# Can you use the chi-squared test when the expected values are not determined?

I have six ratios (females-males). These are the observed values:

1.78 - 1.17 - 0.53 - 1 - 0.85 - 0.56

I need to run a chi-squared test to show that these ratios are significantly different.

I expect all these ratios to be equal. But as for the “equal” value, there is no expectation. It could be any value, providing that this value is the same for all the cases. How can I run a chi-squared test in this condition? What do I plug into the “expected value” of the formula? Thanks

These are the data:

Group    1    2    3    4    5    6        Total

Men :    9   17   13   12   11   19         81
Women:  16    9   11   14   11    5         66

Total:  25   26   24   26   22   23        147

• Do you know the numbers of females and males that produced these ratios? – bdeonovic Oct 3 '13 at 0:47
• A "Chi-squared" test is a very general expression covering many different statistical tests. Please include in your question the formula of the test statistic you intent to use. – Alecos Papadopoulos Oct 3 '13 at 0:57
• chi squared = the sum over [(observed value-expected value)squared /expected value] – Gwen Oct 3 '13 at 1:32
• The expected value comes as a result of your null hypothesis. +1 for posting the data. – Glen_b Oct 3 '13 at 4:48

Testing for equality of ratio of females::males across groups is the same as testing for equality of proportion of females across all groups ($r_i = f_i/m_i$, $p_i = f_i/(f_i+m_i)$, so $p_i = r_i/(1+r_i)$ and $r_i = p_i/(1-p_i)$. If the $p_i$ differ so do the $r_i$.

If you check that the proportions, $p_i$ differ significantly, you can then conclude that the ratios, $r_i$ differ significantly when the $p_i$ do.

So you are testing for equality of proportion across groups.

This is actually the same testing for independence in the two-way table.

As a result, for the data,

Group    1    2    3    4    5    6        Total

Men :    9   17   13   12   11   19         81
Women:  16    9   11   14   11    5         66

Total:  25   26   24   26   22   23        147


your expected values are just $E_i = \text{row total}\times\text{column total}/\text{overall total}$. E.g. the expected values to go with the first group are:

 25 x 81 / 147 = 13.78

25 x 66 / 147 = 11.22


The table of expecteds is:

Group    1     2     3     4     5     6
Men    13.78 14.33 13.22 14.33 12.12 12.67
Women  11.22 11.67 10.78 11.67  9.88 10.33


As a result, you can just calculate the chisquare for the table - just find

(observed - expected)^2/expected


for all $6\times 2$ numbers and add the 12 terms up.

The first column:

(9- 13.78)^2/13.78 = 1.66

(16 - 11.22)^2/11.22 = 2.03


though it's better if you keep more than two decimal places for all the intermediate calculations. If you do it right you should get a chi-square of somewhere close to 11.5 on 5 df. Looking at the Pearson residuals, almost all of that is coming from the first and sixth groups (especially the sixth group).

It depends on your null hypothesis. If your hypothesis is that the ratio of males to females is equal then the expected value is 0.5.

EDIT:

So your data should look something like this:

       1    2    3    4    5    6      TOTAL
MEN                                   | a
WOMEN                                 | b
---------------------------------------
TOTAL  c    d    e    f    g    h


$\chi^2 = \sum_{i=1}^2\sum_{j=1}^6 \dfrac{(O_{i,j} - E_{i,j})^2}{E_{i,j}}$

where the entries in your table are your observed values (e.g. How many men in group 1 is $O_{1,1}$). Under the null hypothesis of independence $E_{1,1} = a*c/N$, $E_{1,2} = a*d/N$, etc. where $N$ is total number of observations.

in R you can do

men <- c( 9  , 17 ,  13 ,  12  , 11  , 19 )
women<- c(16  ,  9  , 11  , 14  , 11  ,  5)

prop.test(x=men, n=(men+women))

6-sample test for equality of proportions without
continuity correction

data:  men out of (men + women)
X-squared = 11.4978, df = 5, p-value = 0.04236
alternative hypothesis: two.sided
sample estimates:
prop 1    prop 2    prop 3    prop 4    prop 5
0.3600000 0.6538462 0.5416667 0.4615385 0.5000000
prop 6
0.7916667


So since the p-value is below 0.05 I would say we have enough evidence to reject the null hypothesis that ALL 6 proportions are equal. At least one of the proportions is not equal to the rest.

• That is the issue, the hypothesis is that the ratio is equal, but there is no hypothesis as for the value of this ratio – Gwen Oct 3 '13 at 0:43
• Ah, my mistake, I did not read the question carefully enough. – bdeonovic Oct 3 '13 at 0:44
• So....is it possible to conduct the test? I have been told it is, but I can't figure out how.... – Gwen Oct 3 '13 at 1:07
• If you know the number of males and females that produced the ratios I think it would be quite easy to do a chi-squared test. – bdeonovic Oct 3 '13 at 1:08
• Yes, I do know the numbers of the females and males that generated the ratios. – Gwen Oct 3 '13 at 1:34

Is it mandatory to run a test based on the chi2 statistics ? If not, I would suggest you use a likelihood ratio test which can properly account for the fact you do not know the expected fraction of females $p$. If I write $p_i = p + \Delta p_i$ ($\Delta p_0 = 0$) the expected fraction of females in category i, the the null hypothesis can be written as:

$$H_0 : p_0 = p_1 = ... = p_6 = p$$

or equivalently

$$H_0 : \Delta p_1 = \Delta p_2 = ... = \Delta p_6 = 0$$

with $p$ an unknown nuisance parameter.

With a likelihood ratio test, you would build your test statistics as

$$D = -2 log(L({\bf n},{\bf N}, {\bf \Delta p} = 0, \hat{\hat{p}})) + 2 log(L({\bf n},{\bf N}, {\bf \hat{\Delta p}}, \hat{p}))$$

with

$$log(L({\bf n},{\bf N}, {\bf Delta p}, p)) = \sum_i log f_i(n_i, N_i, p_i)$$

Now if you believe that your data are normally distributed (I do not think so... I would rather choose a binomial distribution), then $-2 log(L)$ simplifies to

$$-2 log(L) = \sum_i \frac{(n_i - p_i N_i)^2}{n_i} = \chi^2$$

so the test statistics would a be a difference of $\chi^2$

$$D = \chi^2({\bf n},{\bf N}, {\bf \Delta p} = 0, p) - \chi^2({\bf n},{\bf N}, {\bf \Delta p}, p)$$

In one case you run your least-square fit with all parameters free and in the other case you run your least-square fit with constrained ${\bf \Delta p} = 0$. If $H_0$ is satisfied $D$ should be distributed as a $\chi^2$ distribution, but as I said I think it is much better to work with a likelihood function and $f_i(n_i, N_i, p_i) = C_{N_i}^{n_i} p_i^{n_i} (1 - p_i)^{N_i - n_i}$ binomially distributed.