0
$\begingroup$

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?

$\endgroup$

1 Answer 1

1
$\begingroup$

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 $\sigma^2.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.