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A doctor in Cleveland wants to know whether the average life span for heart disease patients at four hospitals in the city differ. The data below represents the life span, in years, of heart disease patients from each hospital. Perform an ANOVA test with a 9% level of significance to test whether the average life span of heart disease patients in Cleveland differs depending on the hospital that treats them Life Span of Patients Treated at Hospital 1: 7.4, 7.8, 7.7, 7.5, 8, 8.2, 7.8, 8.6, 8, 7.8, 8.3, 8.3, 8, 7.6, 8.2, 7.9, 7.3, 8, 8.6, 7.3, 8.3, 8, 7.8, 8, 7.8, 8.1, 8.1, 8, 7.6, 7.6, 7.7, 7.4, 7.7, 7.8, 7.8 Life Span of Patients Treated at Hospital 2: 7.9, 7.9, 8.2, 8, 8.1, 8.5, 8.3, 8.4, 8, 8.2, 7.7, 8, 8, 7.8, 7.9, 8.1, 8.1, 7.8, 7.9, 8, 8.5, 8.3, 8.2, 8.3, 7.8, 7.9 Life Span of Patients Treated at Hospital 3: 8.2, 8.1, 7.4, 8.7, 8.6, 8.2, 7.9, 8.1, 8.1, 8.3, 8.3, 8, 7.6, 8, 7.4, 8.6, 8.2, 8.2, 7.9, 7.7, 8.1, 7.9, 8, 8.3 Life Span of Patients Treated at Hospital 4: 8.1, 8.2, 8.1, 8.3, 7.9, 8.6, 7.2, 7.5, 7.9, 7.8, 8.3, 8.5, 7.8, 7.4, 8.1, 7.9, 7.6, 7.6, 7.9, 8.4, 8.6, 7, 7.7 Step 1: State the null and alternative hypotheses. Ho : u, = U_ = k, =U, 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 =3 distribution with numerator degrees of freedom ✓and denominator degrees of freedom dfwithin = 103 Step 3: Find the p-value of the test statistic. #₁ = X Part 3 of 4

A doctor in Cleveland wants to know whether the average life span for heart disease patients at four hospitals in the city differ. The data below represents the life span, in years, of heart disease patients from each hospital. Perform an ANOVA test with a 9% level of significance to test whether the average life span of heart disease patients in Cleveland differs depending on the hospital that treats them Life Span of Patients Treated at Hospital 1: 7.4, 7.8, 7.7, 7.5, 8, 8.2, 7.8, 8.6, 8, 7.8, 8.3, 8.3, 8, 7.6, 8.2, 7.9, 7.3, 8, 8.6, 7.3, 8.3, 8, 7.8, 8, 7.8, 8.1, 8.1, 8, 7.6, 7.6, 7.7, 7.4, 7.7, 7.8, 7.8 Life Span of Patients Treated at Hospital 2: 7.9, 7.9, 8.2, 8, 8.1, 8.5, 8.3, 8.4, 8, 8.2, 7.7, 8, 8, 7.8, 7.9, 8.1, 8.1, 7.8, 7.9, 8, 8.5, 8.3, 8.2, 8.3, 7.8, 7.9 Life Span of Patients Treated at Hospital 3: 8.2, 8.1, 7.4, 8.7, 8.6, 8.2, 7.9, 8.1, 8.1, 8.3, 8.3, 8, 7.6, 8, 7.4, 8.6, 8.2, 8.2, 7.9, 7.7, 8.1, 7.9, 8, 8.3 Life Span of Patients Treated at Hospital 4: 8.1, 8.2, 8.1, 8.3, 7.9, 8.6, 7.2, 7.5, 7.9, 7.8, 8.3, 8.5, 7.8, 7.4, 8.1, 7.9, 7.6, 7.6, 7.9, 8.4, 8.6, 7, 7.7 Step 1: State the null and alternative hypotheses. Ho : u, = U_ = k, =U, 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 =3 distribution with numerator degrees of freedom ✓and denominator degrees of freedom dfwithin = 103 Step 3: Find the p-value of the test statistic. #₁ = X 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
Section4.5: Correlation And Causation
Problem 2AGP
