3
$\begingroup$

Suppose women may have Blue or Red eyes. We make a first cheap research on a 100 women checking just one eye of each woman. So after the research we have results:

At least one eye Blue have avg height 177 cm (50 women)
At least one eye Red have avg height 164cm (70 women)

Since we noticed that one women may have eyes of different color we decided to make second more accurate research. This time we checked both eyes of each woman. Women with eyes:
Blue Blue have avg height 180 cm (30 women)
Red Red have avg height 160 cm (50 women)
Blue Red (sequence does not matter) have avg height 173 cm (20 women)

Now the question. After making the first research we might expect that women with different colors of eyes will have avg height of (177+164)/2=170 cm. It turned out in the second research that they had 3 cm more. What kind of statistical test can determine that 170=173 if we have 20 women of that kind? I would like to know if the BlueRed combination is something totally different feature or it is determined by Blue and Red.

In the same way we may ask - can we expect that women with double Red eyes will have average height (164+164)/2=164cm? It turned out that they had 160cm.

Should we use Chi-squared test? But the problem is that we have a 100 women and 50 Blue and 70Red (don't sum up). Or maybe should we use a two-sample location test of the null hypothesis that the means of two populations are equal? But then, we use the same sample of a 100 women in both cases. How can we check the independence of Blue and Red?

$\endgroup$
  • $\begingroup$ You are in fact thinking about a regression model where the number of red eyes determines height (or vice versa). This way, you would also model nolens volens how tall you expect a woman with only one eye (e.g. due to an accident) to be? Or a woman with three eyes (e.g. due to an even worse accident)? $\endgroup$ – Horst Grünbusch Oct 15 '14 at 14:36
  • $\begingroup$ I want to check dependence of variables. I do not want to model women height. Regression model does not show dependence of independent variables. $\endgroup$ – Przemyslaw Remin Oct 15 '14 at 17:29
  • $\begingroup$ Do the women have two eyes? $\endgroup$ – Sextus Empiricus Sep 19 '19 at 12:03
2
$\begingroup$

In your question you have been comparing the red-blue value (173 cm) from the second experiment with the minimal one eye blue and minimal one eye red from the first experiment (177 cm and 164 cm). In that case it goes like part 1 and is a simple matter of averaging with incorporating weights.

If however you are comparing the values 180, 173, 160 from the second experiment with each other, then your case is like part 2.

Part one

Your two different experiments are not contradicting.

Let's put the data in a table:

               bb     rb     rr
number         30     20     50
mean height    180    173    160

When you add those groups together for instance bb+rb for at least one blue eye or rb+rr for at least one red eye then the means for those composite groups become:

$$\frac{180 \times 30+173 \times 20}{50} = 177.2 \approx 177$$ and $$\frac{160 \times 50+173 \times20}{70} = 163.7143 \approx 164 $$

where you should note that you need to incorporate the number of each category. For instance the mean of a group with 1 women of height 160 cm and 9 women of height 170 cm should be 169 cm, and not $\frac{160+170}{2}\,cm = 165 \, cm$

See also: two ways to compute average prices

Part two

If you are comparing the values 180, 173, 160 from the second experiment with each other, then you might do this based on some model. For instance you could imagine that the mean height of women is something like a base value plus some additional value for each blue eye (and some error term for random variation among the women with the same colour of eyes).

$$\text{Women height} = a + c \times n_{\text{blue eyes}} + \epsilon $$

or

$$\text{Women height} = \begin{cases} a & + &\epsilon \quad \quad \text{if $0$ blue eyes} \\ a + c& +& \epsilon \quad \quad \text{if $1$ blue eyes} \\ a + 2c & +& \epsilon \quad \quad\text{if $2$ blue eyes} \\ \end{cases}$$

this you could compare with a model in which the second blue eye will add a different value to the mean height than twice the value for the case of 1 blue eye.

$$\text{Women height} = \begin{cases} a & + &\epsilon \quad \quad \text{if $0$ blue eyes} \\ a + c_1& +& \epsilon \quad \quad \text{if $1$ blue eyes} \\ a + c_2 & +& \epsilon \quad \quad\text{if $2$ blue eyes} \\ \end{cases}$$

This can be compared with GLM/ANOVA or other modeling and it requires you to incorporate the error distribution and your data needs to capture not only the mean value of the groups but also the individual values of the women.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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