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I have a rather simple question, which however makes my head ache. I have country A and country B. My dependent variable is "Have you participated in training?" and my independent variable is company size (groups). I compiled contingency tables, which show that 34% of workers of small companies and 66% of workers of large companies have participated in training in country A. 20% of workers of small companies and 40% of workers of large companies have participated in country B. How can I tell whether there is a difference between the countries, i.e. which country has larger inequality in participation rate due to company size? Or in this case there is non?

The difference in participation rate between the company size groups in country A is 32 percentage points, in country B 20 percentage point. So I would assume country A is less equal. However, the participation rate is two times higher in large companies in country A and also two times higher in country B. So I would assume the countries are the same (in)equal.

Thank you!


Thank you! So no way to just look a graph/table and describe to the reader without a statistical test that "we can see there is/is not difference between countries"?.

I have the counts (participation/no participation in each size group). I used the counts to create the crosstab, which gave me the percentages (participation rate). Now I put the table in my thesis and I want to describe it. I say that the participation rate is higher in country A in both of the company size groups. But I'd also like to say whether the countries are similar or different regarding the difference between the size groups. I figured I can just look at the difference between the percentages (participation rate) of the size groups in country A and country B and say something about it without performing further tests.

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  • $\begingroup$ I don't get notified when you edit the question, so it's better to ask clarifying questions in comments. You can say in your thesis that one proportion is higher than another, but if you wish to claim that this property of your random sample reflects on the population parameters, then it's best to do a statistical test. $\endgroup$ Apr 10, 2022 at 16:03

3 Answers 3

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If you know the counts rather than just the proportions, you can do a Chi-squared test, Fisher's exact test, or Barnard's test.

If you don't know the counts, then it's more difficult to do a statistical test, because 60% could be 6 in 10 or it could be 6000 in 10000, and these imply differing levels of certainty.

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  • $\begingroup$ Can you explain why a t-test of proportions is not valid here? $\endgroup$ Apr 11, 2022 at 1:23
  • $\begingroup$ @Learningstatsbyexample a t-test of proportions isn't appropriate because you would only have N of 1 in each group. This means that you have no idea about the level of uncertainty in the mean. It would be like trying to see if 4 is significantly different from 5. Well, they are different, but we can't say anything about uncertainty, because we have no idea what the populations they came from are like. They could easily come from the same population, and the difference could simply be down to chance. $\endgroup$ Apr 11, 2022 at 3:29
  • $\begingroup$ I guess I don't fully understand why that's necessary. Why not simply use the size of the groups as n1, n2, and p1, p2 as the respective proportions who participated in trainings? I understand why the 2x2 contingency table approach makes sense. It just seems like a proportion test is valid here too, though any independence assumptions seem difficult to make. i'm actually thinking a hierarchical model might make sense to account for any correlated errors. but this is the t formula i had in mind: leansigmacorporation.com/wp/wp-content/uploads/2016/01/… $\endgroup$ Apr 11, 2022 at 10:31
  • $\begingroup$ ok, i see, chi square 2x2 and z test of two proportions are identical. stats.stackexchange.com/questions/403912/… so it seems either would work just fine here.. i don't think fisher's exact is a good choice for this number of n as it would just be an approximation, no? $\endgroup$ Apr 11, 2022 at 11:19
  • $\begingroup$ Fisher's exact test is, well, an exact test, so no it's not an approximation. Although it is somewhat conservative relative to Barnard's test, I doubt this would present an issue here. $\endgroup$ Apr 11, 2022 at 17:05
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As @AlanOcallaghan (+1) has said, you need to use counts, not percentages. Also, it is not clear whether you are mainly comparing large vs. small companies or whether you are mainly comparing countries.

Let's take Country A, and compare large vs. small companies there. Suppose 34 employees out of 100 at small companies completed training and 660 employees out of 1000 at big companies completed training.

In these circumstances you want to know whether the proportion $\hat p_s = .34$ is significantly different from $\hat p_b = .66.$

There are several tests for this, based on normal or chi-squared approximations. Some use continuity corrections and some do not. Some use the null hypothesis to find the standard error of $\hat p_s - \hat p_b.$

One such test is implemented in R, as follows (not using continuity correction), it finds a highly significant difference with P-value very near $0.$

prop.test(c(34,660), c(100,1000), cor=FALSE)

    2-sample test for equality of proportions 
    without continuity correction

data:  c(34, 660) out of c(100, 1000)
X-squared = 39.977, df = 1, p-value = 2.57e-10
alternative hypothesis: two.sided
95 percent confidence interval:
 -0.4173769 -0.2226231
sample estimates:
prop 1 prop 2 
  0.34   0.66 

However, if both sample sizes are 50, then you still have a significant difference at the 1% level (P-value below 0.01.) [I used the continuity correction here on account of the small sample sizes.]

 prop.test(c(17,33), c(50,50), cor=TRUE)

        2-sample test for equality of proportions 
        with continuity correction

data:  c(17, 33) out of c(50, 50)
X-squared = 9, df = 1, p-value = 0.0027
alternative hypothesis: two.sided
95 percent confidence interval:
 -0.5256904 -0.1143096
sample estimates:
prop 1 prop 2 
  0.34   0.66 
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I think to compare 2 groups you may go for T test and set the null hypothesis as " There's no difference " and check for p values of they are above or below the critical values to accept or reject the null hypothesis

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  • $\begingroup$ Could you illustrate how you would use a t test here? $\endgroup$
    – BruceET
    Apr 10, 2022 at 22:31

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