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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

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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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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