Difference between revisions of "Univariate data analysis"

From Statistics for Engineering
Jump to navigation Jump to search
 
(3 intermediate revisions by the same user not shown)
Line 327: Line 327:
[http://www.r-fiddle.org/#/fiddle?id=SkursT0M&version=2  Try this code in a web-browser]
[http://www.r-fiddle.org/#/fiddle?id=SkursT0M&version=2  Try this code in a web-browser]


<syntaxhighlight lang="rsplus">
<html><div data-datacamp-exercise data-lang="r">
# Analysis of the data here:
<code data-type="sample-code">
dilution <-  c(11, 26, 18, 16, 20, 12,  8, 26, 12, 17, 14)
dilution <-  c(11, 26, 18, 16, 20, 12,  8, 26, 12, 17, 14)
manometric <- c(25,  3, 27, 30, 33, 16, 28, 27, 12, 32, 16)
manometric <- c(25,  3, 27, 30, 33, 16, 28, 27, 12, 32, 16)
N <- length(dilution)
N <- length(dilution)


mean(manometric)
paste0('The average of the dilution values is = ', mean(dilution))
mean(dilution)
paste0('The average of the manometric values is = ', mean(manometric))


plot(c(dilution, manometric), ylab="BOD values", xaxt='n')
plot(c(dilution, manometric), ylab="BOD values", xaxt='n',
    main='Dilution and manometric values, side by side')
text(5.5,3, "Dilution")
text(5.5,3, "Dilution")
text(18,3, "Manometric")
text(18,3, "Manometric")
abline(v=11.5)
abline(v=11.5)


par(mar=c(4.2, 4.2, 0.2, 0.2))  # (bottom, left, top, right); defaults are par(mar=c(5, 4, 4, 2) + 0.1)
 
plot(dilution, type="p", pch=4,  
plot(dilution, type="p", pch=4,  
     cex=2, cex.lab=1.5, cex.main=1.8, cex.sub=1.8, cex.axis=1.8,  
     cex=2, cex.lab=1.5, cex.main=1.8, cex.sub=1.8,  
    ylab="BOD values", xlab="Sample number",
    cex.axis=1.8, ylab="BOD values", xlab="Sample number",
     ylim=c(0,35), xlim=c(0,11.5), col="darkgreen")
     ylim=c(0,35), xlim=c(0,11.5), col="darkgreen",
    main="Dilution and Manometric values as paired experiments")
lines(manometric, type="p", pch=16, cex=2, col="blue")
lines(manometric, type="p", pch=16, cex=2, col="blue")
lines(rep(0, N), dilution, type="p", pch=4, cex=2, col="darkgreen")
lines(rep(0, N), dilution, type="p", pch=4, cex=2,  
lines(rep(0, N), manometric, type="p", pch=16, cex=2, col="blue")
      col="darkgreen")
lines(rep(0, N), manometric, type="p", pch=16,  
      cex=2, col="blue")
grid()
abline(v=0.5)


abline(v=0.5)
legend(8, 5, pch=c(4, 16), c("Dilution", "Manometric"),
      col=c("darkgreen", "blue"), pt.cex=2)


legend(8, 5, pch=c(4, 16), c("Dilution", "Manometric"), col=c("darkgreen", "blue"), pt.cex=2)


par(mar=c(4.2, 4.2, 0.2, 0.2))  # (bottom, left, top, right); defaults are par(mar=c(5, 4, 4, 2) + 0.1)
plot(dilution-manometric, type="p",  
plot(dilution-manometric, type="p", ylab="Dilution - Manometric", xlab="Sample number",  
    ylab="Dilution - Manometric", xlab="Sample number",  
     cex.lab=1.5, cex.main=1.8, cex.sub=1.8, cex.axis=1.8, cex=2)
     cex.lab=1.5, cex.main=1.8, cex.sub=1.8,  
    cex.axis=1.8, cex=2,
    main="Dilution minus Manometric differences")
abline(h=0, col="grey60")
abline(h=0, col="grey60")
</syntaxhighlight>
</code>
</div></html>

Latest revision as of 08:14, 15 January 2019

Learning outcomes

  • The study of variability important to help answer: "what happened?"
  • Univariate tools such as the histogram, median, MAD, standard deviation, quartiles will be reviewed from prior courses (as a refresher)
  • The normal and t-distribution will be important in our work: what are they, how to interpret them, and use tables of these distributions
  • The central limit theorem will be explained conceptually: you cannot finish a course on stats without knowing the key result from this theorem.
  • Using and interpreting confidence intervals will be crucial in all the modules that follow.

