Performing a Chi-Square Independence Test In ...

Question

Performing a Chi-Square Independence Test In Exercises 13–28, perform the indicated chi-square independence test by performing the steps below.

a. Identify the claim and state H₀ and Hₐ

b. Determine the degrees of freedom, find the critical value, and identify the rejection region.

c. Find the chi-square test statistic.

d. Decide whether to reject or fail to reject the null hypothesis.

e. Interpret the decision in the context of the original claim.

Use the contingency table and expected frequencies from Exercise 11. At α=0.10, test the hypothesis that the variables are independent.

Accepted Answer

Hello everyone and welcome back. Here's the next problem. Survey of 200 college students recorded their preferred study location, library, dorm, coffee shop, and whether they were undergraduate or graduate students. At alpha equals 0.05, can you conclude that student status and preferred study location are related? And we have a table with the numbers for undergraduates, graduates in library, dorm, and coffee shop. We'll look at those in a minute. And our answer choices are A, there is sufficient evidence to conclude a relationship. There is insufficient evidence to conclude a relationship. The test is inconclusive, or D, the data are insufficient for analysis. So we're talking about looking to see relatedness, and in that case, our null hypothesis would be that the student status and preferred study location are independent, so but not related. Well, the alternative hypothesis would say that student status and preferred study location are related. So, the first thing we do is go to our table, and we're going to calculate the row and column totals. So, for undergraduates, we show library 40, dorm 60, coffee shop 30. So, the undergraduate row total. It is going to equal 130, adding those all together. Then graduate students have 50 in the library, 10 in the dorm, 10 in the coffee shop. So that total. Will be 70 And then we need column totals as well. So for the library, we have 40 and 50, so that makes 90 for the total. The dorm was 60 and 10, so that makes 70. And the coffee shop has 30 and 10. Which gives you a total of 40. So we have our observed numbers, and we need our expected counts. So, for expected, we'd start with undergraduate and a library, I'll put UL. For that the expected would be equal to. Will have the undergraduate. Row total was 130. We multiply by the library column total of 90. And divide that by 200, our total number of students. That gives us expected value of 58.5. So now that you see how to do that, rather than talk through each one, I'm just going to paste up here the uh expected value results for all of these different categories, all calculated in the same way. And then we have the graduate, so here's all our undergraduate results, then we have our graduate student frequencies. So we'd have graduates in the library. And then We have a graduate in the dorm. And a graduate student. In the coffee shop. So now that we have all of our Evals, we need to use that chi squared formula. So, chi 2 is equal to the sum of, and then in the numerator. We have observed minus expected. That different squared. Overexpected. So, for example, for an undergraduate in a library. For instance, The observed value is 40. Our expected value that we calculated is 58.5. So we have 40 minus 58.5, that squared. Divided by that expected value of 58.5. And that's going to give you a value of 5.851. Now I'm not going to write that all out for every single one of our values. We'll just indicate the expected value for each one. And go ahead and write out our sum. So our chi squared is going to be the sum of all of those. So, for undergraduate in the library, we got 5.851. And then undergraduate in the dorm. That chi squared value is 4.634. And then go ahead and add the undergraduate. Coffee Shop, and that gives us a chi squared value of 0.615. Then graduate student in the library. That is 10.857. And graduate student. In the dorm, which is 8.580. And finally graduate student in the coffee shop. Which is 1.143. Now I'm going to scoot this up a little to make more room. I add all those values together, and that gives me a chi squared value of. 31 680. So now we need to look up our critical chi squared value, so we need our degrees of freedom. So we know that's the number of categories. -1 for the rows, and then multiplied by that for the columns. So we have for our rose. We have Two categories, undergraduate and graduate, so that's 2 minus 1. And that is multiplied by 4 columns. There are 3 library dorm, coffee shop. 3 minus 1 So that will end up being 1 multiplied by 2, so we have 2 degrees of freedom. And then we know that our alpha is 0.05, that's given to us. So we looked that up in the table. And get a critical value of chi squared. Equal to 5.991. So, now we go to our calculated chi squared value. Our chi squared value of 31.680 is larger than 5.991, which is the critical value. And therefore we reject our null hypothesis. There are no hypothesis is that. The student status and preferred study location are not related, that means. Choice A is our answer that there is sufficient evidence to conclude a relationship. So, again, we're doing a test of whether these two variables are related, student status and preferred study location, at alpha equals 0.05. We have 200 college students. We calculated the chi squared value. Compare that to our critical value of chi squared, it was greater than the critical value, and so we reject the null hypothesis, and we can say the answer is choice A, there is sufficient evidence to conclude a relationship. See you in the next video.

Key Concepts

Chi-Square Test of Independence

Null and Alternative Hypotheses (H₀ and Hₐ)

Degrees of Freedom and Critical Value

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