Biol20N02 2017

Analysis of Biological Data (BIOL 20N02, Spring 2017) Instructor: Dr Weigang Qiu, Associate Professor, Department of Biological Sciences Room: 1001B HN (North Building, 10th Floor, Mac Computer Lab) Hours: Thursdays 11:10-1:40 Office Hours: Belfer Research Building (Google Map) BB-402; Tuesdays 5-7 pm or by appointment Course Website: http://diverge.hunter.cuny.edu/labwiki/Biol20N2_2017

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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.

This 3-credit experimental course fulfills elective requirements for Biology Major I. Hunter pre-requisites are BIOL100, BIOL102 and STAT113.

Learning Goals

• Be able to use R as a plotting tool to visualize large-scale biological data sets
• Be able to use R as a statistical tool to summarize data and make biological inferences
• Be able to use R as a programming language to automate data analysis

Textbooks

• Whitlock & Schluter (2015). Analysis of Biological Data. (2nd edition). Amazon link

Exams & Grading

• Attendance (or a note in case of absence) is required
• In-Class Exercises (50 pts).
• 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).
• Three Mid-term Exams (3 X 30 pts each = 90 pts)
• Comprehensive Final Exam (50 pts)
• Bonus for active participation in classroom discussions

Course Outline

Feb 2. Introduction & R Demo

1. Course overview
2. Tutorial 1: R Demo

• Create a new project by navigating: File | New Project | New Directory. Name it project file "human_genes"
• 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
• 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

dim(hg) # show dimension
head(hg) # show top rows
tail(hg) # show bottom rows
hg.len <- hg$Gene.End - hg$Gene.Start + 1 # create a vector of gene lengths (Tab for auto-completion)
hg.len[1] # show first length
hg.len[10] # show the 10th item
hg.len[1:10] # show items 1 through 10
hg.len[c(1,10)] # show items 1 and 10
hg[1,1] # show item in 1st row, 1 column
hg[1,] # show all values for 1st row
hg[,1] # show all values for 1st column
hg[1:10,] # show rows 1 through 10
hg[,1:7] # show columns 1 through 10
hg[1:10, 1:7] # a subset
hist(hg$Length, br = 200) # plot gene-length distribution. Not normal: mostly genes are short, few very long
hist(log10(hg$Length), br = 200) # log transformation for non-normally distributed variable
gene.cts <- table(hg$Chromosome) # count number of genes, separated by chromosomes
barplot(gene.cts) # show distribution by a barplot
mean(hg$Length) # not representative, super-long genes carry too much weight to the average length
median(hg$Length) # More representative. Use median for a variable not normally distributed
summary(hg$Length) # Show all quartiles
hg$Gene.Length <- hg$Gene.End - hg$Gene.Start + 1 # create a new column named "Gene.Length"
boxplot(log10(Length) ~ Chromosome, data = hg)  # show gene length by chromosomes
write.csv(hg, "hg.csv", row.names = FALSE) # save into a file
hg <- read.csv("hg.csv") # read back into R

• Export a PDF or image
• Open a new R script, name it as "hg.R"
• Select commands and save to script
• Retrieve and edit a command by pressing "up" or "down" arrows
• Retrieve commands by using the search box on the "History" table
• Type q() to quit. Answer "y" to save workspace
• To reload and restore workspace, go to "C:/Users/instructor/Documents/human.genes" and double click on the file "human.gene"

Assignment #1. Due 2/9, Thursday (Finalized)

(2 pts) Install R & R Studio on your own computer: First, download & install R from this website: https://mirrors.nics.utk.edu/cran/ Next, download & install R studio from this website: https://www.rstudio.com/ (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] (4 pts) Vector operations. Create a new vector (named as "gene.length") of gene length Show commands for extracting the first item, first 10 items, items 20 through 30, the 1st, 2nd, and 5th items Apply the following functions: range(), min(), max(), mean(), var(). [Hint: use help(var), help(min) for help]

