imported>Weigang |
imported>Ntino |
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| <center>'''Analysis of Biological Data''' (BIOL 20N02, Spring 2017)</center>
| | #REDIRECT [[Biol425 2017]] |
| <center>'''Instructor:''' Dr Weigang Qiu, Associate Professor, Department of Biological Sciences </center>
| |
| <center>'''Room:''' 1001B HN (North Building, 10th Floor, Mac Computer Lab)</center>
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| <center>'''Hours:''' Thursdays 11:10-1:40</center>
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| <center>'''Office Hours:''' Belfer Research Building ([https://www.google.com/maps/place/413+E+69th+St,+New+York,+NY+10021/@40.7655886,-73.9561743,17z/data=!3m1!4b1!4m2!3m1!1s0x89c258c3d235f76f:0x4f3d0d5d8a78fe6?hl=en Google Map]) BB-402; Tuesdays 5-7 pm or by appointment</center>
| |
| <center>'''Course Website:''' http://diverge.hunter.cuny.edu/labwiki/Biol20N2_2017</center>
| |
| ----
| |
| ==Course Description==
| |
| With rapid accumulation of genome sequences and digital health data, biomedicine is becoming an information science. This course is a hands-on, computer-based workshop on how to visualize and analyze biological data. The course introduces R, a modern statistical computing language and platform. In the first half, students will learn to use R to make scatter plots, bar plots, box plots, and other commonly used data-visualization techniques. In the second half, the course will review & apply statistical hypothesis tests including significance testing of means, association tests, and correlation analysis. Throughout the course, students will apply these methods to the analysis of large biological data sets, such as the human genome, transcriptomes (RNA-SEQ), and human genome variations.
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| | |
| This 3-credit experimental course fulfills elective requirements for Biology Major I. Hunter pre-requisites are BIOL100, BIOL102 and STAT113.
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| | |
| ==Learning Goals==
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| * Be able to use R as a plotting tool to visualize large-scale biological data sets
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| * Be able to use R as a statistical tool to summarize data and make biological inferences
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| * Be able to use R as a programming language to automate data analysis
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| | |
| ==Textbooks==
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| * Whitlock & Schluter (2015). Analysis of Biological Data. (2nd edition). [http://www.amazon.com/Analysis-Biological-Data-Michael-Whitlock/dp/1936221489/ref=sr_1_1?ie=UTF8&qid=1455723611&sr=8-1&keywords=whitlock+and+schluter%27s+analysis+of+biological+data Amazon link]
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| | |
| ==Exams & Grading==
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| * Attendance (or a note in case of absence) is required
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| * In-Class Exercises (50 pts).
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| * Assignments. All assignments should be handed in as hard copies only. Email submission will not be accepted. Late submissions will receive 10% deduction (of the total grade) per day (~100 pts total).
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| * Three Mid-term Exams (3 X 30 pts each = 90 pts)
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| * Comprehensive Final Exam (50 pts)
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| * Bonus for active participation in classroom discussions
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| | |
| ==Course Outline==
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| ===Feb 2. Introduction & R Demo===
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| # Course overview
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| # Tutorial 1: R Demo
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| * Create a new project by navigating: File | New Project | New Directory. Name it project file "human_genes"
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| * Import the human genes data set: File | Import DataSet | CSV, copy & paste this address: http://diverge.hunter.cuny.edu/~weigang/data-sets-for-biostat/hg.tsv2
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| * Click "update". Rename the data set if you wish (short but informative names, e.g., hg, or human.genes). Do not use spaces, use dot or underscore as name delimiters (e.g., "human.genes" or "human_genes", but never "human genes") Same rule for column or row names
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| <syntaxhighlight lang="bash">
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| dim(hg) # show dimension
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| head(hg) # show top rows
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| tail(hg) # show bottom rows
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| hg.len <- hg$Gene.End - hg$Gene.Start + 1 # create a vector of gene lengths (Tab for auto-completion)
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| hg.len[1] # show first length
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| hg.len[10] # show the 10th item
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| hg.len[1:10] # show items 1 through 10
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| hg.len[c(1,10)] # show items 1 and 10
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| hg[1,1] # show item in 1st row, 1 column
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| hg[1,] # show all values for 1st row
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| hg[,1] # show all values for 1st column
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| hg[1:10,] # show rows 1 through 10
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| hg[,1:7] # show columns 1 through 10
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| hg[1:10, 1:7] # a subset
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| hist(hg$Length, br = 200) # plot gene-length distribution. Not normal: mostly genes are short, few very long
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| hist(log10(hg$Length), br = 200) # log transformation for non-normally distributed variable
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| gene.cts <- table(hg$Chromosome) # count number of genes, separated by chromosomes
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| barplot(gene.cts) # show distribution by a barplot
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| mean(hg$Length) # not representative, super-long genes carry too much weight to the average length
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| median(hg$Length) # More representative. Use median for a variable not normally distributed
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| summary(hg$Length) # Show all quartiles
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| hg$Gene.Length <- hg$Gene.End - hg$Gene.Start + 1 # create a new column named "Gene.Length"
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| boxplot(log10(Length) ~ Chromosome, data = hg) # show gene length by chromosomes
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| write.csv(hg, "hg.csv", row.names = FALSE) # save into a file
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| hg <- read.csv("hg.csv") # read back into R
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| </syntaxhighlight>
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| * Export a PDF or image
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| * Open a new R script, name it as "hg.R"
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| * Select commands and save to script
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| * Retrieve and edit a command by pressing "up" or "down" arrows