Resources

Tasks to do first Quiz Solution
Complete steps 10, 11, 12 and 13 of the software tutorial

(also steps 1 through 9)

Quiz 1 Solution 1
Watch videos 1, 2, 3, 4, and 5 Quiz 2

Quiz 3

Solution 2

Solution 3

Watch videos 6, 7, and 8 Quiz 4

Quiz 5

Solution 4

Solution 5

Watch videos 9, 10, 11 and 12 Quiz 6

Quiz 7

Solution 6

Solution 7

Watch videos 13, 14 and 15 Quiz8 Solution 8
Watch video 16 Quiz 9

Quiz 10

Solution 9

Solution 10

Extended readings

  • New Boeing planes will generate 0.5 TB of data per flight. Read about this, and other sources of data: "every piece of that plane has an internet connection, from the engines to the flaps to the landing gear".
  • An interesting move has started to take place over the last few years in academic publishing, but is really accelerating now. Journals are now disallowing the use of "p-values", as described why in this editorial in Basic and Applied Social Psychology: http://dx.doi.org/10.1080/01973533.2015.1012991. I intentionally don't cover p-values in the course, because they can be confusing and counterintuitive for engineers. You see these p-values listed in the R-output though for linear models, and they are very closely related to confidence intervals. This means that future courses will start to de-emphasize confidence intervals and look at the alternatives suggested in the link above. Confidence intervals still have their place though: they are widely used in existing literature, and are still a valid way of interpreting results, as long as you are aware of exactly what its interpretation is. This is important to note for those of you going to grad school and looking at graduate research.
  • All students, but especially the 600-level students should read the article by Peter J. Rousseeuw, Tutorial to Robust Statistics it is easy to read, and contains so much useful content.

Class videos from prior years

Videos from 2015

Watch all these videos in this YouTube playlist

  1. Introduction [05:59]
  2. Histograms [04:50]
  3. Basic terminology [06:41]
  4. Outliers, medians and MAD [04:42]
  5. The central limit theorem [06:56]
  6. The normal distribution, and standardizing variables [05:54]
  7. Normal distribution notation and using tables and R [05:48]
  8. Checking if data are normally distributed [05:57]
  9. Introducing the idea of a confidence interval [covered in class]
  10. Confidence intervals when we don't know the variance [07:59]
  11. Interpreting the confidence interval [07:52]
  12. A worked example: calculating and interpreting the CI [03:37]
  13. A motivating example to see why tests for differences are important [08:29]
  14. The mathematical derivation for a confidence interval for differences [covered in class]
  15. Using the confidence interval to test for differences to solve the motivating example [covered in class]
  16. Confidence intervals for paired tests: theory and an example [11:59]
05:59 | Download video | Download captions | Script
04:50 | Download video | Download captions | Script
06:41 | Download video | Download captions | Script
04:42 | Download video | Download captions | Script
06:56 | Download video | Download captions | Script
05:54 | Download video | Download captions | Script
05:48 | Download video | Download captions | Script
05:57 | Download video | Download captions | Script
Covered in class | No video | Script
07:59 | Download video | Download captions | Script
07:52 | Download video | Download captions | Script
03:37 | Download video | Download captions | Script
08:29 | Download video | Download captions | Script
Audio only | No video | Script
Audio only | No video | Script
11:59 | Download video | Download captions | Script

Videos from 2014

See the webpage from 2014

Videos from 2013

See the webpage from 2013

Software codes for this section

Code to show how to deal with missing values

Try this code in a web-browser

f <- 'http://openmv.net/file/raw-material-properties.csv' data <- read.csv(f) # notice the NAs in the columns: these refer to # missing value (Not Available) summary(data) sd(data$density1) # why NA as the answer? help(sd) # no NA as the answer anymore! sd(data$density1, na.rm=TRUE) help(mad) help(IQR) # etc: all these functions accept and na.rm input

