Need statistics help! @phdstudents @socialscience
My hypothesis is equivalent to:
The height of Brazilian men is more varied than the height of Brazilian women
I have census data on every Brazilian's height.
My colleague is insisting that I do a T test:
For each men calculate the absolute distance from male average height. The same for women. Then put the resulting arrays in a T test.
Why can't I simply compare the standard deviation from the two groups to test my hypothesis?
@Jey_snow @phdstudents @socialscience t test only works when the variable has a normal distribution. I can imagine that an absolute distance from the mean has a gamma or negative exponential distribution of some sort.
Using an F test seems more appropriate here. Curious on what others think.
@Marrekoo @Jey_snow @phdstudents @socialscience you're correct that an F-test is used to compare variances. However, in this case, because we're only comparing two groups (and height is generally considered normally distributed), a t-test can also be used and is in fact closely related to an F-test. This is because both methods involve squaring and summing the differences between group means. In fact, F=t^2 when k=2:
Proof: https://stats.stackexchange.com/questions/55236/prove-f-test-is-equal-to-t-test-squared
I'm glossing over a lot of details, but hopefully this helps.
@sojournTime @Marrekoo @phdstudents @socialscience
My key issue here is that t-test (as far as I know) compares the means of *random samples*. I don't have random samples, I have the full population, therefore I have access to all the population's parameters, such as variance and mean. For this reason the t-test seems inadequate to me.
My (last minute) source:
https://www.statisticshowto.com/probability-and-statistics/t-test/#ttest1
@Jey_snow @Marrekoo @phdstudents @socialscience
I misspoke in my previous toot. The F=t^2 equivalence is for the one-way ANOVA, not F-tests in general (see https://en.m.wikipedia.org/wiki/One-way_analysis_of_variance).
I'm unfamiliar with how the census in Brazil works, but if a person is not legally bound to submit data to the census, then you do not *technically* have the whole population. Granted, censuses try very hard to represent the whole country. In any case, to compare the variances, you can do the following procedure:
1) Perform normality test(s) on both populations (see https://en.m.wikipedia.org/wiki/Normality_test). You usually do not need to do this, especially on height data which is typically normal. If the populations are both normal, go to 2a. Otherwise, 2b.
2a) Compare the variances using an F-test (see https://en.m.wikipedia.org/wiki/F-test_of_equality_of_variances). If the normality tests show relatively weak normality, use Bartlett's test (see https://en.m.wikipedia.org/wiki/Bartlett%27s_test).
2b) If normality fails (or more formally, the null hypothesis that the data sets are normally distributed is rejected), then Levene's test (see https://en.m.wikipedia.org/wiki/Levene%27s_test) or the Brown–Forsythe test (see https://en.m.wikipedia.org/wiki/Brown%E2%80%93Forsythe_test) can be performed with the latter being more robust to non-normal data.
@sojournTime @Marrekoo @phdstudents @socialscience
That is a very complete answer, thank you so much!