title: “WPA#2” author: “Rebekka Herz” date: “23. April 2015” output: html_document

Coding Basics

  1. Find the three errors in this code:
mydata <- rnorm(100, mean = 0, sd = 1)   # Can't have negative SD, can't have space in object name
data.mean <- mean(mydata) # Can't have space in object name
data.mean # Remove quotation marks to refer to object
## [1] -0.0486427
#correct code
mydata <- rnorm(100, mean = 0, sd = 1) 
data.mean <- mean(mydata) 
data.mean
## [1] -0.01412308
  1. Create a vector called “mixed” with the values [“batman”, 20, “superman”, 10]. Now print the vector “mixed” - did anything in the vector change and if so, why?
mixed <- c("batman", 20, "superman", 10) 
mixed
## [1] "batman"   "20"       "superman" "10"
#In the Output there are now quotation marks aournd "20" and "10". It changed because R changed all the numeric scalers to characters. 

#[1] "batman"   "20"       "superman" "10"
  1. The forecast for the high temperatures for the next 5 days is 17, 21, 18, 21, 14. Your American penpal wants to know what the temperatures are in Fahrenheit. Convert these temperatures to Fahrenheit using basic math operations. Call this vector “fahrenheit.” Hint: The formula to convert celcius to Fahrenheit is y = x * 9/5 + 32, where x is the temperature in celcius and y is the temperature in Fahrenheit.
Celsius <- c(17,21,18,21,14)
Celsius * 9/5 + 32
## [1] 62.6 69.8 64.4 69.8 57.2
Fahrenheit <- c(62.6, 69.8, 64.4, 69.8, 57.2) 
Fahrenheit # [1] 62.6 69.8 64.4 69.8 57.2
## [1] 62.6 69.8 64.4 69.8 57.2
  1. Create the following sequence of numbers [10, 20, 30, 40, 50, 12, 13, 14, 97, 107, 117, 127, 137] using a combination of the c(), a:b, and seq() functions.
a <- seq(from = 10, to = 50, by = 10)
a
## [1] 10 20 30 40 50
b <- 12:14
b
## [1] 12 13 14
c <- seq(from = 97, to = 117, by = 10)
c
## [1]  97 107 117
c(a,b,c)
##  [1]  10  20  30  40  50  12  13  14  97 107 117
# [1]  10  20  30  40  50  12  13  14  97 107 117
  1. Create the sequence [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5] using a combination of the rep() and a:b functions.
rep(1:5, each = 2, times = 2) 
##  [1] 1 1 2 2 3 3 4 4 5 5 1 1 2 2 3 3 4 4 5 5
#[1] 1 1 2 2 3 3 4 4 5 5 1 1 2 2 3 3 4 4 5 5
  1. Create a vector called “squares” that contains the squares (x^2) of the values from 1 to 10. What is the mean and median value of the squares from 1 to 10?
"numbers" <- 1:10
numbers 
##  [1]  1  2  3  4  5  6  7  8  9 10
# [1]  1  2  3  4  5  6  7  8  9 10

"squares" <- numbers^2
squares
##  [1]   1   4   9  16  25  36  49  64  81 100
# [1]   1   4   9  16  25  36  49  64  81 100

mean(squares) # [1] 38.5
## [1] 38.5
median(squares) # [1] 30.5
## [1] 30.5
#alternative solution: 
squares <- (1:10) ^ 2
mean(squares)
## [1] 38.5
#[1] 38.5

Sampling Data and Descriptive Statistics

  1. Create a vector called “normal.samp” with 50 values from a Normal distribution with population mean 30 and standard deviation 5
normal.samp <- rnorm(50, mean = 30, sd = 5)
normal.samp
##  [1] 33.37964 27.31284 26.24168 27.93522 29.85343 24.44303 33.18505
##  [8] 28.61569 31.53742 30.77116 18.72977 32.45014 32.89467 26.30186
## [15] 36.63767 30.36546 27.94786 37.33674 28.76847 33.85961 23.71574
## [22] 22.85559 21.17301 37.99507 32.13158 32.60755 22.80834 28.62620
## [29] 30.60299 25.67926 36.77393 34.60707 39.17245 31.00509 30.88363
## [36] 24.65534 27.74486 45.20054 31.67259 31.78011 38.10723 26.42067
## [43] 27.37119 37.04119 20.25604 38.86526 35.85145 32.85908 30.48115
## [50] 31.13338
#  [1] 31.60301 23.15112 31.45615 36.92773 30.84523 35.34461 22.07055 28.08075 34.32417
# [10] 32.80625 22.80081 24.96917 25.29822 30.26726 27.13027 31.10274 25.82742 29.34823
# [19] 32.83665 36.04233 30.63610 25.24295 26.68743 31.72297 24.60073 31.04826 23.48540# 
# [28] 28.15094 26.78531 35.45582 31.68297 28.42584 29.10464 30.26134 33.05441 28.53102
# [37] 16.60142 23.81078 32.88284 35.67945 33.47071 31.43286 20.00966 35.74889 37.17930
# [46] 27.60280 41.44553 14.81109 24.78237 31.23251
  1. Create a vector with 25 values from a Uniform distribution where there minimum and maximum are determined by the results in your previous vector called “normal.samp”. Specifically, make the the minimum value of the distribution to be the minimum value of “normal.samp” and the maximum value of the distribution to be the median of “normal.samp.”
min(normal.samp) 
## [1] 18.72977
#[1] 19.60059

