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1. (50 points) Cholesterol was measured in the serum samples of five randomly selected patients from a pool of patients. Two independently prepared repli- cate tubes were prepared for each patient for...

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1. (50 points) Cholesterol was measured in the serum samples of five randomly selected patients from a pool of patients. Two independently prepared repli- cate tubes were prepared for each patient for each of spectrophotometer from four brands. The objective of the study was to determine whether the relative cholesterol measurements for patients were consistent for four brands. The data mg/dl of cholesterol in the replicate samples from each patient on each brand.

Patient?Brand XXXXXXXXXX

XXXXXXXXXX XXXXXXXXXX XXXXXXXXXX148.5?

XXXXXXXXXX XXXXXXXXXX XXXXXXXXXX154.7?

XXXXXXXXXX XXXXXXXXXX XXXXXXXXXX145.9?

XXXXXXXXXX XXXXXXXXXX XXXXXXXXXX151.0?

(1) From the statement above identify the following experimental design com- ponents:

? Experimental unit?

? Treatments?

? Fixed or random levels of the treatments

. (2) Write the model to study cholesterol measurements on patients and brands. You need to fully specify model components and the distributions of ran- dom terms.?

. (3) Find out the estimates of the parameters in the model specified in part (2).?

. (4) Test if the main effects and interaction effect in the model are significantly different from zero at level a = 0.05.?? Specify hypotheses H0and H1for each effect.?? Test the hypotheses (reject or accept H0) and present the p-values for?each effect.?

. (5) Examine the following assumptions of the error term in the model.?? Independence: Show the test statistic and the decision?? Constant variance: Check with residual plots?? Normality: Test statistics and test under 5% significance level?

2. (50 points) A research specialist for a large seafood company investigated bac- terial growth on oysters and mussels subjected to three different storage tem- peratures. Nine cold storage units were available. Oysters and mussels were stored for two weeks in each of the cold storage units. A bacterial count was made from a sample of oysters and mussels at the end of two weeks. Hence, with 3 replicates, the first fixed treatment applied to a batch of 2 storage units and then the batch has been split into 2 storage units to apply the second fixed

2

treatment. The logarithm of bacterial count for each sample is shown in the following table

Temperature (?C) 0

5

10

Seafood Oyster Mussel XXXXXXXXXX XXXXXXXXXX.5780

XXXXXXXXXX XXXXXXXXXX3861

XXXXXXXXXX XXXXXXXXXX11.0329

. (1) From the statement above identify the following experimental design com- ponents:?? Experimental units? Treatments?

. (2) Write the model to study the bacterial counts on temperature and seafood with full specification of model components.?

. (3) Construct the analysis of variance (ANOVA) table with a column of the expected mean square (EMS). Use the auxiliary table to find the EMS for full credit.?

3

. (4) Test the significances of treatment effects in the model at the level a = 0.01.?? Specify hypotheses H0and H1for each effect.?? Test the hypotheses (reject or accept H0) and present the p-values for?each effect.?

. (5) Theresearchistoinvestigatebacterialgrowthonoystersandmusselsunder different storage temperatures. Briefly write a conclusion with respect to the research object.?

Answered Same DayDec 26, 2021

Solution

David answered on Dec 26 2021
77 Votes
QUESTION 1
1. Identifying the experimental design components.
Experimental unit = Number of blocks = 5
Treatments = Different
ands being used = 4
Levels of treatment = 2
2. Write the Model of study.
Yijk = µ + αk + βi + rj + ϵijk
The distribution of the random terms are as follows
Yijk = t-distribution
µ = Normal distribution
αk , βi and rj = t-distribution
ϵijk = Normal distribution
3. Estimates for the parameters
µ =Overall mean = ̅
αk = ̅̅ ̅̅ ̅ - ̅
βi = ̅̅ ̅̅ - ̅
j = ̅̅̅̅ - ̅
ϵijk = Random term
Anova: Two-Factor With Replication
SUMMARY 1 2 3 4 5 Total
Brand 1
Count 2 2 2 2 2 10
Sum 334 370.9 207.9 429.8 297 1639.6
Average 167 185.45 103.95 214.9 148.5 163.96
Variance 0.18 3.125 31.205 0.32 0 1538.572
Brand 2
Count 2 2 2 2 2 10
Sum 354.9 392.7 226 449.3 313.3 1736.2
Average 177.45 196.35 113 224.65 156.65 173.62
Variance 9.245 13.005 3.92 36.125 7.605 1585.133
Brand 3
Count 2 2 2 2 2 10
Sum 335.3 357 210 415.3 290.6 1608.2
Average 167.65 178.5 105 207.65 145.3 160.82
Variance 6.125 1.62 1.62 0.605 0.72 1314.304
Brand 4
Count 2 2 2 2 2 10
Sum 354.8 382.8 228 440.7 307.1 1713.4
Average 177.4 191.4 114 220.35 153.55 171.34
Variance 0.18 2 0.72 0.845 13.005 1434.187
Total
Count 8 8 8 8 8
Sum 1379 1503.4 871.9 1735.1 1208
Average 172.375 187.925 108.9875 216.8875 151
Variance 31.45357 53.68214 28.92411 51.56411 25.09429
ANOVA
Source of VariationSS df MS F P-value F crit
Sample 1093.371 3 364.457 55.14973 7.48E-10 3.098391
Columns 52608.1 4 13152.03 1990.168 1.07E-25 2.866081
Interaction 109.4865 12 9.123875 1.380627 0.252997 2.277581
Within 132.17 20 6.6085
Total 53943.13 39
4. Testing significance of the main effect and the interaction effect
Specifying Hypothesis for each interaction effect
H0 : β1 = β2 = β3 = β4
Ha : β1 ≠ βi
H0 : β2 = β1 = β3 = β4
Ha : β2 ≠ βi
H0 : β3 = β1 = β2 = β4
Ha : β3 ≠ βi
H0 : β4 = β1 = β2 = β3
Ha : β4 ≠ βi
To test this hypothesis we look at the ANOVA table the effect on interaction
We...
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