Question 534:
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We will use the Chi-Square test of independence here to see if the difference in credits is greater than what we'd expect to see from chance alone. The null hypothesis of the Chi-Square test is that there is no independence between categories (basically they occur at the same frequency).
The Chi-Square test uses the formula
Sum of (O-E)2
---------
E
- O means the observed frequencies
- E means the expected frequencies
We will need to do this for each cell. I'll walk through the first cell example. We have12 people who are Very Certain and have earned between 0-9 credits. This is the observed count. The expected count is found by multiplying the row total and column totals together and then multiplying this by the grand total. We get (23*21)/64 = 7.54 as our expected value. Now we just plug and chug in the formula (12-7.54)2/7.54 = 2.627.
We now repeat this step for all values and then add them all up. This gets us 14.76, which is our test statistic. To see if this is significant, we can use the excel formula =CHIDIST(14.76,4) where the second parameter is the degrees of freedom, found as (row count -1)*(column count-1) = 2*2=4. We get the p-value .0052 which is less than our alpha of .01 so we would reject the null hypothesis and say that there is a difference in the amount of credits earned and the certainty of choosing a major.
In looking at the counts, the biggest influence on the Chi-Square comes from the lack of students who are both very uncertain and have 60 or more credits (only 1 person), which is what we'd expect (and hope) that as you are more certain about your major, you take more credits, and vice-versa.