Feb 16. R Data Structure & Variable Types

• Vector

x <- c(1,2,3,4,5) # construct a vector using the c() function
x # show x
2 * x + 1 # arithmetic operations, applied to each element
exp(x) # exponent function (base e)
x <- 1:5 # alternative way to construct a vector, if consecutive
x <- seq(from = -1, to = 14, by = 2) # use the seq() function to create a numeric series
x <- rep(5, times = 10) # use the rep() function to create a vector of same element
x <- rep(NA, times = 10) # pre-define a vector with unknown elements; Use no quotes
# Apply vector functions
length(x)
sum(x)
mean(x)
range(x)
# Access vector elements with indices
x[1]
x[1:3]
x[-2]
x[c(1,3)]
# Character vectors
gender <- c("male", "female", "female", "male", "female")
gender[3]

• Matrix

BMI <- c(28, 32, 21, 27, 35) # a vector of body-mass index
bp <- c(124, 145, 127, 133, 140) # a vector of blood pressure
data.1 <- cbind(age, BMI, bp) # create a matrix using column bind function cbind(), individuals in rows
data.1
data.2 <- rbind(age, BMI, bp) # create a matrix using row bind function rbind()
t(data.1) # transpose a matrix: columns to rows & rows to columns
dim(data.1) # dimension of the matrix
colnames(data.1)
rownames(data.1) <- c("subject1", "subject2", "subject3", "subject4", "subject5")
data.1
data.1[3,1] # access the element in row 3, column 1
data.1[2,] # access all elements in row 2
data.1[,2] # access all elements in column 2
matrix(data = 1:12, nrow = 3, ncol =4) # create a matrix with three rows and four columns; filled by column
matrix(data = 1:12, nrow = 3, ncol =4, byrow = TRUE) # filled by row
mat <- matrix(data = NA, nrow = 2, ncol = 3) # create an empty matrix
mat[1,3] <- 5 # assign a value to a matrix element

• Dataframe

class(hg) # show object class
hg[1,] # how first row
hg[,1] # show first column
hg[1:3,] # show rows 1 through 3
hg[,1:3] # show columns 1 through 3
hg$Gene.Name # show column "Gene.Name"

Assignment #2. Due 2/23 (Finalized)

(2 pts) Construct a numeric vector of 10 random numbers from the uniform distribution between 0 and 1 (Hint: use the function runif()). Name the resulting vector as "ran.1". Show length, range, mean, and variance. (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 rnorm()). Name the resulting vector as "ran.2". Show length, range, mean, and variance. Compare the variance with the previous one: which is large? (2 pts) Construct a matrix of 10 rows by combining the previous two vectors using the cbind 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". (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. (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. 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) State & explain which variable is the explanatory (i.e., predictive) and which is the response variable.

Feb 23. Data Visualization

• Vector functions returning indices

# The which() function returns the indices of TRUE elements
hg$Gene.Length <- hg$Gene.End - hg$Gene.Start + 1 # add a length column
hg.long.idx <- which(hg$Gene.Length > 1e6) # returns indices
hg.long <- hg[hg.long.idx,] # genes longer than 1 million bases
hg.mt.idx <- which(hg$Chromosome == "MT")
hg.mt <- hg[hg.mt.idx,] # mitochondrial genes
# The grep() function returns the indices of matching a pattern
p53.idx <- grep("P53", hg$Gene.Name)
hg.p53 <- hg[p53.idx,]
# The order() function returns the indices of sorted elements
idx.sorted <- order(hg$Gene.Length)
hg.sorted <- hg[idx.sorted,]

• Textbook: Chapter 1. Statistics and Sample; Chapter 2. Displaying Data
• Lecture slides:
  File:Lecture-slides-part-1.pdf

Assignment #3. Due 3/2 (Finalized)

