introductory statistics with r - ucla statistical...

Prelim. Data Descriptive Statistics Prob. Models Hyp. Test & CI Linear Reg. Resources Upcoming Exercises

UCLA Department of StatisticsStatistical Consulting Center

Introductory Statistics with R

Mine [email protected]

February 1, 2010

Mine Cetinkaya [email protected]

Introductory Statistics with R UCLA SCC


Outline

1 Preliminaries

2 Data sets

3 Descriptive Statistics

4 Probability Models

5 Hypothesis Testing and Confidence Intervals

6 Linear Regression

7 Online Resources for R

8 Upcoming Mini-Courses

9 ExercisesMine Cetinkaya [email protected]



1 PreliminariesSoftware InstallationR Help

2 Data sets




6 Linear Regression






Software Installation

Installing R on a Mac

1 Go tohttp://cran.r-project.org/

and select MacOS X

2 Select to download thelatest version: 2.11.0(2010-04-22)

3 Install and Open. The Rwindow should look like this:




R Help

R Help

For help with any function in R,put a question mark before thefunction name to determine whatarguments to use, examples andbackground information.

1 ?plot




1 Preliminaries

2 Data setsLoading data into RViewing data sets in R




6 Linear Regression






Loading data into R

Loading data into R

Loading a data set into R:

1 survey = read.table("http://www.stat.ucla.edu/~mine/students_survey_2008. txt",header = TRUE , sep = "\t")

Displaying the dimensions of the data set:

1 dim(survey)

[1] 1325 29




Viewing data sets in R


Displaying the first 3 rows and 5 columns of the data set:

1 survey [1:3 ,1:5]

gender hand eyecolor glasses california

1 female left hazel yes yes

2 male right brown no no

3 female right brown yes yes

Displaying the variable names in the data set:

1 names(survey)

[1] "gender" "hand" "eyecolor" "glasses" "california"

[6] "birthmonth" "birthday" "birthyear" "ageinmonths" "height"

[11] "graduate" "oncampus" "time" "walk" "hsclass"

...





Attaching / detaching data frames in RAttaching the variables in a data set::

1 attach(survey)

The following object(s) are masked from package:datasets :

sleep

The warning is telling us that we have attached a data framethat contains a column, whose name is sleep. If you type:

1 sleep

the object with that name in the data frame will be seenbefore another object with the same name that is lower in thesearch() path. Thus, your object is “masking” the other.To detach a data frame, i.e. remove from the search() pathof available R objects - but we won’t do that now.

1 detach(sleep)Mine Cetinkaya [email protected]



1 Preliminaries

2 Data sets

3 Descriptive StatisticsVariable classesDisplaying categorical dataDisplaying quantitative dataDescribing distributions numerically



6 Linear Regression



9 Exercises




Variable classes

Displaying the class of a variable:

1 class(instructor)

[1] "factor"

Changing the class of a variable:

1 instructor = as.character(instructor)2 class(instructor)

[1] "character"




Displaying categorical data

Tables

Tables are useful for displaying the distribution of categoricalvariables.

1 table(gender)

genderfemale male

882 443





Contingency tables

Contingency tables display two categorical variables at a time.

1 table(gender , hand)

handgender ambidextrous left right

female 9 67 806male 11 45 387





Frequency bar plotsDisplay counts of each category next to each other for easycomparison.

1 barplot(table(gender), main = "Barplot ofGender")

female male

Barplot of Gender

0200

600





Relative frequency bar plots

Display relative proportions of each category.

1 barplot(table(gender)/length(gender), main = "Relative Frequency \n Barplot of Gender")

female male

Relative Frequency

Barplot of Gender

0.0

0.3

0.6





Segmented bar chartsDisplays two categorical variables at a time.

1 barplot(table(gender , hand), col = c("skyblue", "blue"), main = "Segmented Bar Plot \nof Gender")

2 legend("topleft", c("females","males"), col =c("skyblue", "blue"), pch = 16, inset =0.05)

ambidextrous left right

Segmented Bar Plot

of Gender

0200400600800

females

males





Pie chartsPie charts display counts as percentages of individuals in eachcategory.

1 pct = round(table(gender) / length(gender) *100)

2 lbls = paste(names(table(gender)), "\n", "%",pct)

3 pie(table(gender), labels = lbls)

female

% 67

male

% 33




Displaying quantitative data

HistogramsDisplay the number of cases in each bin

1 hist(ageinmonths , main = "Histogram of Age inMonths")

Histogram of Age in Months

ageinmonths

Frequency

200 250 300 350

0100

200

300

400





Relative frequency histogramsDisplay the proportion of of cases in each bin.

