▼ Part 1 of 4 A doctor in Oklahoma City 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 10% level of significance to test whether the average life span of heart disease patients in Oklahoma City differs depending on the hospital that treats them Life Span of Patients Treated at Hospital 1: 8.1, 1.2, 7, 11.5, 21, 2, 15.3, 19.2, 10.5, 15.4, 12.7, 15.3, 12.1, 7.5, 11.4, 16.4, 26.7, 14.6, 1.5, 16.3, 12.6, 0.8, 18.2, 4.2, 9.6, 21, 16.3, 2.7 Life Span of Patients Treated at Hospital 2: 17.6, 11.6, 16.8, 5.2, 6.7, 3.2, 13,8, 12.5, 0.9, 6, 14.3, 1.8, 14.8, 10, 5.8, 15.2, 10, 3.7, 11.4, 21.6, 12.9, 0.5, 9.2, 13.9, 10.9 Life Span of Patients Treated at Hospital 3: 8.5, 2.3, 11, 7.4, 14.8, 5.8, 11, 3.4, 10.9, 1.1, 5.9, 21.6, 12.5, 0.1, 15.5, 15.4, 0.8, 7.7, 5.6, 12, 7.1 Life Span of Patients Treated at Hospital 4: 29.7, 6.1, 8.5, 2.8, 9.9, 13.6, 3, 2.2, 4, 9.7, 18, 5.4, 10.4, 10.5, 0.6, 0.6, 1.6, 7.9, 4.8, 12.2, 2.9, 9.9, 8.2, 2.2, 4, 6.8, 14.8, 11.4, 5.8, 8.4, 11.4 Step 1: State the null and alternative hypotheses. Ho: BBP. Ha: At least one mean isn't equal to the other means V✔ 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 ain) F dfbetween 3 Step 3: Find the p-value of the test statistic. Z₁ = Z₂ = Z₁ == Z₁ = E Between Within ✓✓distribution with numerator degrees of freedom ✓and denominator degrees of freedom dfwithin 101 Submit Part SS P(FEV Question Help: D Post to forum ANOVA Table df MS F Part 3 of 4
▼ Part 1 of 4 A doctor in Oklahoma City 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 10% level of significance to test whether the average life span of heart disease patients in Oklahoma City differs depending on the hospital that treats them Life Span of Patients Treated at Hospital 1: 8.1, 1.2, 7, 11.5, 21, 2, 15.3, 19.2, 10.5, 15.4, 12.7, 15.3, 12.1, 7.5, 11.4, 16.4, 26.7, 14.6, 1.5, 16.3, 12.6, 0.8, 18.2, 4.2, 9.6, 21, 16.3, 2.7 Life Span of Patients Treated at Hospital 2: 17.6, 11.6, 16.8, 5.2, 6.7, 3.2, 13,8, 12.5, 0.9, 6, 14.3, 1.8, 14.8, 10, 5.8, 15.2, 10, 3.7, 11.4, 21.6, 12.9, 0.5, 9.2, 13.9, 10.9 Life Span of Patients Treated at Hospital 3: 8.5, 2.3, 11, 7.4, 14.8, 5.8, 11, 3.4, 10.9, 1.1, 5.9, 21.6, 12.5, 0.1, 15.5, 15.4, 0.8, 7.7, 5.6, 12, 7.1 Life Span of Patients Treated at Hospital 4: 29.7, 6.1, 8.5, 2.8, 9.9, 13.6, 3, 2.2, 4, 9.7, 18, 5.4, 10.4, 10.5, 0.6, 0.6, 1.6, 7.9, 4.8, 12.2, 2.9, 9.9, 8.2, 2.2, 4, 6.8, 14.8, 11.4, 5.8, 8.4, 11.4 Step 1: State the null and alternative hypotheses. Ho: BBP. Ha: At least one mean isn't equal to the other means V✔ 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 ain) F dfbetween 3 Step 3: Find the p-value of the test statistic. Z₁ = Z₂ = Z₁ == Z₁ = E Between Within ✓✓distribution with numerator degrees of freedom ✓and denominator degrees of freedom dfwithin 101 Submit Part SS P(FEV Question Help: D Post to forum ANOVA Table df MS F Part 3 of 4
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:Part 1 of 4
A doctor in Oklahoma City 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 10% level of significance to test whether
the average life span of heart disease patients in Oklahoma City differs depending on the hospital
that treats them
Life Span of Patients Treated at Hospital 1:
8.1, 1.2, 7, 11.5, 21, 2, 15.3, 19.2, 10.5, 15.4, 12.7, 15.3, 12.1, 7.5, 11.4, 16.4, 26.7, 14.6, 1.5, 16.3, 12.6,
0.8, 18.2, 4.2, 9.6, 21, 16.3, 2.7
Life Span of Patients Treated at Hospital 2:
17.6, 11.6, 16.8, 5.2, 6.7, 3.2, 13.8, 12.5, 0.9, 6, 14.3, 1.8, 14.8, 10, 5.8, 15.2, 10, 3.7, 11.4, 21.6, 12.9,
0.5, 9.2, 13.9, 10.9
Life Span of Patients Treated at Hospital 3:
8.5, 2.3, 11, 7.4, 14.8, 5.8, 11, 3.4, 10.9, 1.1, 5.9, 21.6, 12.5, 0.1, 15.5, 15.4, 0.8, 7.7, 5.6, 12, 7.1
Life Span of Patients Treated at Hospital 4:
29.7, 6.1, 8.5, 2.8, 9.9, 13.6, 3, 2.2, 4, 9.7, 18, 5.4, 10.4, 10.5, 0.6, 0.6, 1.6, 7.9, 4.8, 12.2, 2.9, 9.9, 8.2,
2.2, 4, 6.8, 14.8, 11.4, 5.8, 8.4, 11.4
Step 1: State the null and alternative hypotheses.
Ho: P. P.
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.
We will use a(n) F
dfbetween = 3
Z₁ =
Step 3: Find the p-value of the test statistic.
#₂ =
#3 =
Z₁ =
Between
Within
Submit Part
✔and denominator degrees of freedom dfwith
f freedom
SS
✓✓distribution with numerator degrees of freedom
dfwithin = 101
P(FEV
Question Help: D Post to forum
V✓
L
ANOVA Table
df
MS
F
MacBook Pro
Part 3 of 4
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