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
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.
- 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
- 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
Feb 2. Introduction & R Demo
- Course overview
- 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 # show first length
hg.len # 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 9. (Class cancelled due to snow storm)
Feb 16. R Data Structure & Variable Types
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
# Access vector elements with indices
# Character vectors
gender <- c("male", "female", "female", "male", "female")
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.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
rownames(data.1) <- c("subject1", "subject2", "subject3", "subject4", "subject5")
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
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,]
|Assignment #3. Due 3/2 (Finalized)
- (1 pt) Load the iris data set by typing
- (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,
- (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 2. Exam 1
March 9. Describing data & Standard Error
- Textbook: Chapters 3 & 4
- 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
- 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. (Finalized)
- (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
- (1 pt) Explain why mean is not a good description of a "typical" human gene length and which statistic is a better?
- (0.5 pt) Sort the human genes by length (using the
order()) & created a sorted data.frame ("hg.sorted") of the original data.frame ("hg").
- (0.5 pt) Obtain human genes longer than 1 million bases by using the
which() function & create a new data.frame of long genes ("hg.long")
- (2 pts) Combine the three simulated-mean vectors into one (see below). Define "Standard error" & "confidence interval (CI)". Does CI quantify accuracy or precision?
sample.combined <- cbind(mean.100, mean.500, mean.5000)
colnames(sample.combined) <- c("samp.100", "samp.500", "samp.5000")
# plot in a single frame
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")
March 16. Hypothesis Testing
- Chapter 6
- 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
- Species 1: 8, 8, 8, 8, 8, 8, 8, 6, 7 ,7, 7, 8, 8, 8, 7, 7
- Species 2: 16, 16, 16, 12, 16, 15, 15, 17, 15, 16, 15, 16
- Species 3: 40, 43, 37, 38, 43, 33, 35, 37, 36, 42, 36, 36, 39, 37, 34, 41
- Use the 2SE rule of thumb, calculate 95% confidence interval.
- Plot standard error & standard deviation
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))
df2 <- data.frame(bm = c(16, 16, 16, 12, 16, 15, 15, 17, 15, 16, 15, 16), species = rep("sp2", 12))
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))
df.bm <- rbind(df1, df2, df3)
boxplot(bm ~ species, data= df.bm)
se.1 <- sd(df1$bm)/sqrt(nrow(df1))
ci.1 <- c(mean(df1$bm) - 2 * se.1, mean(df1$bm) + 2*se.1)
bm.aov <- aov(bm ~ species, data = df.bm)
- In-Class exercise 2. 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
# Binomial test
binom.test(x=14, size=18, p =0.5)
|Assignment #5. Due 3/23 (Finalized)
- (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
rbind() 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.
- (1 pt) Obtain the mean, median, standard deviation, standard error, and 95% confidence intervals of expression for each gene
- (1 pt) Run two-sampled t test
t.test 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
|Assignment #6. Due 3/30 (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
- 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]
- (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
|Assignment #7. 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
- What are the standard errors (i.e., "margins of error" of the pool) of these two estimates? Use the formula
- What are the 95% confidence intervals for these two estimates? Use the approximate formula
- 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
- 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 13. No Class (Spring Break)
April 20. No Class (Monday Schedule)
April 27. Contingency Analysis
|Assignment #8. Due 5/4 (Finalized)
The following table shows results of genotype counts in "Taster" and "non-Taster" individuals.
- 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
|Number of visits by non-parent adults
|Future aggressive behavior
- 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
|Inbreeding coefficient between mating pairs
|Number of pups surviving first winter
- 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
|Assignment #9. Due 5/18 (Finalized)
(Note: Lecture slides has been posted. For help, refer to the slides and 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:
- 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 SLC11A1 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"):
|Duration of settlement (years)
- 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. 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
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|Assignment #10. Due In-Class or 5/25
|(10 pts) Using the "Prairie Home Companion" data set to answer the following questions
- Test the normality of response variable using
- Make scatter plot with axes labels
- Log-transform the response variable and make a new scatter plot
- Perform regression analysis using
- 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