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John Paul 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. John Paul's student's exam scores are listed below. Use an ANOVA test with a level of significance of 5% to test whether the average score is the same for all five sections. Section 1: 62, 76, 70, 66, 94, 59, 76, 65, 72, 79, 52, 46, 74, 43, 68, 61, 75, 65, 63, 58, 62, 76, 70, 61, 76, 73, 53, 57, 73, 78, 51, 54, 42, 66, 58, 72, 44, 87 Section 2: 65, 52, 51, 67, 64, 75, 56, 68, 72, 82, 74, 73, 83, 42, 73, 85, 67, 46, 70, 51, 72, 53, 28, 58, 62, 68, 58, 85, 71, 57, 60, 79, 64, 60, 69, 65, 66 Section 3: 83, 67, 69, 72, 67, 92, 57, 57, 57, 61, 90, 75, 62, 69, 50, 57, 60, 94, 83, 69, 68, 65, 81, 60, 81, 67, 72, 78, 79,65 Section 4: 63, 75, 55, 70, 75, 81, 50, 77, 59, 42, 44, 85, 57, 62, 33, 45, 82, 67, 68, 47, 75, 69, 48, 96, 79,77 Section 5: 30, 82, 83, 54, 54, 59, 75, 43, 68, 58, 58, 45, 72, 67, 58, 82, 50, 63, 55, 50, 84, 71, 68, 80, 85, 40, 68, 85, 73, 84, 47, 48, 57, 56 Step 1: State the null and alternative hypotheses. Ho: P. H. Ha: At least one mean isn't equal to the other means VV Y 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 dfbetween=4 Step 3: Find the p-value of the test statistic. Z₁ = #₂ = #₂ = #₁ = Between Within V distribution with numerator degrees of freedom ✓and denominator degrees of freedom dfwithin = 160 SS P(FEV Question Help: D Post to forum Submit Part ANOVA Table df MS F ✓ Part 3 of 4

John Paul 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. John Paul's student's exam scores are listed below. Use an ANOVA test with a level of significance of 5% to test whether the average score is the same for all five sections. Section 1: 62, 76, 70, 66, 94, 59, 76, 65, 72, 79, 52, 46, 74, 43, 68, 61, 75, 65, 63, 58, 62, 76, 70, 61, 76, 73, 53, 57, 73, 78, 51, 54, 42, 66, 58, 72, 44, 87 Section 2: 65, 52, 51, 67, 64, 75, 56, 68, 72, 82, 74, 73, 83, 42, 73, 85, 67, 46, 70, 51, 72, 53, 28, 58, 62, 68, 58, 85, 71, 57, 60, 79, 64, 60, 69, 65, 66 Section 3: 83, 67, 69, 72, 67, 92, 57, 57, 57, 61, 90, 75, 62, 69, 50, 57, 60, 94, 83, 69, 68, 65, 81, 60, 81, 67, 72, 78, 79,65 Section 4: 63, 75, 55, 70, 75, 81, 50, 77, 59, 42, 44, 85, 57, 62, 33, 45, 82, 67, 68, 47, 75, 69, 48, 96, 79,77 Section 5: 30, 82, 83, 54, 54, 59, 75, 43, 68, 58, 58, 45, 72, 67, 58, 82, 50, 63, 55, 50, 84, 71, 68, 80, 85, 40, 68, 85, 73, 84, 47, 48, 57, 56 Step 1: State the null and alternative hypotheses. Ho: P. H. Ha: At least one mean isn't equal to the other means VV Y 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 dfbetween=4 Step 3: Find the p-value of the test statistic. Z₁ = #₂ = #₂ = #₁ = Between Within V distribution with numerator degrees of freedom ✓and denominator degrees of freedom dfwithin = 160 SS P(FEV Question Help: D Post to forum Submit Part ANOVA Table df MS F ✓ Part 3 of 4