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| * Retrieve commands by using the search box on the "History" table
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| * Type q() to quit. Answer "y" to save workspace
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| * To reload and restore workspace, go to "C:/Users/instructor/Documents/human.genes" and double click on the file "human.gene"
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| {| class="wikitable sortable mw-collapsible"
| |
| |- style="background-color:lightsteelblue;"
| |
| ! Assignment #1. Due 2/9, Thursday (Finalized)
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| |- style="background-color:powderblue;"
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| |
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| # (2 pts) Install R & R Studio on your own computer:
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| ## First, download & install R from this website: https://mirrors.nics.utk.edu/cran/
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| ## Next, download & install R studio from this website: https://www.rstudio.com/
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| # (4 pts) Reproduce the "human.gene" project (follow steps in Demo). Save and print the histogram for gene length & the barplot for number of genes on each chromosome. Label x and y axes. Show all commands. [Hint: combine the commands and figures into a single WORD document]
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| # (4 pts) Vector operations.
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| ## Create a new vector (named as "gene.length") of gene length
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| ## Show commands for extracting the first item, first 10 items, items 20 through 30, the 1st, 2nd, and 5th items
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| ## Apply the following functions: range(), min(), max(), mean(), var(). [Hint: use help(var), help(min) for help]
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| |}
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| | |
| ===Feb 9. (<font color="red">Class cancelled due to snow storm</font>)===
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| | |
| ===Feb 16. R Data Structure & Variable Types===
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| * Vector
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| <syntaxhighlight lang=R">
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| x <- c(1,2,3,4,5) # construct a vector using the c() function
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| x # show x
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| 2 * x + 1 # arithmetic operations, applied to each element
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| exp(x) # exponent function (base e)
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| x <- 1:5 # alternative way to construct a vector, if consecutive
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| x <- seq(from = -1, to = 14, by = 2) # use the seq() function to create a numeric series
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| x <- rep(5, times = 10) # use the rep() function to create a vector of same element
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| x <- rep(NA, times = 10) # pre-define a vector with unknown elements; Use no quotes
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| # Apply vector functions
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| length(x)
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| sum(x)
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| mean(x)
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| range(x)
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| # Access vector elements with indices
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| x[1]
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| x[1:3]
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| x[-2]
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| x[c(1,3)]
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| # Character vectors | |
| gender <- c("male", "female", "female", "male", "female")
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| gender[3]
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| </syntaxhighlight>
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| * Matrix
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| <syntaxhighlight lang=R">
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| BMI <- c(28, 32, 21, 27, 35) # a vector of body-mass index
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| bp <- c(124, 145, 127, 133, 140) # a vector of blood pressure
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| data.1 <- cbind(age, BMI, bp) # create a matrix using column bind function cbind(), individuals in rows
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| data.1
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| data.2 <- rbind(age, BMI, bp) # create a matrix using row bind function rbind()
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| t(data.1) # transpose a matrix: columns to rows & rows to columns
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| dim(data.1) # dimension of the matrix
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| colnames(data.1)
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| rownames(data.1) <- c("subject1", "subject2", "subject3", "subject4", "subject5")
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| data.1
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| data.1[3,1] # access the element in row 3, column 1
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| data.1[2,] # access all elements in row 2
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| data.1[,2] # access all elements in column 2
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| matrix(data = 1:12, nrow = 3, ncol =4) # create a matrix with three rows and four columns; filled by column
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| matrix(data = 1:12, nrow = 3, ncol =4, byrow = TRUE) # filled by row
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| mat <- matrix(data = NA, nrow = 2, ncol = 3) # create an empty matrix
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| mat[1,3] <- 5 # assign a value to a matrix element
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| </syntaxhighlight>
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| * Dataframe
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| <syntaxhighlight lang=R">
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| class(hg) # show object class
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| hg[1,] # how first row
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| hg[,1] # show first column
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| hg[1:3,] # show rows 1 through 3
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| hg[,1:3] # show columns 1 through 3
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| hg$Gene.Name # show column "Gene.Name"
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| </syntaxhighlight>
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| {| class="wikitable sortable mw-collapsible"
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| |- style="background-color:lightsteelblue;"
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| ! Assignment #2. Due 2/23 (Finalized)
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| |- style="background-color:powderblue;"
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| |
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| # (2 pts) Construct a numeric vector of 10 random numbers from the uniform distribution between 0 and 1 (Hint: use the function <code>runif()</code>). Name the resulting vector as "ran.1". Show length, range, mean, and variance.
