BME 530 Assignment 5 Use ? = 0.05 for all hypotheses. For each question, please also answer the...

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BME 530 Assignment 5

Use ? = 0.05 for all hypotheses. For each question, please also answer the following
questions.
1) What is the experimental design used?
2) Describe the assumptions needed for the ANOVA model.
3) Write the hypothesis or hypotheses to be tested.
4) When the null hypothesis or hypotheses are rejected, please perform the post-analysis
to identify which alternative hypothesis is the cause of rejection.
Q1 (5pts). A research group compared the effect of skin temperature on the critical threshold
temperature eliciting heat pain with the effect of skin temperature on the response latency to
the first heat pain sensation. Subjects were healthy adults between the ages of 23 and 54
years. Among the data collected were the latencies (seconds) to the first pain response
induced by radiant heat stimulation at three different skin temperatures (see dataset Q1). Test
the null hypothesis that the skin temperature has no different effect on the latencies to
the first pain response induced by radiant heat stimulation.

Q2 (5pts). A research team compared the effects of short-term treatments with growth
hormone (GH) and insulin-like growth factor I (IGF-I) on biochemical markers of bone
metabolism in men with
idiopathic osteoporosis. Subjects ranged in age from 32 to 57 years. Among the data collected
were the serum concentrations of IGF binding protein-3 at 0 and 7 days after first injection and
1,4, 8, and 12 weeks after last injection with GH and IGF-I (see data set Q2). Test the null
hypothesis that there is no difference between the two treatments on serum
concentrations of IGF binding protein-3.

Q3 (5pts). Research indicates that dietary copper deficiency reduces growth rate in rats. In a
elated study, the researcher assigned weanling male Sprague-Dawley rats to one of three
food groups: copper-deficient (CmuD), copper-adequate (CmuA), and pair-fed (PF). Rats in
the PF group were initially weight matched to rats of the CmuD group and then fed the same
weight of the CmuA diet as that consumed by their CmuD counterparts. After 20 weeks, the
ats were anesthetized, blood samples were drawn, and organs were harvested. As part of the
study the following data were collected. See dataset Q3. The meaning of each column is as
follows.
Rat ID
Diet: 1: CmuD, 2: PF and 3: CmuA
BW: Body weight (g)
HW: Heart weight (g)
LW: Liver weight (g)
KW: Kidney weight (g)
SW: Spleen weight (g)
CERUL: Ceruloplasmin (mg/dl)

Test the null hypothesis that the food groups do not have influence for each measurement.

Q4 (5pts). In a study of pulmonary effects on guinea pigs, the researcher exposed 18
ovalbumin-sensitized guinea pigs and 18 non-sensitized guinea pigs to regular air,
enzaldehyde, andacetaldehyde. At the end of exposure, the guinea pigs were anesthetized,
and allergic responses were assessed in
onchoalveolar lavage (BAL). The dataset Q4
shows the alveolar cell count (x106) by treatment group for the ovalbumin-sensitized and non-
sensitized guinea pigs.

Test for differences (a) between ovalbumin-sensitized and non-sensitized outcomes, (b)
among the three different exposures, and (c) interaction

Submission
Submit a zip file (HW5_name.zip) on Blackboard that includes:
1) both R notebook (.Rmd) and rendered results (.html). Your notebook should include some
annotations for the scripts so that a grader knows what the script is for.
2) report document (.pdf)

RESPONSE,SUBJ,TEMP
6.4,1,25
8.1,2,25
9.4,3,25
6.75,4,25
10,5,25
4.5,6,25
4.5,1,30
5.7,2,30
6.8,3,30
4.6,4,30
6.2,5,30
4.2,6,30
3.6,1,35
6.3,2,35
3.2,3,35
3.9,4,35
6.2,5,35
3.4,6,35

PATIENT,TREAT,DAY0,DAY7,WEEK1,WEEK4,WEEK8,WEEK12
1,1,4507,4072,3036,2484,3540,3480
2,1,2055,4095,2315,1840,2483,2354
3,1,3178,3574,3196,2365,4136,3088
4,1,3464,5874,2929,3903,3367,2938
5,1,4142,4465,3967,4213,4321,4990
6,1,3622,6800,6185,4247,4450,4199
7,1,5390,5188,4788,4602,4926,5793
8,1,3161,4942,3222,2699,3514,2963
9,1,3228,5995,3315,2919,3235,4379
10,1,5628,6152,4415,5251,3334,3910
11,1,2304,4721,3700,3228,2440,2698
1,2,3480,3515,4003,3667,4263,4797
2,2,2354,3570,3630,3666,2700,2782
3,2,3088,3405,3309,3444,2357,3831
4,2,2905,2888,2797,3083,3376,3464
5,2,4990,4590,2989,4081,4806,4435
6,2,3504,3529,4093,4114,4445,3622
7,2,5130,4784,4093,4852,4943,5390
8,2,3074,2691,2614,3003,3145,3161
9,2,4379,3548,3339,2379,2783,3000
10,2,5838,5025,4137,5777,5659,5628
11,2,2698,2621,3072,2383,3075,2822

