Enrique teaches five sections of Introductory Psychology. He wants to know whether or not the average score on the second exam is the same for all five sections. Enrique's student's exam scores are listed below. Use an ANOVA test with a level of significance of 6% to test whether the average Score is the same for all five sections. Section 1: 80, 80, 84, 88, 85, 89, 80, 79, 85, 80, 70, 82, 76, 86, 72, 84, 80, 79, 87, 74, 81, 86, 74, 79, 81, 77, 75, 76, 82, 76, 84, 80, 88, 92 Section 2: 78, 90, 77, 83, 79, 82, 82, 89, 78, 82, 81, 81, 80, 85, 81, 84, 86, 84, 75, 81, 82, 80, 83, 83, 81, 86, 85, 82, 79, 81, 82 Section 3: 82, 80, 78, 81, 82, 83, 83, 81, 76, 81, 84, 78, 75, 86, 83, 89, 79, 84, 80, 80, 77, 76, 80, 81, 79, 76, 79, 87, 77, 85, 82, 85, 81, 85, 75, 83, 87, 84, 79 Section 4: 84, 88, 77, 87, 89, 83, 88, 83, 84, 70, 85, 82, 82, 84, 83, 82, 78, 83, 82, 85, 78, 81, 78, 86, 87 Section 5: 83, 85, 78, 82, 82, 87, 82, 90, 80, 84, 87, 83, 89, 81, 80, 89, 82, 83, 84, 92, 85, 79, 83, 91, 84, 84, 81, 82, 84, 89, 88, 73, 86 78, 77 Step 1: State the null and alternative hypotheses. Ho: P1 = P2 = P3 = H4 = Hs Ha: At least one mean isn't equal to the other means Part 2 of 4 Step 2: Assuming the null hypothesis is true, determine the features of the distribution of test statistics. vv distribution with numerator degrees of freedom We will use a(n) F d fbetween = and denominator degrees of freedom dfwithin = Question Help: D Post to forum

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Enrique teaches five sections of Introductory Psychology. He wants to know whether or not the
average score on the second exam is the same for all five sections. Enrique's student's exam scores
are listed below. Use an ANOVA test with a level of significance of 6% to test whether the average
Score is the same for all five sections.
Section 1:
80, 80, 84, 88, 85, 89, 80, 79, 85, 80, 70, 82, 76, 86, 72, 84, 80, 79, 87, 74, 81, 86, 74, 79, 81, 77, 75, 76,
82, 76, 84, 80, 88, 92
Section 2:
78, 90, 77, 83, 79, 82, 82, 89, 78, 82, 81, 81, 80, 85, 81, 84, 86, 84, 75, 81, 82, 80, 83, 83, 81, 86, 85, 82,
79, 81, 82
Section 3:
82, 80, 78, 81, 82, 83, 83, 81, 76, 81, 84, 78, 75, 86, 83, 89, 79, 84, 80, 80, 77, 76, 80, 81, 79, 76, 79, 87,
77, 85, 82, 85, 81, 85, 75, 83, 87, 84, 79
Section 4:
84, 88, 77, 87, 89, 83, 88, 83, 84, 70, 85, 82, 82, 84, 83, 82, 78, 83, 82, 85, 78, 81, 78, 86, 87
Section 5:
83, 85, 78, 82, 82, 87, 82, 90, 80, 84, 87, 83, 89, 81, 80, 89, 82, 83, 84, 92, 85, 79, 83, 91, 84, 84, 81, 82,
84, 89, 88, 73, 86 78, 77
Step 1: State the null and alternative hypotheses.
Ho: P1 = P2 = µ3 = H4 = Ps
Ha: At least one mean isn't equal to the other means v
Part 2 of 4
Step 2: Assuming the null hypothesis is true, determine the features of the distribution of
test statistics.
We will use a(n) F
vv distribution with numerator degrees of freedom
d fbetween =
and denominator degrees of freedom d fwithin
Question Help: D Post to forum
Transcribed Image Text:Enrique teaches five sections of Introductory Psychology. He wants to know whether or not the average score on the second exam is the same for all five sections. Enrique's student's exam scores are listed below. Use an ANOVA test with a level of significance of 6% to test whether the average Score is the same for all five sections. Section 1: 80, 80, 84, 88, 85, 89, 80, 79, 85, 80, 70, 82, 76, 86, 72, 84, 80, 79, 87, 74, 81, 86, 74, 79, 81, 77, 75, 76, 82, 76, 84, 80, 88, 92 Section 2: 78, 90, 77, 83, 79, 82, 82, 89, 78, 82, 81, 81, 80, 85, 81, 84, 86, 84, 75, 81, 82, 80, 83, 83, 81, 86, 85, 82, 79, 81, 82 Section 3: 82, 80, 78, 81, 82, 83, 83, 81, 76, 81, 84, 78, 75, 86, 83, 89, 79, 84, 80, 80, 77, 76, 80, 81, 79, 76, 79, 87, 77, 85, 82, 85, 81, 85, 75, 83, 87, 84, 79 Section 4: 84, 88, 77, 87, 89, 83, 88, 83, 84, 70, 85, 82, 82, 84, 83, 82, 78, 83, 82, 85, 78, 81, 78, 86, 87 Section 5: 83, 85, 78, 82, 82, 87, 82, 90, 80, 84, 87, 83, 89, 81, 80, 89, 82, 83, 84, 92, 85, 79, 83, 91, 84, 84, 81, 82, 84, 89, 88, 73, 86 78, 77 Step 1: State the null and alternative hypotheses. Ho: P1 = P2 = µ3 = H4 = Ps Ha: At least one mean isn't equal to the other means v Part 2 of 4 Step 2: Assuming the null hypothesis is true, determine the features of the distribution of test statistics. We will use a(n) F vv distribution with numerator degrees of freedom d fbetween = and denominator degrees of freedom d fwithin Question Help: D Post to forum
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