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| # (2 pts) Construct a numeric vector of 10 random numbers from a normal distribution with mean of 0 and variance of 1 (Hint: use the function <code>rnorm()</code>). Name the resulting vector as "ran.2". Show length, range, mean, and variance. Compare the variance with the previous one: which is large?
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| # (2 pts) Construct a matrix of 10 rows by combining the previous two vectors using the <code>cbind</code> function. Name the matrix as "mat". Assign row names as "ind1" .. "ind10". Show row values for ind1, column values for ran.2; transpose the matrix and save it as "mat.t".
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| # (2 pts) Construct a character vector of 10 US States (Hint: use the c() function). Name it "us.states". Use full, case-sensitive names and "_" in place of spaces. Show the first and the fifth states with one command.
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| # (2 pts) Understand variable types. (From Whitlock & Schluter) Researchers randomly assign diabetes patients to two groups. In the first group, the patients receive a new drug while the other group received standard treatment without the new drug. The researchers compared the rate of insulin release in the two groups.
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| ## List the two variables and state whether each is categorical (if so, whether it is nominal or ordinal) or numerical (if so, whether it is discrete or continuous)
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| ## State & explain which variable is the explanatory (i.e., predictive) and which is the response variable.
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| |}
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| | |
| ===Feb 23. Data Visualization===
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| * Vector functions returning indices
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| <syntaxhighlight lang="bash">
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| # The which() function returns the indices of TRUE elements
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| hg$Gene.Length <- hg$Gene.End - hg$Gene.Start + 1 # add a length column
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| hg.long.idx <- which(hg$Gene.Length > 1e6) # returns indices
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| hg.long <- hg[hg.long.idx,] # genes longer than 1 million bases
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| hg.mt.idx <- which(hg$Chromosome == "MT")
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| hg.mt <- hg[hg.mt.idx,] # mitochondrial genes
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| # The grep() function returns the indices of matching a pattern
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| p53.idx <- grep("P53", hg$Gene.Name)
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| hg.p53 <- hg[p53.idx,]
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| # The order() function returns the indices of sorted elements
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| idx.sorted <- order(hg$Gene.Length)
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| hg.sorted <- hg[idx.sorted,]
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| </syntaxhighlight>
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| * Textbook: Chapter 1. Statistics and Sample; Chapter 2. Displaying Data
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| * Lecture slides: [[File:Lecture-slides-part-1.pdf|thumbnail]]
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| {| class="wikitable sortable mw-collapsible"
| |
| |- style="background-color:lightsteelblue;"
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| ! Assignment #3. Due 3/2 (Finalized)
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| |- style="background-color:powderblue;"
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| # (1 pt) Load the iris data set by typing <code>data(iris)</code>
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| # (1 pt) Identity a character variable and obtain frequency counts using the "table()" function
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| # (1 pt) Identity a numerical variable and obtain frequency distribution by a histogram. Use customized x-axis label
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| # (2 pts) Make a boxplot of distribution of each numerical variable with respect to species. For example, <code>boxplot(Sepal.Length ~ Species, data=iris)</code>. Label axes and title.
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| # (2 pt) Make a strip chart of distribution of a numerical variable with respect to species using the <code>stripchart()</code> function. Customize it with axes labels, open circle symbol (pch = 1), the method of "jitter", and being vertical ("vertical=T").
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| # (1 pt) Make a scatter plot to show relations between two numerical variables. For example, <code>plot(iris$Sepal.Length, iris$Sepal.Width)</code>
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| # (2 pts) Among graduate school applicants to a university department, 512 males were admitted, 313 males were rejected, 89 females were admitted, and 19 females were rejected. Explore if there is gender bias in admission by
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| ## Identify the explanatory and response variables, as well as whether the variables are character or numerical
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| ## Make a contingency table using the matrix() function. Add labels to columns and rows using colnames() and rownames() functions
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| ## Plot the contingency table using grouped bar plot with the "barplot()" function, and the "beside = T" option.
