Getting Started
load(url("http://www.openintro.org/stat/data/kobe.RData"))
head(kobe)
## vs game quarter time
## 1 ORL 1 1 9:47
## 2 ORL 1 1 9:07
## 3 ORL 1 1 8:11
## 4 ORL 1 1 7:41
## 5 ORL 1 1 7:03
## 6 ORL 1 1 6:01
## description basket
## 1 Kobe Bryant makes 4-foot two point shot H
## 2 Kobe Bryant misses jumper M
## 3 Kobe Bryant misses 7-foot jumper M
## 4 Kobe Bryant makes 16-foot jumper (Derek Fisher assists) H
## 5 Kobe Bryant makes driving layup H
## 6 Kobe Bryant misses jumper M
kobe$basket[1:9]
## [1] "H" "M" "M" "H" "H" "M" "M" "M" "M"
# Exercise 1 - What does a streak length of 1 mean, i.e. how
# many hits and misses are in a streak of 1? What about a
# streak length of 0?
# Answer: Streak length of 1 is a hit (into the basket)
# followed by a miss Streak length of 0 is a miss (not making
# to the basket) followed by another miss
kobe_streak <- calc_streak(kobe$basket)
kobe_streak_df <- as.data.frame(kobe_streak)
names(kobe_streak_df) <- c("# of Streaks")
# kobe_streak_df
barplot(table(kobe_streak), xlab = "Number of Continuous Streaks",
ylab = "Number of Occurences", cex.axis = par("cex.axis"),
beside = FALSE)
# Exercise 2 - Describe the distribution of Kobe’s streak
# lengths from the 2009 NBA finals. What was his typical
# streak length? How long was his longest streak of baskets?
# Answer:
boxplot(kobe_streak_df)
nrow(as.data.frame(kobe_streak_df[kobe_streak_df$`# of Streaks` ==
"0", ])) # 39
## [1] 39
nrow(as.data.frame(kobe_streak_df[kobe_streak_df$`# of Streaks` ==
"1", ])) # 24
## [1] 24
nrow(as.data.frame(kobe_streak_df[kobe_streak_df$`# of Streaks` ==
"2", ])) # 6
## [1] 6
nrow(as.data.frame(kobe_streak_df[kobe_streak_df$`# of Streaks` ==
"3", ])) # 6
## [1] 6
nrow(as.data.frame(kobe_streak_df[kobe_streak_df$`# of Streaks` ==
"4", ])) # 1
## [1] 1
# The distribution is righly skewed
# Typical Streak Length is 0
max(kobe_streak)
## [1] 4
# Longest streak length: 4
Simulations in R
outcomes <- c("heads", "tails")
sample(outcomes, size = 1, replace = TRUE)
## [1] "heads"
sim_fair_coin <- sample(outcomes, size = 100, replace = TRUE)
sim_fair_coin
## [1] "heads" "heads" "tails" "heads" "tails" "heads" "tails" "tails"
## [9] "heads" "heads" "heads" "heads" "tails" "tails" "heads" "tails"
## [17] "tails" "tails" "heads" "tails" "heads" "heads" "heads" "heads"
## [25] "tails" "tails" "tails" "heads" "heads" "tails" "tails" "heads"
## [33] "heads" "heads" "tails" "tails" "tails" "tails" "heads" "tails"
## [41] "heads" "tails" "heads" "tails" "heads" "tails" "tails" "heads"
## [49] "heads" "heads" "heads" "tails" "tails" "tails" "heads" "tails"
## [57] "heads" "heads" "heads" "tails" "tails" "tails" "tails" "heads"
## [65] "heads" "heads" "heads" "tails" "tails" "heads" "heads" "tails"
## [73] "heads" "tails" "heads" "tails" "heads" "tails" "heads" "tails"
## [81] "tails" "tails" "heads" "heads" "tails" "heads" "heads" "tails"
## [89] "tails" "heads" "heads" "heads" "tails" "heads" "heads" "tails"
## [97] "tails" "tails" "tails" "heads"
table(sim_fair_coin)
## sim_fair_coin
## heads tails
## 51 49
sim_unfair_coin <- sample(outcomes, size = 100, replace = TRUE,
prob = c(0.2, 0.8))
table(sim_unfair_coin)
