Does anyone know how if it's possible to do a Chi Square analysis in SPSS, Excel, or ACCESS. And do I need to have previous data to compare it to or can I just plug in the data I have to run the query?
In Excel 2007 there are three formula functions to calculate chi square.
Two of them make an independence test. In Spanish they are called PRUEBA.CHI (in English it may be TEST.CHI or CHITEST) and PRUEBA.CHI.INV.
The other one uses a chi square distribution. In Spanish it is called DISTR.CHI (in English it may be CHIDIST.
The best way to know how to use them is viewing the Excel help, but you can also find a lot of examples through Google.
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Thanks very much for the step by step guide on how to do the Chi Square. I did it with my data for male and female (see attached), and it seemed to work. I got 6%. Does this mean that my data has an error of about 6%?
It's been a while since I've done extensive statistics, but your approach is the right one: you test whether the result differt a lot from the expected result with the right test....
I believe this to be right but I'm open to be corrected.
Chi^2 is not a percentage. The value obtained in your example, using Yates' correction is 1.71. That value needs to be looked up in a Chi^2 chart using 1 degree of freedom.
As I recall, in order to do a proper Chi-Square, you must first know the distribution you are trying to evaluate, you must then establish the number of intervals you want to test, evaluate the number of expected occurances within each of those intervals, and then compute the difference between the observed and expected value, and test to determine whether that is within specific parameters.
if you are only testing against a uniform distribution, then this is pretty simple. If you are testing against a
different distribution (Bernoulli, binomial, Poisson, geometric, continuous uniform, normal (bell curve), exponential, gamma, ...) this becomes a little more challenging, because you will need the cumulative distribution tables so that you can compute your expected values.
There are two types of chi-square tests typically used in this context: tests of independence and tests of goodness of fit. In this case, you are testing goodness of fit - matching data to a hypothetical model of randomness.
Your model, in this case, is 600 cases expected in each. Your observed data are 568 and 632.
When you run CHITEST, you are producing a p-value, which is a probability. In this case, you got 0.06. This is interpreted like this:
If we were to assume that in the population, 600 cases of each gender were to be expected, there is only a 6% chance that we'd see deviations from 600/600 of at least the magnitude observed in this sample.
In inferential statistics/hypothesis testing, we traditionally compare p-values to 0.05. Since 0.06 > 0.05, we would conclude from your data that the balance of men and women does not deviate from what we would expect from chance.
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