Understanding the central limit theorem with the rolling dice example

Try this code in a web-browser

N = 500 m <- t(matrix(seq(1,6), 3, 2)) layout(m) s1 <- as.integer(runif(N, 1, 7)) s2 <- as.integer(runif(N, 1, 7)) s3 <- as.integer(runif(N, 1, 7)) s4 <- as.integer(runif(N, 1, 7)) s5 <- as.integer(runif(N, 1, 7)) s6 <- as.integer(runif(N, 1, 7)) s7 <- as.integer(runif(N, 1, 7)) s8 <- as.integer(runif(N, 1, 7)) s9 <- as.integer(runif(N, 1, 7)) s10 <- as.integer(runif(N, 1, 7)) hist(s1, main="", xlab="One throw", breaks=seq(0,6)+0.5) bins = 8 hist((s1+s2)/2, breaks=bins, main="", xlab="Average of two throws") hist((s1+s2+s3+s4)/4, breaks=bins, main="", xlab="Average of 4 throws") hist((s1+s2+s3+s4+s5+s6)/6, breaks=bins, main="", xlab="Average of 6 throws") bins=12 hist((s1+s2+s3+s4+s5+s6+s7+s8)/8, breaks=bins, main="", xlab="Average of 8 throws") hist((s1+s2+s3+s4+s5+s6+s7+s8+s9+s10)/10, breaks=bins, main="", xlab="Average of 10 throws")

Code used to illustrate how the q-q plot is constructed

Try this code in a web-browser

N <- 10 # What are the quantiles from the theoretical # normal distribution? index <- seq(1, N) P <- (index - 0.5) / N theoretical.quantity <- qnorm(P) # Our sampled data: yields <- c(86.2, 85.7, 71.9, 95.3, 77.1, 71.4, 68.9, 78.9, 86.9, 78.4) mean.yield <- mean(yields) # 80.0 sd.yield <- sd(yields) # 8.35 # What are the quantiles for the sampled data? yields.z <- (yields - mean.yield)/sd.yield yields.z yields.z.sorted <- sort(yields.z) # Compare the values in text: yields.z.sorted theoretical.quantity # Compare them graphically: plot(theoretical.quantity, yields.z.sorted, asp=1) abline(a=0, b=1) # Built-in R function to do all the above for you: qqnorm(yields) qqline(yields) # A better function: see # http://learnche.org/4C3/Software_tutorial/Extending_R_with_packages library(car) qqPlot(yields)

Code to illustrate the central limit theorem's reduction in variance

Try this code in a web-browser

# Show the 3 plots side by side layout(matrix(c(1,2,3), 1, 3)) # Sample the population: N <- 100 x <- rnorm(N, mean=80, sd=5) mean(x) sd(x) # Plot the raw data x.range <- range(x) plot(x, ylim=x.range, main='Raw data') # Subgroups of 2 subsize <- 2 x.2 <- numeric(N/subsize) for (i in 1:(N/subsize)) { x.2[i] <- mean(x[((i-1)*subsize+1):(i*subsize)]) } plot(x.2, ylim=x.range, main='Subgroups of 2') # Subgroups of 4 subsize <- 4 x.4 <- numeric(N/subsize) for (i in 1:(N/subsize)) { x.4[i] <- mean(x[((i-1)*subsize+1):(i*subsize)]) } plot(x.4, ylim=x.range, main='Subgroups of 4')

Paired test example

Try this code in a web-browser

dilution <- c(11, 26, 18, 16, 20, 12, 8, 26, 12, 17, 14) manometric <- c(25, 3, 27, 30, 33, 16, 28, 27, 12, 32, 16) N <- length(dilution) paste0('The average of the dilution values is = ', mean(dilution)) paste0('The average of the manometric values is = ', mean(manometric)) plot(c(dilution, manometric), ylab="BOD values", xaxt='n', main='Dilution and manometric values, side by side') text(5.5,3, "Dilution") text(18,3, "Manometric") abline(v=11.5) plot(dilution, type="p", pch=4, cex=2, cex.lab=1.5, cex.main=1.8, cex.sub=1.8, cex.axis=1.8, ylab="BOD values", xlab="Sample number", ylim=c(0,35), xlim=c(0,11.5), col="darkgreen", main="Dilution and Manometric values as paired experiments") lines(manometric, type="p", pch=16, cex=2, col="blue") lines(rep(0, N), dilution, type="p", pch=4, cex=2, col="darkgreen") lines(rep(0, N), manometric, type="p", pch=16, cex=2, col="blue") grid() abline(v=0.5) legend(8, 5, pch=c(4, 16), c("Dilution", "Manometric"), col=c("darkgreen", "blue"), pt.cex=2) plot(dilution-manometric, type="p", ylab="Dilution - Manometric", xlab="Sample number", cex.lab=1.5, cex.main=1.8, cex.sub=1.8, cex.axis=1.8, cex=2, main="Dilution minus Manometric differences") abline(h=0, col="grey60")