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| ## Plot the contingency table using the mosaicplot() function. Based on the plot, explain if there is evidence for gender bias. [Hint: try matrix transposition]
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| |}
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| | |
| ===March 2. Exam 1===
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| | |
| ===March 9. Describing data & Standard Error===
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| * Textbook: Chapters 3 & 4
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| * Population and sample
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| <syntaxhighlight lang="bash">
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| x <- rnorm(1000)
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| x.sample <- sample(x, size = 100)
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| n.genes <- nrow(hg) # number of rows
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| sampled.rows <- sample(1:n.genes, size = 100)
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| hg.sample <- hg[sampled.rows,] # a random sample of 100 genes
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| </syntaxhighlight>
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| * Explore variable distributions
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| <syntaxhighlight lang=R">
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| x <- rnorm(1000)
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| hist(x, breaks = 100) # distribution for continuous variable
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| hist(hg$Gene.Length, br = 100)
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| hist(log10(hg$Gene.Length), br = 100)
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| gene.cts <- table(hg$Chromosomes) # distribution for a categorical vector
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| barplot(gene.cts)
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| </syntaxhighlight>
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| * In-Class exercise: A study of human gene length
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| <syntaxhighlight lang="bash">
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| hg.len <- hg$Gene.End - hg$Gene.Start + 1 # calculate gene length
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| hist(hg.len, br = 200) # plot gene-length distribution. Not normal: mostly genes are short, few very long
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| mean(hg.len) # not representative, super-long genes carry too much weight to the average length
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| median(hg.len) # More representative. Use median for a variable not normally distributed
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| summary(hg.len) # Show all quartiles
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| IQR(hg.len) # 3rd Quartile - 1st Quartile, the range of majority data points, even for skewed distribution
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| log.len <- log10(hg.len); hist(log.len, br=200) # Log of gene length is more normally distributed
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| mean(log.len); median(log.len) # They should be similar, since log.len is normal
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| # The next block is intend to show that the "mean length" of samples is normally distributed, although the length itself is not
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| samp.len <- sample(hg.len, 100) # take a random sample of 100 length
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| mean(samp.len) # a sample mean
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| # Repeat the above 1000 times, so we could study the distribution of "mean length" (not "length" itself)
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| mean.sample.100 <- sapply(1:1000, function(x) mean(sample(hg$Gene.Length, size = 100)))
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| hist(mean.sample.100, br=100) # you should see a more normally distributed histogram
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| # The above exercise is a demonstration of the "Central Limit Theorem"
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| </syntaxhighlight>
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| {| class="wikitable sortable mw-collapsible"
| |
| |- style="background-color:lightsteelblue;"
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| ! Assignment #4. Due 3/16. (Finalized)
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| |- style="background-color:powderblue;"
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| |
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| # (2 pts) Repeat the above sampling experiment (sample size N =100). Save results to a vector called "mean.100" (which is a vector of means, with a length of 1000 elements). Show histogram. Show mean. Show standard deviation.
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| # (2 pts) Repeat the above sampling experiment (sample size N =500). Save results to a vector called "mean.500". Show histogram. Show mean. Show standard deviation
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| # (2 pts) Repeat the above sampling experiment (sample size N =5000). Save results to a vector called "mean.5000". Show histogram. Show mean. Show standard deviation
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| # (1 pt) Explain why mean is not a good description of a "typical" human gene length and which statistic is a better?
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| # (0.5 pt) Sort the human genes by length (using the <code>order()</code>) & created a sorted data.frame ("hg.sorted") of the original data.frame ("hg").
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| # (0.5 pt) Obtain human genes longer than 1 million bases by using the <code>which()</code> function & create a new data.frame of long genes ("hg.long")
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| # (2 pts) Combine the three simulated-mean vectors into one (see below). Define "Standard error" & "confidence interval (CI)". Does CI quantify accuracy or precision?
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| <syntaxhighlight lang="bash">
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| # Combine
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| sample.combined <- cbind(mean.100, mean.500, mean.5000)
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| colnames(sample.combined) <- c("samp.100", "samp.500", "samp.5000")
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| # plot in a single frame
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| par(mfrow=c(3,1))
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| hist(sample.combined[,1], br=100, xlim=c(1e4, 2e5), main="sample size 20", xlab = "mean gene length")
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| hist(sample.combined[,2], br=100, xlim=c(1e4, 2e5), main = "sample size 100", xlab = "mean gene length")
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| hist(sample.combined[,3], br=100, xlim=c(1e4, 2e5), main = "sample size 500", xlab = "mean gene length")
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| par(mfrow =c(1,1))
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| </syntaxhighlight>
| |
| |}
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| | |
| ===March 16. Hypothesis Testing===
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| * Chapter 6
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| * In-Class exercise 1. The following are measurements of body mass (in grams) of three species of finches in Africa, calculate mean, standard deviation, and CV of each species. Make a boxplot and a strip chart separated by species
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| #Species 1: 8, 8, 8, 8, 8, 8, 8, 6, 7 ,7, 7, 8, 8, 8, 7, 7
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| #Species 2: 16, 16, 16, 12, 16, 15, 15, 17, 15, 16, 15, 16
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| #Species 3: 40, 43, 37, 38, 43, 33, 35, 37, 36, 42, 36, 36, 39, 37, 34, 41
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| # Use the 2SE rule of thumb, calculate 95% confidence interval.
