2. CREATININE LEVEL PROBLEMTo compare the performance of the two machines a technician took eight specimens ofblood and measured the creatinine level (in micromoles per litre) of each specimenusing each machine. The results were as follows.SpecimAnalyzerAnalyzer 10enAB9611133نياev2Test Procedure:17150301083499p-value:Conclusion: At alpha= 0.05,955Justify your choice of test procedure:77691361212784848The 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.73d. A point estimate of the average difference in the creatinine level (inmicromoles per litre) of each specimen using each machine is67e. At 5% level of significance, test if there is difference in the measuredcreatinine level (in micromoles per litre) of each specimen using eachmachine.Ho: (in symbols): (in words) The average creatinine level (in micromolesper litre) of the specimen using Analyzer A isthe averagecreatinine level (in micromoles per litre) of the specimen using Analyzer A.Ha: (in symbols)___________ (in words) The average creatinine level (in micromolesper litre) of the specimen using Analyzer A isthe averagecreatinine level (in micromoles per litre) of the specimen using Analyzer B. 2. CREATININE LEVEL PROBLEMSome statisticsAnalyzer A: mean = 105.75; median = 99.5Analyzer B: mean = 97.5; median = 89.5Let difference = Analyzer A - Analyzer Bd=8.25Md = 7data: Analyzer AW = 0.88028p-value = 0.1895Wilk-Shapiro Test for Normalitydata: Analyzer BW = 0.91541p-value = 0.3937Ha: not equal to 0t = 2.9652p-value = 0.02095data: differenceW = 0.95788p-value 0.7897Student's t-test on Paired Samples: Analyzer A vs Analyzer BHa: less than 0t = 2.9652p-value = 0.9895Ha: greater than 0t = 2.9652p-value = 0.01048Wilcoxon Matched-Pairs Signed Ranks Test: Analyzer A vs Analyzer BHa: not equal to 0Ha: less than 0Ha: greater than 0V = 27V = 27V = 27p-value = 0.03461p-value = 0.9888p-value0.01731F-test for Equality of VariancesF = 1.1311p-value = 0.856Student's t-test on Two Independent Pop'n Means: Analyzer A vs Analyzer BHa: not equal to 0Ha: less than 0.Ha: greater than 0t = 9.2109t = 9.2109t = 9.2109p-value 0.00003668 p-value=1p-value = 0.00001834Welch's t-test on Two Independent Pop'n Means: Analyzer A vs Analyzer BHa: not equal to 0Ha: less than 0Ha: greater than 0t = 9.4637t = 9.4637t = 9.4637p-value = 1p-value = 0.00003073p-value = 0.00001537Mann-Whitney Test on Two Independent Pop'n Means: Analyzer A vs Analyzer BHa: not equal to 0.Ha: greater than 0Ha: less than 0W = 36W = 36W = 36p-value = 0.01427p-value = 0.9952p-value0.007133

Question

2. CREATININE LEVEL PROBLEMTo compare the performance of the two machines a technician took eight specimens ofblood and measured the creatinine level (in micromoles per litre) of each specimenusing each machine. The results were as follows.SpecimAnalyzerAnalyzer 10enAB9611133نياev2Test Procedure:17150301083499p-value:Conclusion: At alpha= 0.05,955Justify your choice of test procedure:77691361212784848The data can also be found in the worksheet &#34;creatinine_level&#34; of excel file&#34;exer3_data.xlsx&#34;. R outputs can be found in the file &#34;exer3_outputs.pdf&#34;.Let difference be Analyzer A-Analyzer B.73d. A point estimate of the average difference in the creatinine level (inmicromoles per litre) of each specimen using each machine is67e. At 5% level of significance, test if there is difference in the measuredcreatinine level (in micromoles per litre) of each specimen using eachmachine.Ho: (in symbols): (in words) The average creatinine level (in micromolesper litre) of the specimen using Analyzer A isthe averagecreatinine level (in micromoles per litre) of the specimen using Analyzer A.Ha: (in symbols)___________ (in words) The average creatinine level (in micromolesper litre) of the specimen using Analyzer A isthe averagecreatinine level (in micromoles per litre) of the specimen using Analyzer B. 2. CREATININE LEVEL PROBLEMSome statisticsAnalyzer A: mean = 105.75; median = 99.5Analyzer B: mean = 97.5; median = 89.5Let difference = Analyzer A - Analyzer Bd=8.25Md = 7data: Analyzer AW = 0.88028p-value = 0.1895Wilk-Shapiro Test for Normalitydata: Analyzer BW = 0.91541p-value = 0.3937Ha: not equal to 0t = 2.9652p-value = 0.02095data: differenceW = 0.95788p-value 0.7897Student's t-test on Paired Samples: Analyzer A vs Analyzer BHa: less than 0t = 2.9652p-value = 0.9895Ha: greater than 0t = 2.9652p-value = 0.01048Wilcoxon Matched-Pairs Signed Ranks Test: Analyzer A vs Analyzer BHa: not equal to 0Ha: less than 0Ha: greater than 0V = 27V = 27V = 27p-value = 0.03461p-value = 0.9888p-value0.01731F-test for Equality of VariancesF = 1.1311p-value = 0.856Student's t-test on Two Independent Pop'n Means: Analyzer A vs Analyzer BHa: not equal to 0Ha: less than 0.Ha: greater than 0t = 9.2109t = 9.2109t = 9.2109p-value 0.00003668 p-value=1p-value = 0.00001834Welch's t-test on Two Independent Pop'n Means: Analyzer A vs Analyzer BHa: not equal to 0Ha: less than 0Ha: greater than 0t = 9.4637t = 9.4637t = 9.4637p-value = 1p-value = 0.00003073p-value = 0.00001537Mann-Whitney Test on Two Independent Pop'n Means: Analyzer A vs Analyzer BHa: not equal to 0.Ha: greater than 0Ha: less than 0W = 36W = 36W = 36p-value = 0.01427p-value = 0.9952p-value0.007133

Accepted Answer

The problem is to test if there is a difference in the measured creatinine level b/w the two…

d. A point estimate of the average difference in the creatinine level (in micromoles per litre) of each specimen using each machine is e. At 5% level of significance, test if there is difference in the measured creatinine level (in micromoles per litre) of each specimen using each machine.