Unit 2 Assignments #8-#16 - R Syntax
Brad Thiessen
July 10, 2014
Loading packages
Let’s load some packages we’ll use in these assignments:
require(ggplot2)
require(dplyr)
require(mosaic)
require(parallel)Assignment #8: Minority Coaches
Likelihood of 6 or fewer minority coaches at 3rd base.
Here’s our dataset:
| Observed | Caucasian | Minority | Total |
|---|---|---|---|
| 1st Base | 15 | 15 | 30 |
| 3rd Base | 34 | 6 | 30 |
| Total | 49 | 21 | 60 |
Our sample of 60 coaches includes 30 3rd base coaches. Under our null hypothesis, these 30 people would have coached 3rd base regardless of their race.
We can now simulate this study.
## Create 60 coaches (30 who will coach 3rd base)
## 3rd Base = 1 and 1st Base = 0
Coaches <- c(rep(1, 30), rep(0, 30))
## Let's select 21 of these people at random to "be minorities"
## We'll simulate this process 50,000 times
ThirdBase <- do(50000) * sample(Coaches, 21, replace=FALSE)
## Now, let's count the number of 3rd base coaches who were selected in each replication
ThirdBase$sum <- rowSums(ThirdBase)
## Let's create a histogram showing the number of people who improved
histogram( ~sum, width=1, main="Minority 3rd Base Coaches", v=6, col="lightgrey", data=ThirdBase)
## Find the p-value -- P(6 or fewer 3rd base coaches | race has no effect)
## The answer should be close to 1.5%.
prop( ~(sum<=6), data=ThirdBase)## TRUE
## 0.0139
Assignment #8b: Practice Problems
Rather than simulating each scenario, let’s just use R’s built-in probability distributions.
## Question #1: Geometric - P(1st win on 3rd trial)
# Remember the first input is the number of trials BEFORE the first win
dgeom(2, .1, log=FALSE)## [1] 0.081## Question #2: Binomial - P(X >= 15 | p=.75)
# Same as 1 - P(X < 14)
1-pbinom(14, 16, 0.750, lower.tail=TRUE, log.p=FALSE)## [1] 0.06348## Question #3: Geometric - P(1st win on 1st or 2nd trial)
# P(X = 1) + P(X = 2)
dgeom(0, .75, log=FALSE) + dgeom(1, .75, log=FALSE)## [1] 0.9375## Question #4: Hypergeometric - P(0, 1, 2, 3, or 4)
# P(X = 0)
dhyper(0, 10, 40, 4, log = FALSE)## [1] 0.3968# P(X = 1)
dhyper(1, 10, 40, 4, log = FALSE)## [1] 0.429# P(X = 2)
dhyper(2, 10, 40, 4, log = FALSE)## [1] 0.1524# P(X = 3)
dhyper(3, 10, 40, 4, log = FALSE)## [1] 0.02084# P(X = 4)
dhyper(4, 10, 40, 4, log = FALSE)## [1] 0.0009119## Question #5: NegBin - P(X = 3 in 4 or fewer trials)
pnbinom(1, 3, .6, lower.tail = TRUE, log.p = FALSE)## [1] 0.4752## Question #6: Poisson - P(X = 0 | lambda = .25)
dpois(x=0, lambda=.25)## [1] 0.7788# Part b: P(X <= 1 | lambda = .25)
ppois(q=1, lambda=.25, lower.tail=TRUE, log.p=FALSE)## [1] 0.9735## Question #7: Hypergeometric - P(X=0)
dhyper(0, 6, 10, 5, log = FALSE)## [1] 0.05769## Question #8: Poisson - P(X <= 3 | lambda = 47/7)
ppois(q=3, lambda=47/7, lower.tail=TRUE, log.p=FALSE)## [1] 0.09793## Question #10: Exponential - P(X <= 7 | lambda = 1/20)
pexp(7, rate=1/20, lower.tail=TRUE, log.p=FALSE)## [1] 0.2953
Assignment #9a: Industrial Robot
Questions 2-3: Find area under curve; sketch cdf.