RAT,DIET,BW,HW,LW,KW,SW,CERUL
1,1,253.66,0.89,2.82,1.49,0.41,
2,1,400.93,1.41,3.98,2.15,0.76,5.27
3,1,355.89,1.24,5.15,2.27,0.69,4.8
4,1,404.7,2.18,4.77,2.99,0.76,4.97
6,2,397.28,0.99,2.34,1.84,0.5,35.3
7,2,421.88,1.2,3.26,2.32,0.79,39
8,2,386.87,0.88,3.05,1.86,0.84,28
9,2,401.74,1.02,2.8,2.06,0.76,34.2
10,2,437.56,1.22,3.94,2.25,0.75,45.2
11,3,490.56,1.21,4.51,2.3,0.78,34.6
12,3,528.51,1.34,4.38,2.75,0.76,39
13,3,485.51,1.36,4.4,2.46,0.82,37.1
14,3,509.5,1.27,4.67,2.5,0.79,33.4
15,3,489.62,1.31,5.83,2.74,0.81,37.3

Sens,Treat,Count
No,Act,49.9
No,Act,50.6
No,Act,50.35
No,Act,44.1
No,Act,36.3
No,Act,39.15
No,Air,24.15
No,Air,24.6
No,Air,22.55
No,Air,25.1
No,Air,22.65
No,Air,26.85
No,Benz,31.1
No,Benz,18.3
No,Benz,19.35
No,Benz,15.4
No,Benz,27.1
No,Benz,21.9
Yes,Act,90.3
Yes,Act,72.95
Yes,Act,138.6
Yes,Act,80.05
Yes,Act,69.25
Yes,Act,31.7
Yes,Air,40.2
Yes,Air,63.2
Yes,Air,59.1
Yes,Air,79.6
Yes,Air,102.45
Yes,Air,64.6
Yes,Benz,22.15
Yes,Benz,22.75
Yes,Benz,22.15
Yes,Benz,37.85
Yes,Benz,19.35
Yes,Benz,66.7

bme-530-assignment5-3qcs4d1b.pdf q1-qkfn3ssp.csv q2-xxayuqio.csv q3-wboq5ftc.csv q4-fw3li2cc.csv

Atreye · Accepted Answer

Solution 1: 
Part 1: 
There is one factor present in the dataset namely temperature having three levels. The response 
variable here is the latencies on the first pain response. We need to investigate the effect of the 
temperature factor on first pain response. One–way ANOVA will be appropriate here to use. 
Part 2: 
Check Assumptions: 
a. Outliers detection: 
Code: 
p  p 
Output:
Interpretation: 
There are no outliers present in the dataset. 
b. Normality assumption  
It is done by analyzing the model residuals. QQ plot and Shapiro-Wilk test of normality are used. QQ 
plot draws the correlation between a given data and the normal distribution. 
Code: 
model                    
1 residuals(model)     0.925   0.159
Interpretation: 
From the QQ plot it is observed that, as all the points fall approximately along the reference line, so 
normality assumption is not violated. This conclusion is also supported by the Shapiro-Wilk test. The 
p-value is not significant (p = 0.159), so we can assume normality. 
Normality test by groups: 
CODE 
data %>%group_by(TEMP) %>%shapiro_test(RESPONSE) 
 
Output: 
 
A tibble: 3 x 4 
  TEMP  variable statistic      p 
              
1 25    RESPONSE     0.964 0.848  
2 30    RESPONSE     0.908 0.426  
3 35    RESPONSE     0.774 0.0337 

Interpretation: 
After computing Shapiro test for each group it is observed that the responses were normally 
distributed (p > 0.05) for each group, as assessed by Shapiro-Wilk’s test of normality. 
c. Homogeneity of variance assumption: 
Code: 
plot(model, 1)
Output:
Interpretation: 
From the plot it is observed that, there is no evident relationships between residuals and fitted 
values (the mean of each groups), which implies that the homogeneity of variances assumption is 
valid. 
Part 3: 
The appropriate null and alternative hypothesis is stated as below: 
0 :

BME 530 Assignment 5 Use ? = 0.05 for all hypotheses. For each question, please also answer the following questions. 1) What is the experimental design used? 2) Describe the assumptions needed for the...

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