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| # Plot standard error & standard deviation
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| <syntaxhighlight lang=R">
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| df1 <- data.frame(bm = c(8, 8, 8, 8, 8, 8, 8, 6, 7 ,7, 7, 8, 8, 8, 7, 7), species = rep("sp1", 16))
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| df2 <- data.frame(bm = c(16, 16, 16, 12, 16, 15, 15, 17, 15, 16, 15, 16), species = rep("sp2", 12))
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| df3 <- data.frame(bm = c(40, 43, 37, 38, 43, 33, 35, 37, 36, 42, 36, 36, 39, 37, 34, 41), species = rep("sp3", 16))
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| df.bm <- rbind(df1, df2, df3)
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| boxplot(bm ~ species, data= df.bm)
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| sd(df1$bm)
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| mean.1 <-mean(df1$bm)
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| se.1 <- sd(df1$bm)/sqrt(nrow(df1))
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| ci.1 <- c(mean(df1$bm) - 2 * se.1, mean(df1$bm) + 2*se.1)
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| bm.aov <- aov(bm ~ species, data = df.bm)
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| </syntaxhighlight>
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| * In-Class exercise 2. Hypothesis testing through simulation
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| <syntaxhighlight lang=R">
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| # coin-flipping experiments
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| runif(1) # take a random sample from 0-1, uniformly distributed
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| rbinom(n = 1, size =100, prob = 0.5) # flipping 100 (size) fair (prob) coin, one (n=1) time
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| rbinom(n = 1000, size =100, prob = 0.5) # repeat above 1000 times
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| num.success <- rbinom(n = 1000, size =100, prob = 0.5) # save
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| barplot(table(num.success)) # distribution of number of successes
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| length(which(num.success<=40))/1000 # probability of success less than or equal to 40
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| # test if toads are right-handed: observation 14/18 are right-handed
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| right.handed.toads.by.chance <- rbinom(n = 1000, size = 18, prob = 0.5) # null distribution, 1000 times
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| barplot(table(right.handed.toads.by.chance)) # plot
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| length(which(right.handed.toads.by.chance >= 14))/1000 # probability of getting a value equal or higher than 14
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| # Binomial test
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| binom.test(x=14, size=18, p =0.5)
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| </syntaxhighlight>
| |
| {| class="wikitable sortable mw-collapsible"
| |
| |- style="background-color:lightsteelblue;"
| |
| ! Assignment #5. Due 3/23 (Finalized)
| |
| |- style="background-color:powderblue;"
| |
| |
| |
| # (1 pt) Make a single data frame containing the gene expression values for two genes (Hint: create two dataframes and name two columns as "expression" and "gene", then use <code>rbind()</code> to combine two dataframes into one)
| |
| ## "MED1": 12.38918, 9.084664, 9.48416, 9.363928, 8.194495, 8.694836, 8.771101, 9.998151, 12.66877, 8.684064, 8.944236, 11.8491, 8.40968, 8.990329, 9.782376, 8.58243, 12.00455, 8.580401, 9.161046, 9.047977, 8.672018, 8.811856, 8.354933, 8.763175
| |
| ## "ZBTB42": 8.377784, 7.832712, 8.65289, 4.59474, 5.598869, 4.912963, 5.24125, 7.688584, 7.36693, 4.463853, 5.646581, 6.830076, 4.485883, 6.741698, 6.967342, 5.307032, 6.80991, 7.612475, 5.795508, 5.033554, 5.032286, 4.979937, 8.315718, 5.801263, 7.136532, 4.722164, 5.416593, 4.456056, 6.253954, 5.684245, 8.255962, 8.629676, 8.348159, 8.114049, 6.786746, 7.893434, 7.836647, 4.733391, 6.895385, 7.123281, 4.75207
| |
| # (1 pt) Make a stripchart of expression values separated by genes.
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| # (1 pt) Obtain the mean, median, standard deviation, standard error, and 95% confidence intervals of expression for each gene
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| # (1 pt) Run two-sampled t test <code>t.test</code> to test if two genes differ significantly in expression levels. Specify what is the null hypothesis and what is the alternative hypothesis
| |
| # (2 pts) Identify whether each of the following statements is more appropriate as the null (H0) or alternative (H1) hypothesis. State appropriate H0 and H1 for each question.
| |
| ## Most genetic mutations are deleterious
| |
| ## A diet has no effect on liver function
| |
| # (4 pts) Can parents distinguish their own children by smell alone? In a study, eight of nine mothers identified their children correctly based on the smell of T-shirts children wore for three consecutive nights.
| |
| ## State the null & alternative hypotheses.
| |
| ## Simulate the null distribution using the "rbinom()" function (with n = 1 million). Plot the distribution using "barplot"
| |
| ## Verify your result with the "binom.test()" function.
| |
| ## Conclude by explain the p-value.
| |
| |}
| |
| | |
| ===March 23. One-sampled & Two-sampled t-test===
| |
| * Chapters 10-12
| |
| * Lecture Slides: [[File:Lecture-slides-part-2.pdf|thumbnail]]
| |
| {| class="wikitable sortable mw-collapsible"
| |
| |- style="background-color:lightsteelblue;"
| |
| ! Assignment #6. Due 3/30 (Finalized)
| |
| |- style="background-color:powderblue;"
| |
| |
| |
| * (5 pts) Ostriches live in hot environments. To test if ostriches could reduce brain temperature relative to body temperature, the mean body and brain temperature (in Celsius) of six ostriches was recorded in the following table.