1 hist(ageinmonths , freq = FALSE , main = "Relative Frequency \n Histogram of Age inMonths", xlab = "Age in Months")

Relative Frequency


Age in Months

Density

200 250 300 350

0.000

0.010

0.020

0.030





Stem-and-Leaf Plots

Preserve individual data values.

1 stem(ageinmonths)

The decimal point is 1 digit(s) to the right of the |

20 | 48

21 | 004444555566666666666666666777777777778888888888889999999999999999

22 | 00000000000000000000000000111111111111111122222222222222222222333333+258

23 | 00000000000000000000000000000000000000000000001111111111111111111111+379

24 | 00000000000000000000000000000000000000000000111111111111111111111111+170

25 | 00000000000001111111111111112222222222222222222223333333344444444445+24

26 | 000000000001111111111222222333334444444444556666778889

27 | 00111222222344566789

28 | 01334558888

29 | 0004569

30 | 267

31 | 02257

32 | 44

33 | 5

34 | 89

35 | 3





Boxplots1 boxplot(ageinmonths , main = "Boxplot of Age in

Months")

200

250

300

350

Boxplot of Age in Months

Five Number Summary (Min, Q1, Median, Q3, Max):

1 fivenum(ageinmonths)

[1] 204 228 235 243 353




Describing distributions numerically

Summary

Categorical variables:

1 summary(hand)

ambidextrous left right

20 112 1193

Quantitative variables:

1 summary(ageinmonths)

Min. 1st Qu. Median Mean 3rd Qu. Max.

204.0 228.0 235.0 237.8 243.0 353.0





Measures of center

Mean (arithmetic average):

1 mean(ageinmonths)

[1] 237.8309

Median (value that divides the histogram into two equalareas):

1 median(ageinmonths)

[1] 235

Mode (the most frequent value): for discrete data

1 as.numeric(names(sort(table(ageinmonths),decreasing = TRUE))[1])

[1] 228





Mode (alternative)

To find the mode, you may also use the Mode function in theprettyR package.

1 install.packages("prettyR")2 library(prettyR)3 Mode(ageinmonths)

[1] "228"





Adding measures to plotsAdding mean and median to a histogram.

1 hist(ageinmonths , main = "Histogram of Age inMonths")

2 abline(v = mean(ageinmonths), col = "blue")3 abline(v = median(ageinmonths), col = "green")4 legend("topright", c("Mean", "Median"), pch =

16, col = c("blue", "green"))


ageinmonths

Frequency

200 250 300 350

0100

200

300

400

Mean

Median





Measures of spread

Range (Min, Max):

1 range(ageinmonths)

[1] 204 353

IQR:

1 IQR(ageinmonths)

[1] 15

Standard deviation:

1 sd(ageinmonths)

[1] 16.03965




1 Preliminaries

2 Data sets


4 Probability ModelsGeometricBinomialPoissonNormal


6 Linear Regression



9 Exercises




Geometric

Geometric distribution

If the probability of success is 0.35, what is the probability that thefirst success will be on the 5th trial?

1 dgeom (4 ,0.35)

[1] 0.06247719

Note: dgeom gives the density (or probability mass function for discrete

variables), pgeom gives the distribution function, qgeom gives the

quantile function, and rgeom generates random deviates. This is true for

the functions used for Binomial, Poisson and Normal calculations as well.




Binomial

Binomial distribution

If the probability of success is 0.35, what is the probability of

3 successes in 5 trials?

1 dbinom (3 ,5 ,0.35)

[1] 0.1811469

at least 3 successes in 5 trials?

1 sum(dbinom (3:5 ,5 ,0.35))

[1] 0.2351694




Poisson

Poisson distribution

The number of traffic accidents per week in a small city hasPoisson distribution with mean equal to 3. What is the probabilityof

two accidents in a week?

1 dpois (2,3)

[1] 0.2240418

at most one accident in a week?

1 sum(dpois (0:1 ,3))

[1] 0.1991483




Normal

Normal distribution

Scores on an exam are distributed normally with a mean of 65 anda standard deviation of 12. What percentage of the students havescores

below 50?

1 pnorm (50 ,65 ,12)

[1] 0.1056498

between 50 and 70?

1 pnorm (70 ,65 ,12)-pnorm (50 ,65 ,12)

[1] 0.5558891




Normal

Normal distribution (cont.)

What is the 90th percentile of the score distribution?

