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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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 related 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 rats 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, benzaldehyde, andacetaldehyde. At the end of exposure, the guinea pigs were anesthetized, and allergic responses were assessed in bronchoalveolar 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
Answered 1 days AfterNov 10, 2021

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

Atreye answered on Nov 11 2021
126 Votes
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 <- ggplot(data, aes(x=TEMP, y=RESPONSE)) +
+ geom_boxplot()
> 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<- lm(RESPONSE ~ TEMP, data = data)
ggqqplot(residuals(model))
OUTPUT
CODE:
shapiro_test(residuals(model))
# A tibble: 1 x 3
variable statistic p.value

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 :H There is no significant difference in the first pain responses between the skin temperatures.
1 :H There is significant difference in the first pain responses between the skin temperatures.
Part 4:
ANOVA TEST:
CODE:
res.aov <- data %>% anova_test(RESPONSE ~ TEMP)

res.aov

OUTPUT:

ANOVA Table (type II tests)

Effect DFn DFd F p p<.05 ges
1 TEMP 2 15 6.198 0.011 * 0.452

INTERPETATION:
From the ANOVA results, it is observed that the p-value is less than 0.05 which implies that the null
hypothesis is rejected and it can be concluded that there is significant difference in the first pain
responses between the skin temperatures.
POST HOC TEST
As There is significant difference in the first pain responses between the skin temperatures, now
post hoc test will be conducted to know which group differs significantly.
Code:

> pwc <- data %>% tukey_hsd(RESPONSE ~ TEMP)
> pwc
OUTPUT:

# A tibble: 3 x 9
term group1 group2 null.value estimate conf.low conf.high p.adj p.adj
.signif
*
1 TEMP 25 30 0 -2.19 -4.54 0.155 0.0688 ns
2 TEMP 25 35 0 -3.09 -5.44 -0.745 0.00994 **
3 TEMP 30 35 0 -0.9 -3.25 1.45 0.59 ns
Interpretation:
From the p-values, it is observed that the pair of temperature 25 and 35 differs significantly as the p-
value 0.00994 is less than 0.05.
Solution 2:
Part 1:
There is one factor present in the dataset namely the treatment having two levels. The response
variable here is the serum concentration. We need to investigate the effect of this treatment on
serum concentration. One–way ANOVA will be appropriate here to use.
Part 2:
Check Assumptions:
a. Outliers detection:
Code:
ggboxplot(df, x = "TREAT", y = "CONCENTRATION")
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<- lm(CONCENTRATION ~ TREAT, data = df)
ggqqplot(residuals(model))
OUTPUT
CODE:
shapiro_test(residuals(model))
A tibble: 1 x 3
variable statistic p.value

1 residuals(model) 0.967 0.00293
Interpretation:
From the QQ plot it is observed that, as all the points fall approximately along the reference line, so...
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