(1 pt) Load the iris data set by typing data(iris) (1 pt) Identity a character variable and obtain frequency counts using the "table()" function (1 pt) Identity a numerical variable and obtain frequency distribution by a histogram. Use customized x-axis label (2 pts) Make a boxplot of distribution of each numerical variable with respect to species. For example, boxplot(Sepal.Length ~ Species, data=iris). Label axes and title. (2 pt) Make a strip chart of distribution of a numerical variable with respect to species using the stripchart() function. Customize it with axes labels, open circle symbol (pch = 1), the method of "jitter", and being vertical ("vertical=T"). (1 pt) Make a scatter plot to show relations between two numerical variables. For example, plot(iris$Sepal.Length, iris$Sepal.Width) (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 Identify the explanatory and response variables, as well as whether the variables are character or numerical Make a contingency table using the matrix() function. Add labels to columns and rows using colnames() and rownames() functions Plot the contingency table using grouped bar plot with the "barplot()" function, and the "beside = T" option. Plot the contingency table using the mosaicplot() function. Based on the plot, explain if there is evidence for gender bias. [Hint: try matrix transposition]

March 9. Describing data

• Textbook: Chapter 3
• Population and sample

x <- rnorm(1000)
x.sample <- sample(x, size = 100)
n.genes <- nrow(hg) # number of rows
sampled.rows <- sample(1:n.genes, size = 100)
hg.sample <- hg[sampled.rows,] # a random sample of 100 genes

• Explore variable distributions

x <- rnorm(1000)
hist(x, breaks = 100) # distribution for continuous variable
hist(hg$Gene.Length, br = 100)
hist(log10(hg$Gene.Length), br = 100)
gene.cts <- table(hg$Chromosomes) # distribution for a categorical vector
barplot(gene.cts)

• In-Class exercise: A study of human gene length

hg.len <- hg$Gene.End - hg$Gene.Start + 1 # calculate gene length
hist(hg.len, br = 200) # plot gene-length distribution. Not normal: mostly genes are short, few very long
mean(hg.len) # not representative, super-long genes carry too much weight to the average length
median(hg.len) # More representative. Use median for a variable not normally distributed
summary(hg.len) # Show all quartiles
IQR(hg.len) # 3rd Quartile - 1st Quartile, the range of majority data points, even for skewed distribution
log.len <- log10(hg.len); hist(log.len, br=200) # Log of gene length is more normally distributed
mean(log.len); median(log.len) # They should be similar, since log.len is normal
# The next block is intend to show that the "mean length" of samples is normally distributed, although the length itself is not
samp.len <- sample(hg.len, 100) # take a random sample of 100 length
mean(samp.len) # a sample mean
# Repeat the above 1000 times, so we could study the distribution of "mean length" (not "length" itself)
mean.sample.100 <- sapply(1:1000, function(x) mean(sample(hg$Gene.Length, size = 100)))
hist(mean.sample.100, br=100) # you should see a more normally distributed histogram
# The above exercise is a demonstration of the "Central Limit Theorem"

Assignment #4. Due 3/16. (Tentative)

(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. (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 (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 (2 pts) Explain why mean is not a good description of a "typical" human gene length (2 pts) Combine the three vectors into one (see below). Define "Standard error" & "confidence interval (CI)". Does CI quantify accuracy or precision? # Combine sample.combined <- cbind(mean.100, mean.500, mean.5000) colnames(sample.combined) <- c("samp.100", "samp.500", "samp.5000") # plot in a single frame par(mfrow=c(3,1)) hist(sample.combined[,1], br=100, xlim=c(1e4, 2e5), main="sample size 20", xlab = "mean gene length") hist(sample.combined[,2], br=100, xlim=c(1e4, 2e5), main = "sample size 100", xlab = "mean gene length") hist(sample.combined[,3], br=100, xlim=c(1e4, 2e5), main = "sample size 500", xlab = "mean gene length") par(mfrow =c(1,1))

