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In this question, we have a randomized double -blind study that asked u .s.
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College students to solve 15 anagrams in five minutes.
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Just before they did this test, students received information about the study written either in red, green, or black ink.
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So we are going to test whether performance, which is measured in the number of anagrams solved, is different based on ink color.
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The first thing that we'll do is take a look at the table shown here below where we have.
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Have the three groups, and it's important to note that we have three groups, red, green, and black.
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And we're now looking at the size of each group, 19, 27, and 25.
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That gives us a total of 71 students.
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Let's go ahead and answer part a by writing the null and alternative hypotheses.
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So here we're going to state that the means are equal.
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So mu 1 equals, well, let's change that.
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How about if we call it mu r for red? mu r equals mu g equals mu b for black.
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So that is our null hypothesis that all three means are equal.
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So then the alternative would be that some two means are not equal, and i'll write it this way.
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Sum mu i is not equal to mu j all right let's go ahead and complete the anova table now the anova table was partially filled in for us and what we see is that the sum of squares for the groups is 27 .7 and the sum of squares for the total is 84 .7 that means that if we take the difference that will be the sum of squares for the error, which is within each group.
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So that works out to be 57 when we do that subtraction.
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The next thing that we can fill in is the degrees of freedom.
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For the group, remember, the degrees of freedom is k minus one.
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So here we have k, which was three groups.
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So we have three minus one, which is two.
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So three minus one is two degrees of freedom.
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And then for the total, the number of degrees of freedom is n minus 1, and our total number of students was 71.
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So here we have 71 minus 1 equals 70...