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?


1 Answer 1


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 $\sigma^2.$


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