March 16. Standard Error of Mean & Hypothesis Testing

• Chapters 4 & 6
• In-Class exercise 1. Descriptive statistics

1. Make a vector of the following blood pressure measurements (in mmHg): 112, 128, 108, 129, 125, 153, 155, 132, 137. Calculate sample size, sum, mean, variance, coefficient of variation (CV), and median
2. Take a sample of 100 human gene lengths. Calculate median, IQR, 1.5*IQR; Make a boxplot
3. 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
   1. Species 1: 8, 8, 8, 8, 8, 8, 8, 6, 7 ,7, 7, 8, 8, 8, 7, 7
   2. Species 2: 16, 16, 16, 12, 16, 15, 15, 17, 15, 16, 15, 16
   3. Species 3: 40, 43, 37, 38, 43, 33, 35, 37, 36, 42, 36, 36, 39, 37, 34, 41

• In-Class exercise 2. standard error & confidence interval

1. Blood pressure: What is the standard deviation of the above blood pressure?
2. What is the sample size? Calculate standard error of the mean.
3. Use the 2SE rule of thumb, calculate 95% confidence interval.
4. Plot standard error & standard deviation

• In-Class exercise 3. Hypothesis testing through simulation

# coin-flipping experiments
runif(1) # take a random sample from 0-1, uniformly distributed
rbinom(n = 1, size =100, prob = 0.5) # flipping 100 (size) fair (prob) coin, one (n=1) time
rbinom(n = 1000, size =100, prob = 0.5) # repeat above 1000 times
num.success <- rbinom(n = 1000, size =100, prob = 0.5) # save
barplot(table(num.success)) # distribution of number of successes
length(which(num.success<=40))/1000 # probability of success less than or equal to 40
# test if toads are right-handed: observation 14/18 are right-handed
right.handed.toads.by.chance <- rbinom(n = 1000, size = 18, prob = 0.5) # null distribution, 1000 times
barplot(table(right.handed.toads.by.chance)) # plot
length(which(right.handed.toads.by.chance >= 14))/1000 # probability of getting a value equal or higher than 14
# If the observation is 10/18
right.handed.toads.by.chance <- rbinom(n = 1e6, size = 18, prob = 0.5)
length(which(right.handed.toads.by.chance <= 8 | right.handed.toads.by.chance>=10))/1e6

Assignment #6. Due 3/29 (Finalized)

A study of expression levels of human genes (2 pts) Make a single data frame containing the gene expression values for 3 genes (Hint: name two columns as "expression" and "gene") "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 "TCF24": 3.427531, 3.383114, 3.574041, 3.449132, 3.784881, 3.686278, 3.545466, 3.624868, 3.377987, 3.256293, 3.381218, 3.79938, 3.419852, 3.292284 (2 pts) Make a boxplot of expression values separated by genes. Explain which gene has the highest level of expression & which has the lowest (2 pts) Make a histogram of all expression values. Are they normally distributed? Test normality by qqnorm() and qqline() plots (2 pts) Obtain the mean, median, and standard deviation of expression separated by genes (Use the tapply() function to get full credits) (2 pts) Explain what is "standard error of mean" and how is it different from standard deviation (Hint: explain sampling]. In taking a sample from a population, what can you do to make the standard error of the mean smaller?

March 23. Hypothesis Testing

• In-Class exercise 3. Hypothesis testing through simulation
• Lecture Slides: to be posted

Assignment #7. Due 4/5 (Finalized)

(2 pts) Identify whether each of the following statements is more appropriate as the null or alternative hypothesis. State appropriate H0 and H1 for each question. Most genetic mutations are deleterious A diet has no effect on liver function (4 pts) The following table lists results of a study on the impact of tourism to species density. Explain what the p-value means Which species show a statistically significant reduction in mean density near heavily visited ruins? Which species show a statistically significant increase in mean density near heavily visited ruins? Which species provide no significant evidence in reduction or in increase? Species Mean density near ruins Mean density far from ruins P-value Agouti 160.2 14.5 0.03 Coatimundi 99.4 1.0 0.01 Collared precany 4.6 1.8 0.79 Deppes squirrel 32.3 2.2 0.54 Howler monkey 7.3 1.9 0.03 Spider monkey 170.8 15.0 0.88 Crested guan 0 49.4 0.001 Great curassow 10.8 72.0 0.048 Ocellated turkey 47.0 0 0.02 Tinamos 0 4.9 0.049 (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" Calculate from the null distribution the probability of obtaining 8 or 1 correct out of 9 by chance, which would be the p-value of this experiment. Is the result significant? Verify your result with the "binom.test()" function.