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Ost watched Ani... Question 2 Y Part 1 of 4 A doctor in Cleveland wants to know whether the average life span for heart disease patients at four hospitals in the city differ. The data below represents the life span, in years, of heart disease patients from each hospital. Perform an ANOVA test with a 9% level of significance to test whether the average life span of heart disease patients in Cleveland differs depending on the hospital that treats them Life Span of Patients Treated at Hospital 1: 7.4, 7.8, 7.7, 7.5, 8, 8.2, 7.8, 8.6, 8, 7.8, 8.3, 8.3, 8, 7.6, 8.2, 7.9, 7.3, 8, 8.6, 7.3, 8.3, 8, 7.8, 8, 7.8, 8.1, 8.1, 8, 7.6, 7.6, 7.7, 7.4, 7.7, 7.8, 7.8 Life Span of Patients Treated at Hospital 2: 7.9, 7.9, 8.2, 8, 8.1, 8.5, 8.3, 8.4, 8, 8.2, 7.7, 8, 8, 7.8, 7.9, 8.1, 8.1, 7.8, 7.9, 8, 8.5, 8.3, 8.2, 8.3, 7.8, 7.9 Life Span of Patients Treated at Hospital 3: 8.2, 8.1, 7.4, 8.7, 8.6, 8.2, 7.9, 8.1, 8.1, 8.3, 8.3, 8, 7.6, 8, 7.4, 8.6, 8.2, 8.2, 7.9, 7.7, 8.1, 7.9, 8, 8.3 Life Span of Patients Treated at Hospital 4: 8.1, 8.2, 8.1, 8.3, 7.9, 8.6, 7.2, 7.5, 7.9, 7.8, 8.3, 8.5, 7.8, 7.4, 8.1, 7.9, 7.6, 7.6, 7.9, 8.4, 8.6, 7,7.7 Step 1: State the null and alternative hypotheses. Ho: u = H, = H, =, ✓✓ Ha: At least one mean isn't equal to the other means ✔ We will use a(n) F dfbetween 3 = Y Part 2 of 4 Step 2: Assuming the null hypothesis is true, determine the features of the distribution of test statistics. #₁ = < ₂ = Step 3: Find the p-value of the test statistic. I3 = > Z4= Between Within P(FZV✔ Submit Part distribution with numerator degrees of freedom ✓and denominator degrees of freedom dfwithin = 103 SS D- ANOVA Table df MS F X Part 3 of 4 5/2
Transcribed Image Text:Ost watched Ani... Question 2 Y Part 1 of 4 A doctor in Cleveland wants to know whether the average life span for heart disease patients at four hospitals in the city differ. The data below represents the life span, in years, of heart disease patients from each hospital. Perform an ANOVA test with a 9% level of significance to test whether the average life span of heart disease patients in Cleveland differs depending on the hospital that treats them Life Span of Patients Treated at Hospital 1: 7.4, 7.8, 7.7, 7.5, 8, 8.2, 7.8, 8.6, 8, 7.8, 8.3, 8.3, 8, 7.6, 8.2, 7.9, 7.3, 8, 8.6, 7.3, 8.3, 8, 7.8, 8, 7.8, 8.1, 8.1, 8, 7.6, 7.6, 7.7, 7.4, 7.7, 7.8, 7.8 Life Span of Patients Treated at Hospital 2: 7.9, 7.9, 8.2, 8, 8.1, 8.5, 8.3, 8.4, 8, 8.2, 7.7, 8, 8, 7.8, 7.9, 8.1, 8.1, 7.8, 7.9, 8, 8.5, 8.3, 8.2, 8.3, 7.8, 7.9 Life Span of Patients Treated at Hospital 3: 8.2, 8.1, 7.4, 8.7, 8.6, 8.2, 7.9, 8.1, 8.1, 8.3, 8.3, 8, 7.6, 8, 7.4, 8.6, 8.2, 8.2, 7.9, 7.7, 8.1, 7.9, 8, 8.3 Life Span of Patients Treated at Hospital 4: 8.1, 8.2, 8.1, 8.3, 7.9, 8.6, 7.2, 7.5, 7.9, 7.8, 8.3, 8.5, 7.8, 7.4, 8.1, 7.9, 7.6, 7.6, 7.9, 8.4, 8.6, 7,7.7 Step 1: State the null and alternative hypotheses. Ho: u = H, = H, =, ✓✓ Ha: At least one mean isn't equal to the other means ✔ We will use a(n) F dfbetween 3 = Y Part 2 of 4 Step 2: Assuming the null hypothesis is true, determine the features of the distribution of test statistics. #₁ = < ₂ = Step 3: Find the p-value of the test statistic. I3 = > Z4= Between Within P(FZV✔ Submit Part distribution with numerator degrees of freedom ✓and denominator degrees of freedom dfwithin = 103 SS D- ANOVA Table df MS F X Part 3 of 4 5/2
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