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Consider the table below that gives the proportions of a sample from each of two populations that fall into one of three categories (table edited after @whuber comment).

                            1   2   3   **4**
First Population Sampled  13% 19% 28% **40%**
Second Population Sampled 7%  11% 22% **60%** 

Test the hypothesis that the population proportions are equal with a column (category) by calculating the p-value of the $\chi^2$ test statistic. Test assuming the size of the sample selected from each population that gave the above proportions was 100 people at level 0.10.

Answer: I have established the following hypotheses:

By the statement of the problem, I am assuming that this is a test for homogeneity. So the hypotheses would be:

  1. $H_0:$ The populations are homogeneous with respect to the three categories. That is, $p_{1j} = p_{2j}$ for $j = 1, 2, 3$.

  2. $H_a:$ The populations are not homogeneous with respect to the three categories.

However, I then realized that the row totals do not sum to 1, which has me concerned.

Is the homogeneity approach correct? I thought it may be a test of independence, but there are two populations.

Based on the comment below, the observed counts are:

                            1   2   3   4
First Population Sampled   13  19  28  40
Second Population Sampled   7  11  22  60 

So we can proceed with the homogeneity test with $n = 200$.

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  • $\begingroup$ Hint: there's an invisible fourth column that you can (easily) recreate from the information given you. $\endgroup$
    – whuber
    Feb 18 at 14:23
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    $\begingroup$ Since each row must sum to 1 = 100%, I should created the fourth column as in my Edit to my post, and then am able to proceed with a test of homogeneity? $\endgroup$
    – Chesso
    Feb 18 at 14:51
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    $\begingroup$ You are given that each sample was 100 people, so just use that to recreate a table of counts! $\endgroup$ Feb 18 at 14:56
  • $\begingroup$ Understood. So this now matches a homogeneity test with $n = 200$. $\endgroup$
    – Chesso
    Feb 18 at 15:06
  • $\begingroup$ You can now answer your Q yourself! Please do so, so the Q do not linger on as unresolved! $\endgroup$ Feb 18 at 15:07
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"Anti-lingering" Comment:

Your stated contingency table is TBL as below:

TBL = rbind(c(13,19,28,40), c(7,11,22,60))
TBL
     [,1] [,2] [,3] [,4]
[1,]   13   19   28   40
[2,]    7   11   22   60

Verifying that both rows sum to $100,$ as stated:

rowSums(TBL)
[1] 100 100

A chi-squared test in R, shows significant departure from homogeneity at 5% level with P-value $0.034 < 0.05 = 5\%.$

chisq.test(TBL)

        Pearson's Chi-squared test

data:  TBL
X-squared = 8.6533, df = 3, p-value = 0.03427

The Pearson residuals reveal explicitly that the 2nd sample showed lower counts for categories 1 and 2 than the first sample, while the first sample showed a lower count for category 4. [Taken together, contributions to the chi-squared statistic from these six cells accounted for almost all of the total chi-squared statistic of about 8.6.]

chisq.test(TBL)$res
           [,1]      [,2] [,3]      [,4]
[1,]  0.9486833  1.032796  0.6 -1.414214
[2,] -0.9486833 -1.032796 -0.6  1.414214

The squares of the Pearson residuals sum to the significantly large chi-squared statistic.

chisq.test(TBL)$res^2
     [,1]     [,2] [,3] [,4]
[1,]  0.9 1.066667 0.36    2
[2,]  0.9 1.066667 0.36    2
sum(chisq.test(TBL)$res^2)
[1] 8.653333
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