| |
| # State the null and alternative hypotheses
| |
| # Make a data frame and visualize with a boxplot combined with a stripchart
| |
| # Test if there is significant difference between the body and brain temperature. [Hint: Choose the most appropriate t-test among the one-sample, paired, or two-sample t-test]
| |
| {| class="wikitable"
| |
| |-
| |
| ! Ostrich !! Body Temperature !! Brain Temperature
| |
| |-
| |
| | 1 || 38.51 || 39.32
| |
| |-
| |
| | 2 || 38.45 || 39.21
| |
| |-
| |
| | 3 || 38.27 || 39.2
| |
| |-
| |
| | 4 || 38.52 || 38.68
| |
| |-
| |
| | 5 || 38.62 || 39.09
| |
| |-
| |
| | 6 || 38.18 || 38.94
| |
| |}
| |
| * (5 pts) Could mosquitoes detect odor of what people drink? The following data points are the relative activation time (the smaller the quicker) of mosquitoes flying toward volunteers, divided into beer- and water-drinking groups:
| |
| ** Beer group: 0.36, 0.46, 0.06, 0.18, 0.25, 0.18, -0.06, -0.14, 0.12, 0.39, 0.17, -0.16, -0.05, 0.19, 0.25, 0.31, 0.17, -0.03, 0.23, -0.03, 0.26, 0.30, 0.11, 0.13, 0.21
| |
| ** Water group: 0.04, 0, -0.08, -0.12, 0.201, -0.039, 0.10, 0.041, 0.02, 0.236, 0.05, 0.097, 0.122, -0.019, 0.021, -0.08, -0.165, -0.28
| |
| # State the null and alternative hypotheses
| |
| # Make a data frame and visualize with a boxplot combined with a stripchart
| |
| # Test normality of data points by showing qqnorm() and qqline() plots (separate for two column)
| |
| # Test if there is significant difference between the two groups using a t-test.
| |
| |}
| |
| | |
| ===March 30. Exam 2===
| |
| | |
| ===April 6. Analyzing Proportions===
| |
| * Chapter 7 & 8
| |
| {| class="wikitable sortable mw-collapsible"
| |
| |- style="background-color:lightsteelblue;"
| |
| ! Assignment #7. Due 4/19 (Finalized)
| |
| |- style="background-color:powderblue;"
| |
| |
| |
| # (Analysis of frequency data. 5 pts) A Siena College Poll has been conducted between April 6-11, 2016 by telephone calls. Based on a sample of 538 likely Democratic primary voters in New York State (which votes next Tuesday on 4/19/2016), the poll concludes that "Vermont Senator Bernie Sanders has narrowed the lead against former New York Senator Hillary Clinton among likely Democratic voters, however, she still has a double digit lead 52-42 percent".
| |
| ## What are the estimates of proportion of voters supporting Clinton and Sanders, respectively? Name the two variables <code>p.clinton</code> & <code>p.sanders</code>
| |
| ## What are the standard errors (i.e., "margins of error" of the pool) of these two estimates? Use the formula <code>SE[p]=sqrt(p*(1-p)/n)</code>
| |
| ## What are the 95% confidence intervals for these two estimates? Use the approximate formula <code>p-2*SE[p], p+2*SE[p]</code>
| |
| ## Assuming that the number of Sanders supporters is 226 out of 538 in this pool, use the <code>binom.test()</code> function to test the null hypothesis that the proportion of Sanders supporter is <code>p=0.5</code>. Could the null hypothesis be rejected at p value of 0.05?
| |
| ## Would you predict based on this pool that Clinton would win the New York Primary?
| |
| # (Test of goodness-of-fit. 5 pts) Variation in flower color of a plant species is determined by a single gene, with ''RR'' individuals being red, ''Rr'' individuals pink, and ''rr'' individuals white. The expected color ratio of crossing F1 individuals is 1:2:1.
| |
| ## In one cross experiment, the results were 10 red, 21 pink, and 9 white-flowered offspring. Do these results differ significantly from the expected frequencies (at 5% level of significance)? Create a data frame with observed and expected counts as the two columns. Calculate the chi-square value and degree of freedom. Obtain p value by using the <code>pchisq()</code> function
| |
| ## Validate your above test using the <code>chisq.test(x=c(observed counts), p=c(0.25, 0.5, 0.25))</code> function. Save test results and show expected counts
| |
| ## In another, larger experiment, 1000 red, 2100 pink, and 900 white flowers were observed. Do these results differ significantly from the expected ratio?
| |
| ## How would you explain the difference between the two results?