April 6. Analyzing Proportions

Assignment #8. Due 4/19 (Finalized)

(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 p.clinton & p.sanders What are the standard errors (i.e., "margins of error" of the pool) of these two estimates? Use the formula SE[p]=sqrt(p*(1-p)/n) What are the 95% confidence intervals for these two estimates? Use the approximate formula p-2*SE[p], p+2*SE[p] Assuming that the number of Sanders supporters is 226 out of 538 in this pool, use the binom.test() function to test the null hypothesis that the proportion of Sanders supporter is p=0.5. 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 pchisq() function Validate your above test using the chisq.test(x=c(observed counts), p=c(0.25, 0.5, 0.25)) 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 27. Contingency Analysis

Lecture Slides:

File:Part-3-frequency.pptx

Assignment #9. Due 5/3 (Finalized)

The following table shows results of genotype counts in "Taster" and "non-Taster" individuals. CC CT TT Taster 79 82 9 Non-Taster 4 6 35 (1 pt) State the explanatory and response variables (1 pt) State the null & alternative hypotheses (2 pts) 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 (1 pt) Test the genotype-phenotype association with a chi-square test, with simulated p value. Conclude if there is significant association (1 pt) Save the above test result as "taste.chisq". Show observed & expected counts (1 pt) Identify over- and under-represented genotypes (3 pts) Test if there is association between the alleles (i.e., "C" or "T") and the Taster/non-Taster phenotypes Calculate allele counts in each phenotypic group State null and alternative hypotheses; Create a matrix called "taster.alleles". Make a mosaic plot Perform chi-square test, draw your conclusions, and identify which allele ("C" or "T") is associated with the "Taster" phenotype

May 4. Exam 3

• Review lecture slides (part 3)
• Review 2 previous exams

May 11. One-sample t-test

• Motivating example: weights of ticks

1. Import data set: http://diverge.hunter.cuny.edu/~weigang/data-sets-for-biostat/tick.tsv
2. How to visualize data? What kinds of plots to make?
3. Is there sexual dimorphism?

• No homework assignment (to be combined with the next one)

May 18. Paired & Two sample t-tests

• Import data set: http://diverge.hunter.cuny.edu/~weigang/data-sets-for-biostat/ahus.csv
• PLEASE FILL IN TEACHER EVALUATIONS: Teacher's evaluation
• All lecture slides
  • Part 1.
    File:Biostat-part-1.pdf
  • Part 2.
    File:Part-2-distribution.pptx
  • Part 3.
    File:Part-3-frequency.pptx
  • Part 4.
    File:Part-4-t-tests.pptx

Assignment #10. Due 5/24 (Finalized)

(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 H0: there is no difference in body and brain temperature; H1: the brain temperature is different from the body temperature Make a data frame and visualize with a boxplot combined with a stripchart x <- data.frame(id = 1:6, body = c(), brain = c()) Test normality of data points by showing qqnorm() and qqline() plots (separate for two column) qqnorm(x$body); qqline(x$body) Test if there is significant difference between the body and brain temperature t.test(x$body, x$brain, paired = T) 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 H0: there is no difference in activation time between the two groups; H1: reaction time differs between the two groups Make a data frame and visualize with a boxplot combined with a stripchartx <- data.frame(time = c(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, 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), group = c(rep("beer", 25), rep("water",18))) Test normality of data points by showing qqnorm() and qqline() plots (separate for two column) qqnorm(x$time[1:25]); qqline(x$time[1:25]) Test if there is significant difference between the two groupst.test(time ~ group, data = x)