Using an ANOVA table for hypothesis testing

Intro

Howdy!  I'm Professor Curtis of Aspire Mountain Academy here with more statistics homework help.  Today, we're going to learn how to use an ANOVA table for hypothesis testing.  Here's our problem statement: A sample of colored candies was obtained to determine the weights of different colors.  The ANOVA table is shown below.  It is known that the population distribution are approximately normal and the variances do not differ greatly.  Use a 0.025 significance level to test that claim that the mean weight of different colored candies is the same.  If the candy maker wants the different color populations to have the same mean weight, do these results suggest that the company has a problem requiring corrective action?

Source

DF

SS

MS

Test Stat F

Critical F

P-value

Treatment

7

0.021

0.003

0.7555

2.4502

0.6260

Error

80

0.320

0.004

Total

87

0.341

Part 1

OK, the first part of this problem asks, “Should the null hypothesis that all the colors have the same mean weight be rejected?”  Well, we have the ANOVA table here, and notice how here at the end we have our P-value, so we can use this and compare this with our significance level and determine the result of the test.  So a P-value of 0.6260 is definitely greater than our significance level of 0.025.  Therefore, we can’t fit the area of the P-value into the area of the significance level, and we are therefore outside the area of rejection.  Therefore we are going to fail the null hypothesis.  So we should not reject the null hypothesis because the P-value is going to be greater than our significance level.  Excellent!

Part 2

Now, the second part of this problem asks, “Does the company have a problem requiring corrective action?”  Well, here in the problem statement it says that “the candy maker wants the different color populations to have the same mean weight.”  That is the null hypothesis, that all of the colors of the same mean weight.  We failed to reject the null hypothesis, which means it could be true.  And if it’s true, then the candy maker is getting what the candy maker wants.  And so therefore there is no problem requiring corrective action.

So the answer is going to be No, no corrective action is required because — let’s see here.  It is likely that the candies do not have equal mean weights.  No, it is likely that they do.  So we’re going to select answer A, even though it’s got this awkward double negative — “not likely that the candies do not have equal mean weights”!  That’s like saying, yeah, because they do have the same weight.  Excellent!

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