| |
| |}
| |
| | |
| ===April 13. No Class (Spring Break)===
| |
| ===April 20. No Class (Monday Schedule)===
| |
| ===April 27. Contingency Analysis===
| |
| * Chapter 9.
| |
| * Lecture Slides: [[File:Part-3-frequency-2017.pdf|thumbnail]]
| |
| {| class="wikitable sortable mw-collapsible"
| |
| |- style="background-color:lightsteelblue;"
| |
| ! Assignment #8. Due 5/4 (Finalized)
| |
| |- style="background-color:powderblue;"
| |
| |
| |
| The following table shows results of genotype counts in "Taster" and "non-Taster" individuals.
| |
| <center>
| |
| {| class="wikitable"
| |
| |-
| |
| ! !! CC !! CT !! TT
| |
| |-
| |
| | Taster || 44 || 20 || 40
| |
| |-
| |
| | Non-Taster || 19 || 10 || 39
| |
| |}
| |
| </center>
| |
| # Hardy-Weinberg Analysis
| |
| ## (1 pt) Calculate overall genotype counts (regardless of phenotype)
| |
| ## (1 pt) Test the observed genotype counts against Hardy-Weinberg expectations (consult lecture slides)
| |
| ## (2 pt) Perform a chi-square test using observed counts and expected proportions. Is the result significant? State for each genotype if it is under- or over-presented. What assumptions of Hardy-Weinberg equilibrium are not met, if your test result is significant?
| |
| # Genotype-phenotype association analysis
| |
| ## (1 pt) State the null & alternative hypotheses
| |
| ## (1 pt) Create a matrix called "taster.genotypes" and display the data as a mosaic plot. Make sure that the explanatory variable is on the ''X''-axis
| |
| ## (2 pts) Test the genotype-phenotype association with a chi-square test, with simulated p value. Conclude if there is significant association between the genotypes and phenotype
| |
| ## (2 pts) Save the above test result as "taste.chisq". Show observed & expected counts and identify over- and under-represented for each genotype/phenotype combination
| |
| |}
| |
| | |
| ===May 4. Exam 3===
| |
| * Review lecture slides (part 3)
| |
| * Review two assignments
| |
| | |
| ===May 11. Correlation===
| |
| * Chapter 16. [http://whitlockschluter.zoology.ubc.ca/r-code/rcode16 R Data & Code on Publisher's Website]
| |
| * Lecture slides: [[File:Part-4-correlation-spring-2017.pdf|thumbnail]]
| |
| * In-class Exercise 1. Aggression behavior in birds. (1) Make a data frame with two columns; (2) Make a scatter plot with title and axes labels
| |
| {| class="wikitable"
| |
| |-
| |
| ! Variable !! Values
| |
| |-
| |
| | Number of visits by non-parent adults || 1,7,15,4,11,14,23,14,9,5,4,10,13,13,14,12,13,9,8,18,22,22,23,31
| |
| |-
| |
| | Future aggressive behavior || -0.8,-0.92,-0.8,-0.46,-0.47,-0.46,-0.23,-0.16,-0.23,-0.23,-0.16,-0.10,-0.10,0.04,0.13,0.19,0.25,0.23,0.15,0.23,0.31,0.18,0.17,0.39
| |
| |}
| |
| * In-class Exercise 2. Inbreeding in wolfs. (1) Make a data frame with two columns; (2) Make a scatter plot with title and axes labels; (3) Calculate correlation coefficient and test its significance by using the <code>cor.test()</code> function
| |
| {| class="wikitable"
| |
| |-
| |
| ! Variable !! Values
| |
| |-
| |
| | Inbreeding coefficient between mating pairs || 0,0,0.13,0.13,0.13,0.19,0.19,0.19,0.22,0.24,0.24,0.24,0.24,0.24,0.24,0.25,0.27,0.30,0.30,0.30,0.30,0.36,0.37,0.40
| |
| |-
| |
| | Number of pups surviving first winter || 6,6,7,5,4,8,7,4,3,3,3,3,2,2,6,3,5,3,2,1,3,2,3
| |
| |}
| |
| * In-class Exercise 3. Miracles of memory. (1) Make a data frame with two columns; (2) Make a scatter plot with title and axes labels; (3) Run a rank correlation using <code>cor.test(method="spearman")</code> function
| |
| {| class="wikitable"
| |
| |-
| |
| ! Variable !! Values
| |
| |-
| |
| | Years elapsed || 2,5,5,4,17,17,31,20,22,25,28,29,34,43,44,46,34,28,39,50,50
| |
| |-
| |
| |Impressive score || 1,1,1,2,2,2,3,4,4,4,4,4,4,4,4,4,4,5,5,5,5
| |
| |}
| |
| {| class="wikitable sortable mw-collapsible"
| |
| |- style="background-color:lightsteelblue;"
| |
| ! Assignment #9. Due 5/18 (Finalized)
| |
| |- style="background-color:powderblue;"
| |
| |
| |
| (Note: Lecture slides has been posted. For help, refer to the slides and [http://whitlockschluter.zoology.ubc.ca/r-code/rcode16 R Data & Code on Publisher's Website])
| |
| | |
| Question 1 (6 pts). In their study of hyena laughter (or "giggle"), Mathevon et al (2010) asked whether sound spectral properties of the giggles are associated with age. Perform correlation analysis using data in the following table:
| |
| {| class="wikitable"
| |
| |-
| |
| | Age (years) || 2,2,2,6,9,10,13,10,14,14,12,7,11,11,14,20
| |
| |-
| |
| | Frequency (Hz) || 840,670,580,470,540,660,510,520,500,480,400,650,460,500,580,500
| |
| |}
| |
| * Identify explanatory and response variables and their data types.