max(normal.samp)
## [1] 45.20054
# 41.83516

runif(n=25, min = 19.60059, max = 41.83516)
##  [1] 26.19445 22.08059 29.23311 40.34769 38.59218 29.98552 23.98050
##  [8] 25.34036 24.19495 23.94771 26.66968 29.60815 27.05277 21.90252
## [15] 26.84882 39.12309 39.99840 29.62902 33.73654 25.73451 25.82408
## [22] 29.58814 28.50130 23.35433 19.69603
# [1] 36.48034 20.04680 28.06982 20.73189 37.34162 40.13864 31.66503 38.55255 32.57586
# [10] 34.46048 30.96945 36.56002 39.68646 37.84349 21.18885 19.68722 20.76120 38.86818
# [19] 32.41315 26.57874 40.93390 32.74553 31.41625 28.13726 26.70572

#alternative solution: 
normal.samp <- rnorm(50, mean = 30, sd = 5)
runif(n = 25, min = min(normal.samp), max = median(normal.samp))
##  [1] 18.30733 23.66720 20.47834 18.93364 28.57387 27.58160 29.68644
##  [8] 23.92159 30.67031 31.65012 24.58033 25.83212 29.27750 30.31890
## [15] 22.52411 18.34373 23.14209 28.40405 21.16719 18.32079 25.17423
## [22] 25.78646 23.56804 28.08079 29.41126
# [1] 31.12966 19.21864 24.51438 19.56310 21.39904 23.57787 27.27583 29.42631 15.22104
# [10] 22.06297 29.24186 29.03857 17.40550 30.76306 17.89672 19.66210 19.94212 24.77739
#[19] 21.13504 26.23833 24.28667 26.43533 23.56719 17.68813 26.03520
  1. What should the mean of the sum of a standard normal distribution (normal with mean of 0 and standard deviation of 1) and a uniform distribution with minimum of 0 and maximum of 10 be? Test your prediction by generating the relevant simulation data with a very large sample size.
# It should be 5. The mean of the sum of two random variables should be the sum of the means. In this case, that's 0 + 5.

sample.1 <- rnorm(1000, mean = 0, sd = 1)
sample.2 <- runif(1000, min = 0, max = 10)
sum.sample <- sample.1 + sample.2
mean(sum.sample)
## [1] 5.017553
# [1] 5.129353
  1. Simulate 100 flips of a fair coin. Do this once using the numeric values 0 and 1 (for heads and tails respectively), and once with string values “Heads” and “Tails”.
sample(x = c(0, 1),
       size = 100,
       replace = T,
       prob = c(.5, .5)
       )
##   [1] 0 1 1 0 1 0 1 1 1 0 0 1 0 0 1 0 0 1 1 1 0 1 0 0 0 0 1 0 1 1 1 1 1 1 0
##  [36] 1 1 0 0 0 1 0 0 0 1 1 1 1 0 1 0 0 0 1 0 0 1 1 0 1 0 1 0 0 0 0 0 0 1 0
##  [71] 1 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 1 0 1 1 1 0
# [1] 1 0 0 0 0 1 0 0 0 1 0 0 0 1 1 0 0 1 1 0 0 1 1 0 1 0 0 1 1 1 0 1 1 1 0 0 0 1 0 1 1 1 1 0
# [45] 1 1 0 1 0 1 1 1 0 1 0 0 0 1 1 1 1 1 0 1 1 1 0 1 1 1 1 0 0 1 0 0 0 1 1 1 1 0 0 1 0 1 1 0
# [89] 1 0 1 0 1 0 0 1 1 1 1 1