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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▾ Part 1 of 4 John Paul 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. John Paul's student's exam scores are listed below. Use an ANOVA test with a level of significance of 5% to test whether the average score is the same for all five sections. Section 1: 62, 76, 70, 66, 94, 59, 76, 65, 72, 79, 52, 46, 74, 43, 68, 61, 75, 65, 63, 58, 62, 76, 70, 61, 76, 73, 53, 57, 73, 78, 51, 54, 42, 66, 58, 72, 44, 87 Section 2: 65, 52, 51, 67, 64, 75, 56, 68, 72, 82, 74, 73, 83, 42, 73, 85, 67, 46, 70, 51, 72, 53, 28, 58, 62, 68, 58, 85, 71, 57, 60, 79, 64, 60, 69, 65, 66 Section 3: 83, 67, 69, 72, 67, 92, 57, 57, 57, 61, 90, 75, 62, 69, 50, 57, 60, 94, 83, 69, 68, 65, 81, 60, 81, 67, 72, 78, 79,65 Section 4: 63, 75, 55, 70, 75, 81, 50, 77, 59, 42, 44, 85, 57, 62, 33, 45, 82, 67, 68, 47, 75, 69, 48, 96, 79, 77 Section 5: 30, 82, 83, 54, 54, 59, 75, 43, 68, 58, 58, 45, 72, 67, 58, 82, 50, 63, 55, 50, 84, 71, 68, 80, 85, 40, 68, 85, 73, 84, 47, 48, 57, 56 Step 1: State the null and alternative hypotheses. Ho: P.H. Ha: At least one mean isn't equal to the other means ✓✓ Y 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 dfbetween = 4 Step 3: Find the p-value of the test statistic. ₁ = #₂ = Es= Z₁ = #₁ = MI Between Within ✓✓distribution with numerator degrees of freedom ✓and denominator degrees of freedom dfwithin 160 SS P(FVV Question Help: D Post to forum Submit Part ANOVA Table df MS F 1 MacBook Pro Part 3 of 4
Transcribed Image Text:▾ Part 1 of 4 John Paul 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. John Paul's student's exam scores are listed below. Use an ANOVA test with a level of significance of 5% to test whether the average score is the same for all five sections. Section 1: 62, 76, 70, 66, 94, 59, 76, 65, 72, 79, 52, 46, 74, 43, 68, 61, 75, 65, 63, 58, 62, 76, 70, 61, 76, 73, 53, 57, 73, 78, 51, 54, 42, 66, 58, 72, 44, 87 Section 2: 65, 52, 51, 67, 64, 75, 56, 68, 72, 82, 74, 73, 83, 42, 73, 85, 67, 46, 70, 51, 72, 53, 28, 58, 62, 68, 58, 85, 71, 57, 60, 79, 64, 60, 69, 65, 66 Section 3: 83, 67, 69, 72, 67, 92, 57, 57, 57, 61, 90, 75, 62, 69, 50, 57, 60, 94, 83, 69, 68, 65, 81, 60, 81, 67, 72, 78, 79,65 Section 4: 63, 75, 55, 70, 75, 81, 50, 77, 59, 42, 44, 85, 57, 62, 33, 45, 82, 67, 68, 47, 75, 69, 48, 96, 79, 77 Section 5: 30, 82, 83, 54, 54, 59, 75, 43, 68, 58, 58, 45, 72, 67, 58, 82, 50, 63, 55, 50, 84, 71, 68, 80, 85, 40, 68, 85, 73, 84, 47, 48, 57, 56 Step 1: State the null and alternative hypotheses. Ho: P.H. Ha: At least one mean isn't equal to the other means ✓✓ Y 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 dfbetween = 4 Step 3: Find the p-value of the test statistic. ₁ = #₂ = Es= Z₁ = #₁ = MI Between Within ✓✓distribution with numerator degrees of freedom ✓and denominator degrees of freedom dfwithin 160 SS P(FVV Question Help: D Post to forum Submit Part ANOVA Table df MS F 1 MacBook Pro Part 3 of 4
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