Using the TI 84 Calculator to Perform an ANOVA test.

Using the TI 84 Calculator to Perform an ANOVA test.


This is a video on ANOVA The question states: A study was done to determine if the average amount spent on
textbooks per student is the same for LTCC, SAC City College, Solano College and De Anza College. What can be concluded at the 0.05 level of significance. So here’s the data and notice that there were four
students each for L TCC, Sac City, Salano, and De Anza, and we have the amount they spent on books. I want to perform an ANOVA test, so for the null
hypothesis on ANOVA, that means all the means
are the same. In particular H0 is that mu1 equals mu2 equals mu3 equals mu4. The alternative hypothesis is the opposite which means at least two of the
means are different from each other. so in order to perform an ANOVA test I will use a
calculator. So here’s a calculator. For ANOVA, I needed to first put in my data so I
go to STAT and then EDIT. I need a clear out the four lists: L1, L2, L3, L4
because I have four different colleges with data. So, let’s clear them out. OK, I’ve cleared my data. Now I enter my data.
For L1 I’ll put the LTCC Data for money spent on
books which is 248, ENTER, and then I put in the rest. and for L2 that’s the Sac City data. That’s 292, ENTER and I put in the rest for SAC City. for L3 I put in the Salano data. That’s 164. ENTER, and then I put in the rest. and then finally, let’s put in the De Anza data. That’s 106 ENTER and the rest. And finally 162. OK, now they have all my data entered let’s go to
the ANOVA test. I use STAT and then TESTS. I go to the UP arrow and the very last item is
ANOVA so I hit enter on ANOVA. and then I need a tell it that I’m looking at L1, L2,
L3, and L4 for my four lists. So 2nd L1 , 2nd L2, 2nd L3, 2nd L4 I end the parentheses and hit ENTER. OK, here is my the F statistic which is 1.65 about and the P-value which is about 0.229698
etcetera. and that’s important thing is the P-value. so let’s go back to the PowerPoint So here’s the PowerPoint. Our P-value was about 0.2297 and that’s greater than the level of significance
0.05 which is Alpha. When the P-value is greater than
the level of significance. That means we fail to reject the null hypothesis. The statement is that there is insufficient evidence to conclude that there is a difference between the mean book
expenditures at the four community colleges. so I’m done concluding. Let’s look at the requirements that we need in
order to perform an ANOVA test. The first one is that the distribution at each college must be normal. The second is that the standard deviations are
the same at all four colleges and the third is that the samples are randomly
taken and independent from each other. OK, these assumptions may or may not be true, but if they’re true then we can go ahead and
perform an ANOVA test. And I’m done with the problem.

8 thoughts to “Using the TI 84 Calculator to Perform an ANOVA test.”

  1. thanks 🙂 but i wonder why my professor teaches a different assumption, that the smallest Std. Dev. times 2 is greater than the largest standard deviation? I dont even know how to find that on my ti-84.. :/

  2. Oh my gosh. I was absent during this lesson and I couldn't figure out the calculator commands! Thank you so much!

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