2. CREATININE LEVEL PROBLEM To compare the performance of the two machines a technician took eight specimens of blood and measured the creatinine level (in micromoles per litre) of each specimen using each machine. The results were as follows. Specim Analyzer Analyzer en A B 9 6 1 11 10 3 3 2 17 15 0 Ho (in words): Ha (in words): Test Procedure: p-value: Conclusion: Provide answers to the following: a. The samples above are 3 10 83 4 99 95 5 77 69 1 3 6 12 12 7 84 84 8 73 The data can also be found in the worksheet "creatinine_level" of excel file "exer3_data.xlsx". R outputs can be found in the file "exer3_outputs.pdf". Let difference be Analyzer A - Analyzer B. 67 (related/independent) samples. b. At 5% alpha, perform test(s) to determine which test procedure should be used to compare the two populations involved. For EACH test, indicate the following:
2. CREATININE LEVEL PROBLEM To compare the performance of the two machines a technician took eight specimens of blood and measured the creatinine level (in micromoles per litre) of each specimen using each machine. The results were as follows. Specim Analyzer Analyzer en A B 9 6 1 11 10 3 3 2 17 15 0 Ho (in words): Ha (in words): Test Procedure: p-value: Conclusion: Provide answers to the following: a. The samples above are 3 10 83 4 99 95 5 77 69 1 3 6 12 12 7 84 84 8 73 The data can also be found in the worksheet "creatinine_level" of excel file "exer3_data.xlsx". R outputs can be found in the file "exer3_outputs.pdf". Let difference be Analyzer A - Analyzer B. 67 (related/independent) samples. b. At 5% alpha, perform test(s) to determine which test procedure should be used to compare the two populations involved. For EACH test, indicate the following:
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
The first photo are the tests needed to answer letter b.
Transcribed Image Text:2. CREATININE LEVEL PROBLEM
To compare the performance of the two machines a technician took eight specimens of
blood and measured the creatinine level (in micromoles per litre) of each specimen
using each machine. The results were as follows.
en
A
B
Specim
Analyzer
Analyzer
9
6
1
11
10
3
3
2
17
Ho (in words):
Ha (in words):
Test Procedure:
p-value:
Conclusion:
15
0
Provide answers to the following:
a. The samples above are
3
10
83
4
99
95
5
77
69
1
3
12
12
7
84
84
8
73
The data can also be found in the worksheet "creatinine_level" of excel file
"exer3_data.xlsx". R outputs can be found in the file "exer3_outputs.pdf".
Let difference be Analyzer A - Analyzer B.
67
(related/independent) samples.
b. At 5% alpha, perform test(s) to determine which test procedure should be
used to compare the two populations involved. For EACH test, indicate the
following:
Transcribed Image Text:2. CREATININE LEVEL PROBLEM
Some statistics
Analyzer A: mean = 105.75; median = 99.5
Analyzer B: mean = 97.5; median = 89.5
Let difference = Analyzer A - Analyzer B
d=8.25
Md = 7
data: Analyzer A
W = 0.88028
p-value = 0.1895
Wilk-Shapiro Test for Normality
data: Analyzer B
W = 0.91541
p-value = 0.3937
Ha: not equal to 0
t = 2.9652
p-value = 0.02095
data: difference
W = 0.95788
p-value 0.7897
Student's t-test on Paired Samples: Analyzer A vs Analyzer B
Ha: less than 0
t = 2.9652
p-value = 0.9895
Ha: greater than 0
t = 2.9652
p-value = 0.01048
Wilcoxon Matched-Pairs Signed Ranks Test: Analyzer A vs Analyzer B
Ha: not equal to 0
Ha: less than 0
Ha: greater than 0
V = 27
V = 27
V = 27
p-value = 0.03461
p-value = 0.9888
p-value
0.01731
F-test for Equality of Variances
F = 1.1311
p-value = 0.856
Student's t-test on Two Independent Pop'n Means: Analyzer A vs Analyzer B
Ha: not equal to 0
Ha: less than 0.
Ha: greater than 0
t = 9.2109
t = 9.2109
t = 9.2109
p-value 0.00003668 p-value=1
p-value = 0.00001834
Welch's t-test on Two Independent Pop'n Means: Analyzer A vs Analyzer B
Ha: not equal to 0
Ha: less than 0
Ha: greater than 0
t = 9.4637
t = 9.4637
t = 9.4637
p-value = 1
p-value = 0.00003073
p-value = 0.00001537
Mann-Whitney Test on Two Independent Pop'n Means: Analyzer A vs Analyzer B
Ha: not equal to 0.
Ha: greater than 0
Ha: less than 0
W = 36
W = 36
W = 36
p-value = 0.01427
p-value = 0.9952
p-value
0.007133
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