| |
| * Make a data frame and test normality of both variables
| |
| * Make a scatterplot including axes labels and a main title
| |
| * Calculate correlation coefficient between the two variables and show test results
| |
| * State null and alternative hypotheses of the correlation test
| |
| * Conclude if there is a correlation between the age and "giggle", if it is statistically significant, and whether positive or negative if the relationship is significant.
| |
| | |
| Question 2(4 pts). A mutation in the <i>SLC11A1</i> gene in humans causes resistance to tuberculosis. Barnes et al (2011) examined the frequency of the resistant allele in different towns in Europe and Asia and compared it to how long humans had been settled in the site ("duration of settlement"):
| |
| {| class="wikitable"
| |
| |-
| |
| | Duration of settlement (years) || 8010,5260,4735,4010,3710,2810,2730,2310,2110,1955,1910,1300,378,194,130,110,91
| |
| |-
| |
| | Allele frequency || 0.99,1.0,0.948,0.885,0.947,0.854,1.0,0.769,0.9956,0.979,0.865,0.922,0.821,0.842,0.734,0.766,0.772
| |
| |}
| |
| * Make a data frame and test variable normality.
| |
| * Examine data with a scatterplot (with axes labels).
| |
| * Since the relationship appears curvilinear, use Spearman's rank correlation to test the association
| |
| * Show test results and state your conclusions (if there is a correlation, if it is significant, and whether the relation is positive or negative if significant).
| |
| |}
| |
| | |
| ===May 18. Regression===
| |
| * Chapter 16. [http://whitlockschluter.zoology.ubc.ca/r-code/rcode17 R Data & Code on Publisher's Website]
| |
| * In-class Exercise 1. Lion noses. Linear regression with regression line & confidence intervals
| |
| * In-class Exercise 2. Prairie Home Companion. Linear regression with test of non-zero slope
| |
| * In-class Exercise 3. Iris pollination. Square-root transformation
| |
| * In-class Exercise 4. Iron and phytoplankton growth. Non-linear regression
| |
| * In-class Exercise 5. Guppy cold death. Logistic regression
| |
| * Teacher's Evaluation
| |
| ** Students can complete them now in 3 simple steps:
| |
| *** Visit [http://www.hunter.cuny.edu/te www.hunter.cuny.edu/te] OR [http://www.hunter.cuny.edu/mobilete www.hunter.cuny.edu/mobilete] (for smartphones)
| |
| *** Sign in with your net ID and net ID password (forgot your password? Click here: https://netid.hunter.cuny.edu/verify-identity)
| |
| *** Complete the evaluation for your instructor(s)
| |
| ** Notes to students:
| |
| *** Your responses are completely anonymous, and that instructors can only see results after grades are released.
| |
| *** Teacher evaluations serve a number of important functions such as: improving classes by providing instructors feedback of their teaching; and, serving as supporting documentation in a faculty’s reappointment, tenure and promotion.
| |
| *** Teacher evaluations also help the student make decisions about what courses and instructors are right for them. Please let the students know that the teacher evaluation results are readily accessible to them at www.hunter.cuny.edu/myprof.
| |
| {| class="wikitable sortable mw-collapsible"
| |
| |- style="background-color:lightsteelblue;"
| |
| ! Assignment #10. Due In-Class or 5/25
| |
| |- style="background-color:powderblue;"
| |
| | (10 pts) Using the "Prairie Home Companion" data set to answer the following questions
| |
| * Test the normality of response variable using <code>qqnorm</code> and <code>qqline</code>
| |
| * Make scatter plot with axes labels
| |
| * Log-transform the response variable and make a new scatter plot
| |
| * Perform regression analysis using <code>lm</code>
| |
| * What is the regression formula (intercept and slope? significance of each?)
| |
| * Calculate confidence interval of the population slope
| |
| * Obtain R-squared value and explain its meaning and significance
| |
| * Add confidence band line
| |
| * Add confidence interval line
| |
| |}
| |
| | |
| ===May 25. Final Exam (11:10-1:40; Comprehensive)===
| |
| | |
| ===May 31. Grades submitted to Registrar Office===
| |