sample(x = c("Heads", "Tails"),
       size = 100,
       replace = T,
       prob = c(.5, .5)
       )
##   [1] "Tails" "Heads" "Tails" "Tails" "Heads" "Tails" "Tails" "Heads"
##   [9] "Heads" "Heads" "Tails" "Tails" "Tails" "Tails" "Tails" "Tails"
##  [17] "Heads" "Tails" "Tails" "Heads" "Heads" "Heads" "Tails" "Tails"
##  [25] "Tails" "Tails" "Heads" "Tails" "Heads" "Tails" "Heads" "Heads"
##  [33] "Tails" "Tails" "Tails" "Tails" "Tails" "Heads" "Tails" "Heads"
##  [41] "Heads" "Heads" "Tails" "Heads" "Tails" "Heads" "Tails" "Heads"
##  [49] "Heads" "Tails" "Tails" "Tails" "Tails" "Tails" "Tails" "Heads"
##  [57] "Tails" "Heads" "Tails" "Heads" "Heads" "Heads" "Tails" "Tails"
##  [65] "Heads" "Tails" "Tails" "Heads" "Tails" "Heads" "Heads" "Heads"
##  [73] "Tails" "Heads" "Tails" "Heads" "Heads" "Tails" "Heads" "Tails"
##  [81] "Heads" "Heads" "Tails" "Heads" "Tails" "Tails" "Tails" "Tails"
##  [89] "Tails" "Tails" "Heads" "Tails" "Heads" "Tails" "Tails" "Tails"
##  [97] "Tails" "Tails" "Heads" "Tails"
#  [1] "Tails" "Heads" "Tails" "Heads" "Tails" "Heads" "Heads" "Tails" "Tails" "Heads" #"Tails"
# [12] "Heads" "Heads" "Heads" "Tails" "Heads" "Heads" "Heads" "Tails" "Tails" "Tails" #"Tails"
# [23] "Tails" "Heads" "Heads" "Heads" "Tails" "Tails" "Heads" "Heads" "Heads" "Tails" #"Heads"
# [34] "Tails" "Heads" "Tails" "Heads" "Tails" "Tails" "Tails" "Heads" "Heads" "Tails" #"Heads"
# [45] "Heads" "Heads" "Heads" "Heads" "Tails" "Tails" "Heads" "Tails" "Tails" "Tails"  #"Tails"
# [56] "Heads" "Tails" "Tails" "Tails" "Tails" "Heads" "Heads" "Tails" "Heads" "Tails"  #"Tails"
# [67] "Heads" "Tails" "Heads" "Tails" "Tails" "Heads" "Tails" "Tails" "Tails" "Heads" #"Tails"
# [78] "Tails" "Tails" "Tails" "Heads" "Heads" "Heads" "Tails" "Tails" "Heads" "Tails" #"Heads"
# [89] "Heads" "Heads" "Tails" "Tails" "Tails" "Tails" "Heads" "Tails" "Heads" "Heads" #"Tails"
# [100] "Heads"
  1. Imagine you have a bag containing 30 red balls, 50 green balls, and 20 yellow balls. Simulate a sample of 20 balls drawn from a this bag. Do this once with replacement (call the vector: “balls.replacement”), and once without replacement (call the vector: “balls.noreplacement”).
# vector representing the bag of balls
bag <- c(rep("red", 30), rep("green", 50), rep("yellow", 20))

# vector representing the probability of selecting each of the 100 balls
balls.prob <- rep(1/100, times = 100)

# Now, sample from the bag!

# With replacement
balls.replacement <-sample(x = bag,
                           size = 20,
                           replace = T,
                           prob = balls.prob
                           )
balls.replacement
##  [1] "green"  "green"  "green"  "green"  "green"  "green"  "green" 
##  [8] "yellow" "green"  "green"  "yellow" "red"    "green"  "yellow"
## [15] "green"  "yellow" "yellow" "red"    "yellow" "green"
# [1] "red"    "yellow" "yellow" "green"  "green"  "red"    "green"  "green"  "green" 
#[10] "green"  "green"  "green"  "green"  "red"    "green"  "red"    "red"    "green" 
# [19] "green"  "green" 


# Without replacement
balls.noreplacement <- sample(x = bag,
                              size = 20,
                              replace = F,
                              prob = balls.prob
                              )
balls.noreplacement
##  [1] "yellow" "green"  "yellow" "red"    "green"  "green"  "green" 
##  [8] "green"  "red"    "green"  "green"  "red"    "green"  "red"   
## [15] "green"  "red"    "yellow" "red"    "yellow" "yellow"
# [1] "green"  "green"  "red"    "red"    "red"    "green"  "red"    "red"    "red"   
#[10] "red"    "yellow" "red"    "green"  "green"  "yellow" "red"    "green"  "green" 
#[19] "green"  "red"
  1. The “Expected Value” of a gamble is the average outcome of a gamble played infinitely many times. To calculate the expected value of a gamble, you calculate the sum of the products of each outcome and its probability. For example, the expected value of a gamble that gives 100EU with probability .5 and -100EU with probability .5 is given by 100 * .5 + (-100) * .5 = 0. This means that, on average, this gamble won’t result in any gains or losses. Now, imagine that there is a gamble that gives 2,000EU with probability .1 and -100EU with probability .9.

Calculate the expected value of this gamble using the formula for expected values. Hint: Use vector multiplication and the function sum() Conduct a simulation using the functions sample() and mean() to test your calculation.

#Expected value
2000 * .1 + -100 * .9
## [1] 110
#[1] 110

#vector notation

outcomes <- c(2000, -100)
probabilities <- c(.1, .9)

sum(outcomes * probabilities)
## [1] 110
#[1] 110

# simulation
simulation <- sample(x = c(2000, -100),
                     replace = T,
                     size = 1e6,
                     prob = c(.1, .9)
                     )

mean(simulation)
## [1] 110.5124
#[1] 110.2478