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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.

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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?

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  • $\begingroup$ A high p-value means you cannot reject the null hypothesis. $\endgroup$
    – whuber ♦
    Commented May 13, 2020 at 17:04

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To get the right sample size, you might need to use G-power for having a more accurate sample size. Your data seems to meet the chi-square assumptions. But generally in each cell, you must have a value more than 5 at least 80% of the cells must have a value more than 5.

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  • $\begingroup$ -1 This echoes a common misconception: the rule you are referring to concerns the expected values in the cells, not the actual values. $\endgroup$
    – whuber ♦
    Commented May 13, 2020 at 17:03

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