## Find the antiderivative
cdf <- antiD(-(x^4) + (6/5) ~ x)##
## Attaching package: 'MASS'
##
## The following object is masked from 'package:dplyr':
##
## select## Find definite integral from 0 to 1
cdf(x=1) - cdf(x=0)## [1] 1## Plot cdf
plotFun(cdf, x.lim=range(0,1))
Questions 4-7: Find probabilities; median; mean
## 4. Find P(X<0.5)
cdf(x=0.5) - cdf(x=0)## [1] 0.5938## 5. Find P(X>0.75)
cdf(x=1) - cdf(x=0.75)## [1] 0.1475## 6. Find the median
# First, let's get a visual estimate
plotFun(cdf, x.lim=range(0,1))
plotFun(0.5 ~ x, add=TRUE, col="black")
# Now, let's find the solution (between 0.3 and 0.5)
# We set 0.5 = -(x^5 / 5) + (6x / 5)
# or 0 = -(x^5 / 5) + (6x / 5) - 0.5
# We tell the computer to look somewhere between 0.3 and 0.5
findZeros(-(x^5/5)+(6*x/5)-(0.5) ~ x, x.lim = range(0.3, 0.5))## x
## 1 0.4188## 7. Find the expected value
# Find the definite integral from 0 to 1 of xf(x)
expectedIntegral <- antiD(-(x^5) + ((6 * x)/5) ~ x)
## Find definite integral from 0 to 1
expectedIntegral(x=1) - expectedIntegral(x=0)## [1] 0.4333
Assignment #10: Fishing Problem
All questions appear in the comments.
## A: Exponential - P(X > 60 | lambda = 1/30)
1-pexp(60, rate=1/30, lower.tail=TRUE, log.p=FALSE)## [1] 0.1353## B: Exponential - P(X <10 | lambda = 1/30)
pexp(10, rate=1/30, lower.tail=TRUE, log.p=FALSE)## [1] 0.2835## C: Exponential - P(20 < X < 40 | lambda = 1/30)
pexp(40, rate=1/30, lower.tail=TRUE, log.p=FALSE) -
pexp(20, rate=1/30, lower.tail=TRUE, log.p=FALSE)## [1] 0.2498## D: Normal - P(12 < X < 18 | lambda = 1/30)
xpnorm(18, mean = 15, sd = 3.5, plot=FALSE, verbose=FALSE) -
xpnorm(12, mean = 15, sd = 3.5, plot=FALSE, verbose=FALSE)## [1] 0.6086## E: Geometric - P(1st success on 1st, 2nd, or 3rd trial)
pgeom(2, .6086341, log=FALSE)## [1] 0.9401## F: Binomial - P(4 successes in 4 trials)
dbinom(4, 4, 0.6086341)## [1] 0.1372## G: Negative Binomial - P(4th win on 7th trial)
dnbinom(3, 4, .6086341)## [1] 0.1645
Assignment #11: Normal Probability Practice
All questions appear in the comments.
## 1: P(X < 10)
xpnorm(10, mean = 11.4, sd = 1.8)##
## If X ~ N(11.4,1.8), then
##
## P(X <= 10) = P(Z <= -0.778) = 0.2184
## P(X > 10) = P(Z > -0.778) = 0.7816
## [1] 0.2184## 2: P(X > 14)
xpnorm(14, mean = 11.4, sd = 1.8, lower.tail=FALSE)##
## If X ~ N(11.4,1.8), then
##
## P(X <= 14) = P(Z <= 1.444) = 0.9257
## P(X > 14) = P(Z > 1.444) = 0.0743
## [1] 0.07431## 3: P(9 < X < 12)
xpnorm(12, mean = 11.4, sd = 1.8, verbose=FALSE, plot=FALSE) -
xpnorm(9, mean = 11.4, sd = 1.8, verbose=FALSE, plot=FALSE)## [1] 0.5393# Plot curve for question #3
curve(dnorm(x,11.4,1.8), xlim=c(5, 18), col="steelblue", lwd=3)
cord.x <- c(9,seq(9,12,0.01),12)
cord.y <- c(0,dnorm(seq(9, 12, 0.01), mean=11.4, sd=1.8),0)
polygon(cord.x,cord.y,col='skyblue')
## 4: P(X>30)
xpnorm(30, mean = 11.4, sd = 1.8, lower.tail=FALSE, plot=FALSE, verbose=FALSE)## [1] 2.49e-25## 5: P(11 < X < 12)
xpnorm(12, mean = 11.4, sd = 1.8, verbose=FALSE, plot=FALSE) -
xpnorm(11, mean = 11.4, sd = 1.8, verbose=FALSE, plot=FALSE)## [1] 0.2185## 7: 95th percentile
xqnorm(.95, mean = 11.4, sd = 1.8)## P(X <= 14.3607365285126) = 0.95
## P(X > 14.3607365285126) = 0.05
## [1] 14.36## 8: 5th percentile
xqnorm(.05, mean = 11.4, sd = 1.8)## P(X <= 8.43926347148735) = 0.05
## P(X > 8.43926347148735) = 0.95
## [1] 8.439## 9: 97.5th percentile
xqnorm(.975, mean = 11.4, sd = 1.8)## P(X <= 14.9279351721721) = 0.975
## P(X > 14.9279351721721) = 0.025
## [1] 14.93