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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 = H2 = P3 = Ha = Hs 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. v distribution with numerator degrees of freedom We will use a(n) F 159 dfbetween and denominator degrees of freedom dfwithin %3D 4 Part 3 of 4

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 = H2 = P3 = Ha = Hs 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. v distribution with numerator degrees of freedom We will use a(n) F 159 dfbetween and denominator degrees of freedom dfwithin %3D 4 Part 3 of 4

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter4: Equations Of Linear Functions
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test statistics. We will use a(n) F vv distribution with numerator degrees of freedom v and denominator degrees of freedom dfwithin = d fpetween 4 159 Part 3 of 4 Step 3: Find the p-value of the test statistic. I1 = 80.91176471/ I2 = 82 81 T4 = 82.76 83.82352941X T = 82.10494118 X ANOVA Table SS df MS F 49.7567796 × 2.986335399 X Between 199.0271184 X 4 153 16.66148404 X Within 2549.207059 X P(F2♥ 0.02079
Transcribed Image Text:test statistics. We will use a(n) F vv distribution with numerator degrees of freedom v and denominator degrees of freedom dfwithin = d fpetween 4 159 Part 3 of 4 Step 3: Find the p-value of the test statistic. I1 = 80.91176471/ I2 = 82 81 T4 = 82.76 83.82352941X T = 82.10494118 X ANOVA Table SS df MS F 49.7567796 × 2.986335399 X Between 199.0271184 X 4 153 16.66148404 X Within 2549.207059 X P(F2♥ 0.02079
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 = P4 = 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. vv distribution with numerator degrees of freedom 159 We will use a(n) F dfpetween = v and denominator degrees of freedom d fwithin = 4 Part 3 of 4 homualue of the test statistic.
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 = P3 = P4 = 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. vv distribution with numerator degrees of freedom 159 We will use a(n) F dfpetween = v and denominator degrees of freedom d fwithin = 4 Part 3 of 4 homualue of the test statistic.
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