174 lines
5.3 KiB
Plaintext
174 lines
5.3 KiB
Plaintext
---
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title: "Case Study Sleep"
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author: "Jonathan Herrewijnen"
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date: "September 17, 2018"
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output: word_document
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---
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```{r setup, include=FALSE}
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knitr::opts_chunk$set(echo = TRUE)
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```
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## R Markdown
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This is an R Markdown document. Markdown is a simple formatting syntax for authoring HTML, PDF, and MS Word documents. For more details on using R Markdown see <http://rmarkdown.rstudio.com>.
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When you click the **Knit** button a document will be generated that includes both content as well as the output of any embedded R code chunks within the document. You can embed an R code chunk like this:
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```{r cars}
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summary(cars)
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```
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## Including Plots
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You can also embed plots, for example:
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```{r pressure, echo=FALSE}
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plot(pressure)
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#For knitting deactivated
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#install.packages("data")
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data(sleep)
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head(sleep)
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sleep
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```
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Note that the `echo = FALSE` parameter was added to the code chunk to prevent printing of the R code that generated the plot.
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We found two groups in our data, but we are not sure whether the second group is our control group or a second medicine. This is not clarified clearly in the data description. So we are assuming that the second group is the control group, but we are not certain.
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//OPDRACHTEN - EXERCISES
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Question 1)
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Describe the dataset by explaining the variables. If any what are the replicates and/or controls?
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```{r}
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summary(sleep)
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sleep
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?sleep
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```
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See the code above, the table "sleep" contains 3 variables, namely: extra, group, ID.
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-"extra" stands for the extra hours of sleep a student who took a certain drug
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-"group" stands for the drug a student was given, with 2 different types of drugs
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-"ID" stands for the patient, one ID per patient/student
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20 observations and 3 variables are contained within the table. Second group is the control group.
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Question 2)
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Describe the performed experiment/ analysis. What was of interest/ what was the research question?
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Students were given a drug, two seperate groups were created, in which the first group got a sleep prolongation drug, and second group was used as control group. The researchers were interested in the amount of prolonged sleep in hours.
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Using ?sleep.
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Question 3)
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Summarize the data statistically.
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See the summary and t.test below in the next code snippet box
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Question 4)
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Are there any clear outliers or missing values? If so, which ones?
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The boxplot is not showing any outliers, so no outliers are visible.
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```{r}
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require(stats)
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#QUESTION THREE
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summary(sleep)
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#Simple t.test
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with(sleep, t.test(extra[group == 1], extra[group == 2], paired = TRUE))
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?sleep
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?boxplot
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#QUESTION FOUR
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#Individual plots
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boxplot(with(sleep, extra[group==1]))
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boxplot(with(sleep, extra[group==2]), add=TRUE)
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boxplot(with(sleep, extra[group==1], extra[group==2]), paired=TRUE)
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#plot(with(sleep, extra[group==2]), add = TRUE)
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boxplot(sleep$extra ~ sleep$group, xlab="Groups", ylab="Sleep prolongation (h)")
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points(sleep$extra ~ sleep$group, pch = 1)
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```
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Question 5)
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Summarize the data graphically using only 1 figure. Choose a figure that summarizes the data adequately. Use the graphical package ggplot2. Explain your choice.
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We have two groups and only one variable. So a boxplot is a logical choice for comparing both datasets.
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```{r}
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library(ggplot2)
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ggplot(sleep, aes(x=group, y=extra)) + geom_boxplot()
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```
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Question 6)
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Explore the data, include check for distribution/ normality range of variables mention number and type of variables and observations
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```{r}
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#Checking for distribution using a QQplot
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par(mfrow=c(2, 1)) #plots 2 graphs
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with(sleep, hist(extra[group==1]))
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with(sleep, hist(extra[group==2]))
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```
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No distribution seems to be visible (if so, then a bell curve would be expected)
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Question 7)
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Calculate the mean sleep prolongation of the two drugs using the aggregate function on the sleep data frame.
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```{r}
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#Calculate mean sleep
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aggregate(formula = extra~group, data = sleep, FUN = mean)
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```
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See function above, results are 0.75 hours and 2.33 hours extra sleep for drug one and two respectively
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Question 8)
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Calculate the difference in sleep prolongation based on drug type for each patient.
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Store this new dataset in a new data frame including patient number. Do this by writing a function with as input argument the sleep data frame and as output the new data frame.
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```{r}
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my_dataframe <- data.frame(with(sleep, extra[group==1]-extra[group==2]))
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my_dataframe
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```
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See above.
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Question 9)
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Is the sleep prolongation difference between the drugs (more) normaly distributed?
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```{r}
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shapiro.test(sleep$extra)
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hist(sleep$extra)
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```
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So the p-value here is 0.3114, meaning that it differs signifcantly from a normal distribution. We can not assume that the data is normaly distributed.
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Question 10)
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Was there a significant difference in sleep prolongation between the drugs? Mention the effect size, is this of a relevant magnitude? Explain why.
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Also explain why you chose this specific test.
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```{r}
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ThisData <- data.frame(with(sleep, mean(extra[group==1])), with(sleep, mean(extra[group==2])))
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ThisData
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with(sleep, t.test(extra[group == 1], extra[group == 2], paired = TRUE))
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```
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The mean between both drugs is very different, 0.75 to 2.33, implying a difference.
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The t.test shows a difference of -1.58 between the test with a reliability of 0.002833.
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