1 qnorm (.90 ,65 ,12)

[1] 80.37862




1 Preliminaries

2 Data sets



5 Hypothesis Testing and Confidence IntervalsOne sample meansTwo sample meansOne sample proportionsTwo sample proportions

6 Linear Regression



9 Exercises




One sample means

Hypothesis testing for one sample means

Is there evidence to suggest that the average age in months forStats 10 students is more than 235 months? Use α = 0.05.

1 sample100 = sample (1:1325 , 100, replace =FALSE)

2 survey.sub = survey[sample100 ,]3 t.test(survey.sub$ageinmonths , alternative = "

greater", mu = 235, conf.level = 0.95)

One Sample t-test

data: survey.sub$ageinmonths

t = 1.5922, df = 99, p-value = 0.05726

alternative hypothesis: true mean is greater than 235

95 percent confidence interval:

234.9118 Inf

sample estimates:

mean of x

237.06




One sample means

Confidence intervals for one sample means

The t.test function prints out a confidence interval as well.

However this function returns a one-sided interval when thealternative is "greater" or "less".

When alternative = "greater" is chosen the lowerconfidence bound is calculated and the upper bound is givenas Inf by default.

When alternative = "less" is chosen the upperconfidence bound is calculated and the lower bound is givenas -Inf by default.

When alternative = "two.sided" is chosen both theupper and the lower confidence bounds are calculated.




One sample means

Confidence intervals for one sample means (cont.)

1 t.test(survey.sub$ageinmonths , alternative = "two.sided", mu = 235, conf.level = 0.90)

One Sample t-test

data: survey.sub$ageinmonths

t = 1.5922, df = 99, p-value = 0.1145

alternative hypothesis: true mean is not equal to 235


234.9118 239.2082

sample estimates:

mean of x

237.06

Note that we changed the confidence level to 0.90 in order tocorrespond to a one-sided hypothesis test with α = 0.05.




One sample means

Confidence intervals for one sample means (cont.)

Alternative calculation of confidence interval:

1 onesample.mean.ci = function(x, conf.level){2 tstar = -qt(p = ((1 - conf.level)/2), df = (

length(x) - 1))3 xbar = mean(x)4 sexbar = sd(x) / sqrt(length(x))5 cilower = xbar - tstar * sexbar6 ciupper = xbar + tstar * sexbar7 return(list = c(cilower , ciupper))8 }9 onesample.mean.ci(survey.sub$ageinmonths ,

0.90)

[1] 234.9118 239.2082




Two sample means

Hypothesis testing and CI for two sample meansIs there a difference between the ages of females and males?Construct a 95% confidence interval for the difference between theaverage ages of females and males.

1 t.test(survey.sub$ageinmonths[survey.sub$gender == "female"], survey.sub$ageinmonths[survey.sub$gender == "male"], alternative = "two.sided", conf.level= 0.95)

Welch Two Sample t-test

data: survey.sub$ageinmonths[survey.sub$gender == "female"] and

survey.sub$ageinmonths[survey.sub$gender == "male"]

t = 1.25, df = 95.736, p-value = 0.2143

alternative hypothesis: true difference in means is not equal to 0


-1.765100 7.768572

sample estimates:

mean of x mean of y

238.1406 235.1389




One sample proportions

Hypothesis testing for one sample proportions

64 out of 100 students in a random sample are females. Is thereevidence to suggest that the population proportion of females isless than 65%? Use a 90% confidence level.

1 prop.test(64, 100, p = 0.65, alternative = "less", conf.level = 0.90)

1-sample proportions test with continuity correction

data: 64 out of 100, null probability 0.65

X-squared = 0.011, df = 1, p-value = 0.4583

alternative hypothesis: true p is less than 0.65


0.0000000 0.7035286

sample estimates:

p

0.64




One sample proportions

Confidence intervals for one sample proportions

Just like the t.test, the prop.test function will calculate boththe upper and the lower bounds of the confidence interval onlywhen alternative = "two.sided" is chosen. Otherwise a lowerbound of 0 or an upper bound of 1 is produced.

1 prop.test(64, 100, p = 0.65, alternative = "two.sided", conf.level = 0.80)

1-sample proportions test with continuity correction

data: 64 out of 100, null probability 0.65


alternative hypothesis: true p is not equal to 0.65


0.5715825 0.7035286

sample estimates:

p

0.64




Two sample proportions

Hypothesis testing and CI for two sample proportions

54 out of 64 females and 32 out of 36 males are right handed. Isthere evidence to suggest that proportions of males and femaleswho are right handed are different?

