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▾ Part hn Paul teaches five sections of Introductory Psychology. He wants to know whether or not the werage score on the second exam is the same for all five sections. John Paul's student's exam cores are listed below. Use an ANOVA test with a level of significance of 5% to test whether the werage score is the same for all five sections. ection 1: 2, 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, 5 3, 78, 51, 54, 42, 66, 58, 72, 44, 87 Section 2: 55, 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

▾ Part hn Paul teaches five sections of Introductory Psychology. He wants to know whether or not the werage score on the second exam is the same for all five sections. John Paul's student's exam cores are listed below. Use an ANOVA test with a level of significance of 5% to test whether the werage score is the same for all five sections. ection 1: 2, 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, 5 3, 78, 51, 54, 42, 66, 58, 72, 44, 87 Section 2: 55, 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

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.3: Measures Of Spread
Problem 11PPS
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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 ▼ Part 1 of 4 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: μ. = μ. = μ = μ = με 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. We will use a(n) F dfbetween= Question Help: Post to forum Submit Part distribution with numerator degrees of freedom and denominator degrees of freedom dfwithin = MacBook Pro
Transcribed Image Text: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 ▼ Part 1 of 4 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: μ. = μ. = μ = μ = με 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. We will use a(n) F dfbetween= Question Help: Post to forum Submit Part distribution with numerator degrees of freedom and denominator degrees of freedom dfwithin = MacBook Pro
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