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I want to use the Chi-square test of independence to test the following two variables: Student knowledge v.s. course attendance

The null hypothesis is: student knowledge and course attendance (X and Y) are independent

Members in each student knowledge group: Low (12), average(29), high(9)

The results show that there are two degrees of freedom, the chi-square statistic is 0.20, and the p-value is 0.90, and we cannot accept the null hypothesis. I added an image of my test.

click to see the image of the test

I have little doubts regarding the following two issues: the student knowledge groups have an unequal number of participants, the number of participated students in each course is fewer than 10.

My question is: does this test fit for my data?

In case, this test cannot be used for my data, what statistical test I should use instead?

Thomas milley
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1 Answers1

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Welcome to stack exchange. Using the Chi-Square test for independence can be an issue with small cell sizes (ie G3, course Y which has a cell count of 2). This has to do with the use of Chi-Square Distribution as an approximation.

I would recommend Fisher's Exact Test. It's usually designated as a tool for small sample sizes, but it is still effective for large samples.

jros
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  • I saw Fisher's exact test only works with 2x2 contingency table, is that ok if i have more than it , like 3x5? – Thomas milley May 14 '20 at 09:16
  • You're correct that Fisher's is for 2x2. If you're looking for evidence of any association, use the Chi-Sq test. If you're looking to differentiate between two groups (more interpretable evidence of association), then you should try Fisher's – jros May 14 '20 at 20:59