1 prop.test(c(54 ,32), c(64 ,36))

2-sample test for equality of proportions with continuity correction

data: c(54, 32) out of c(64, 36)


alternative hypothesis: two.sided


-0.2026789 0.1124012

sample estimates:

prop 1 prop 2

0.8437500 0.8888889




1 Preliminaries

2 Data sets




6 Linear RegressionScatterplots, Association, and CorrelationSimple Linear Regression






Scatterplots, Association, and Correlation

ScatterplotsIs there an association between amount of alcohol consumed andmaximum speed?

1 plot(speed ~ alcohol , main = "Scatterplot ofSpeed vs. Alcohol", pch = 20, cex = 0.5)

0 20 40 60 80

050

100

150

Scatterplot of Speed vs. Alcohol

alcohol

speed




Scatterplots, Association, and Correlation

Correlation

1 cor(alcohol , speed , use = "pairwise.complete.obs")

[1] 0.2309745




Simple Linear Regression

Simple Linear Regression

Build a linear regression model predicting speed from alcohol.

1 summary(lm(speed~alcohol))

Call:

lm(formula = speed ~ alcohol)

Residuals:

Min 1Q Median 3Q Max

-90.769 -8.725 1.275 11.275 91.541

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 88.7248 0.6511 136.261 <2e-16 ***

alcohol 0.9469 0.1108 8.549 <2e-16 ***

---

Signif. codes: 0 ’***’ 0.001 ’**’ 0.01 ’*’ 0.05 ’.’ 0.1 ’ ’ 1

Residual standard error: 21.83 on 1297 degrees of freedom

(26 observations deleted due to missingness)

Multiple R-squared: 0.05335, Adjusted R-squared: 0.05262

F-statistic: 73.09 on 1 and 1297 DF, p-value: < 2.2e-16




1 Preliminaries

2 Data sets




6 Linear Regression






Online Resources for R

Download R: http://cran.stat.ucla.edu/

Search Engine for R: rseek.org

R Reference Card:http://cran.r-project.org/doc/contrib/Short-refcard.pdf

UCLA Statistics Information Portal: http:// info.stat.ucla.edu/grad/

UCLA Statistical Consulting Center: http:// scc.stat.ucla.edu



http://cran.stat.ucla.edu/

rseek.org

http://cran.r-project.org/doc/contrib/Short-refcard.pdf

http://info.stat.ucla.edu/grad/

http://scc.stat.ucla.edu


1 Preliminaries

2 Data sets




6 Linear Regression






Upcoming Mini-Courses

May 5, R Stats II: Linear Regression

May 10, R Stats III: Nonlinear Regression

May 12, LaTeX V: Creating Vector Graphics in LaTeX

For a schedule of all mini-courses offered please visithttp:// scc.stat.ucla.edu/mini-courses .



http://scc.stat.ucla.edu/mini-courses


Thank youAny questions?




1 Preliminaries

2 Data sets




6 Linear Regression






Exercises

1 Construct side-by-side box plots for the distribution of amountof time it takes students to get to class (time) by their meansof transportation (walk).

2 Usually younger students live on campus and older studentslive off campus. Is there evidence to suggest this trend in thisdata set? (Use a random sample of 100 students andα = 0.05.)

3 Calculate a 90% confidence interval for the difference betweenthe average ages of students who live on campus and offcampus.




Solution to Exercise 1

1 boxplot(time ~ walk , main = "Time to get toclass \n by type of transportation")

bicycle bus car (by yourself) carpool motorcycle other segway skateboard walk

050

100

150

Time to get to class

by type of transportation

Minutes





1 t.test(survey.sub$ageinmonths[survey.sub$oncampus == "yes"], survey.sub$ageinmonths[survey.sub$oncampus == "no"], alternative = "less", conf.level =0.95)


data: survey.sub$ageinmonths[survey.sub$oncampus == "yes"] and

survey.sub$ageinmonths[survey.sub$oncampus == "no"]

t = -5.3322, df = 34.867, p-value = 2.964e-06

alternative hypothesis: true difference in means is less than 0


-Inf -10.85376

sample estimates:

mean of x mean of y

232.6111 248.5000





1 t.test(survey.sub$ageinmonths[survey.sub$oncampus == "yes"], survey.sub$ageinmonths[survey.sub$oncampus == "no"], alternative = "two.sided", conf.level= 0.90)


data: survey.sub$ageinmonths[survey.sub$oncampus == "yes"] and

survey.sub$ageinmonths[survey.sub$oncampus == "no"]

t = -5.3322, df = 34.867, p-value = 5.929e-06

alternative hypothesis: true difference in means is not equal to 0


-20.92402 -10.85376

sample estimates:

mean of x mean of y

232.6111 248.5000



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