025), 19)
[1] 2.656027 9.796947


A one-sided 95% interval---essentially upper bound, for $$\sigma^2$$ is shown below. You might write it as $$(0, 8.62).$$

19*var(x)/qchisq(.05, 19)
[1] 8.624746


You can take square roots of endpoints to get CIs for population SD $$\sigma.$$ Here is a 95% 2-sided interval $$(1.63, 3.13).$$

sqrt(19*var(x)/qchisq(c(.975,.025), 19))
[1] 1.629732 3.130008


Note: It is customary to use probability-symmetric 2-sided CIs for $$\sigma^2$$ and $$\sigma.$$ This means that the same probability $$0.025$$ is cut from each tail of the chi-squared distribution to make the 95% CI. However, these intervals are not of the style 'point estimate $$\pm$$ margin of error', so the point estimate $$S^2$$ of $$\sigma^2$$ is not exactly at the center of a CI for $$\sigma^2.$$