To perform a Wilcoxon Signed Ranks Test using R, one can use wilcox.test() function. Data, R code and its output is given below. Data # Data in two numeric vectors # +++++++ # Weight of the mice before treatment before <-c(200.1, 190.9, 192.7, 213, 241.4, 196.9, 172.2, 185.5, 205.2, 193.7) * Weight of the mice after treatment after <-c(392.9, 393.2, 345.1, 393, 434, 427.9, 422, 383.9, 392.3, 352.2) # Create a data frame my_data <- data.frame( +++++++ group = rep(c ("before", "after"), each = 10), weight = c(before, after) R code res <- wilcox.test (before, after, paired = TRUE)

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To perform a Wilcoxon Signed Ranks Test using R, one can use wilcox.test () function. Data, R code and its output is
given below.
Data
# Data in two numeric vectors
# ++++++++++++++++++++++++++
# Weight of the mice before treatment
before <-c(200.1, 190.9, 192.7, 213, 241.4, 196.9, 172.2, 185.5, 205.2, 193.7)
# weight of the mice after treatment
after <-c(392.9, 393.2, 345.1, 393, 434, 427.9, 422, 383.9, 392.3, 352.2)
# create a data frame
my_data <- data.frame (
group = rep (c ("before", "after"), each =
weight = c (before, after)
10),
R code
res <- wilcox.test (before, after, paired = TRUE)
res
R output
wilcoxon signed rank test
data: before and after
v = 0, p-value = 0.001953
alternative hypothesis: true location shift is not equal to 0
Note: The dataset is about the weight of 10 mice before and after the treatment.
Which of the following is the correct interpretation of the data analysis using R?
I. We can conclude that the mean weight of the mice before treatment is significantly different from the mean weight after
treatment with a p-value = 0.001953.
II. We can conclude that the mean weight of the mice before treatment is the same with the mean weight after treatment with
a p-value = 0.001953.
III. We can conclude that the median weight of the mice before treatment is the same with the median weight after treatment
with a p-value = 0.001953.
IV. We can conclude that the median weight of the mice before treatment is significantly different from the median weight
after treatment with a p-value = 0.001953.
a. I only
ь. П оnly
с. Ш оnly
d. IV only
В
Transcribed Image Text:To perform a Wilcoxon Signed Ranks Test using R, one can use wilcox.test () function. Data, R code and its output is given below. Data # Data in two numeric vectors # ++++++++++++++++++++++++++ # Weight of the mice before treatment before <-c(200.1, 190.9, 192.7, 213, 241.4, 196.9, 172.2, 185.5, 205.2, 193.7) # weight of the mice after treatment after <-c(392.9, 393.2, 345.1, 393, 434, 427.9, 422, 383.9, 392.3, 352.2) # create a data frame my_data <- data.frame ( group = rep (c ("before", "after"), each = weight = c (before, after) 10), R code res <- wilcox.test (before, after, paired = TRUE) res R output wilcoxon signed rank test data: before and after v = 0, p-value = 0.001953 alternative hypothesis: true location shift is not equal to 0 Note: The dataset is about the weight of 10 mice before and after the treatment. Which of the following is the correct interpretation of the data analysis using R? I. We can conclude that the mean weight of the mice before treatment is significantly different from the mean weight after treatment with a p-value = 0.001953. II. We can conclude that the mean weight of the mice before treatment is the same with the mean weight after treatment with a p-value = 0.001953. III. We can conclude that the median weight of the mice before treatment is the same with the median weight after treatment with a p-value = 0.001953. IV. We can conclude that the median weight of the mice before treatment is significantly different from the median weight after treatment with a p-value = 0.001953. a. I only ь. П оnly с. Ш оnly d. IV only В
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