## sim_unfair_coin
## heads tails
## 22 78
# Exercise 3 - In your simulation of flipping the unfair coin
# 100 times, how many flips came up heads?
# Answer:
flip_coin_df <- as.data.frame(sample(outcomes, size = 100, replace = TRUE))
names(flip_coin_df) <- c("Outcome")
nrow(as.data.frame(flip_coin_df[flip_coin_df$Outcome == "heads",
])) # 54 in my R studio but it may change during RMD execution
## [1] 46
Simulating the Independent Shooter
outcomes <- c("H", "M")
sim_basket <- sample(outcomes, size = 1000, replace = TRUE)
table(sim_basket)
## sim_basket
## H M
## 485 515
# Exercise 4 - What change needs to be made to the sample
# function so that it reflects a shooting percentage of 45%?
# Make this adjustment, then run a simulation to sample 133
# shots. Assign the output of this simulation to a new
# object called sim_basket.
# Answer:
sim_basket <- sample(outcomes, size = 133, replace = TRUE, prob = c(0.45,
0.55))
table(sim_basket)
## sim_basket
## H M
## 43 90
On your own
Comparing Kobe Bryant to the Independent Shooter
# Using `calc_streak`, compute the streak lengths of
# `sim_basket`.
calc_streak(sim_basket)
## [1] 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 1 0 3 0 0 1 0 0 0 0 0 1
## [36] 0 0 1 0 1 1 0 0 0 1 1 0 0 0 0 0 2 1 1 0 0 3 0 1 0 0 3 0 2 0 0 0 0 0 1
## [71] 0 2 0 0 1 0 0 0 4 0 0 1 0 0 1 0 1 0 2 0 1
# Describe the distribution of streak lengths. What is the
# typical streak length for this simulated independent
# shooter with a 45% shooting percentage? How long is the
# player's longest streak of baskets in 133 shots?
sim_basket_df <- as.data.frame(calc_streak(sim_basket))
names(sim_basket_df) <- c("# of Streaks")
barplot(table(sim_basket_df), xlab = "Number of Continuous Streaks",
ylab = "Number of Occurences", cex.axis = par("cex.axis"),
beside = FALSE)
nrow(as.data.frame(sim_basket_df[sim_basket_df$`# of Streaks` ==
"0", ])) # 39
## [1] 61
nrow(as.data.frame(sim_basket_df[sim_basket_df$`# of Streaks` ==
"1", ])) # 24
## [1] 22
nrow(as.data.frame(sim_basket_df[sim_basket_df$`# of Streaks` ==
"2", ])) # 6
## [1] 4
nrow(as.data.frame(sim_basket_df[sim_basket_df$`# of Streaks` ==
"3", ])) # 6
## [1] 3
nrow(as.data.frame(sim_basket_df[sim_basket_df$`# of Streaks` ==
"4", ])) # 1
## [1] 1
nrow(as.data.frame(sim_basket_df[sim_basket_df$`# of Streaks` ==
"5", ])) # 1
## [1] 0
nrow(as.data.frame(sim_basket_df[sim_basket_df$`# of Streaks` ==
"6", ])) # 1
## [1] 0
# 1 - Describe the distribution of streak lengths The
# distribution is righly skewed
# 2 - What is the typical streak length for this simulated
# independent shooter with a 45% shooting percentage?
# Typical streak length is 0
# 3 - How long is the player's longest streak of baskets in
# 133 shots? Longest streak is 6 and it occurred one time
# If you were to run the simulation of the independent
# shooter a second time, how would you expect its streak
# distribution to compare to the distribution from the
# question above? Exactly the same? Somewhat similar? Totally
# different? Explain your reasoning.
# Answer - The outcome will not be fixed because the
# simulation is picking the Hit or Miss randomly so I did not
# get the exact results
# How does Kobe Bryant's distribution of streak lengths
# compare to the distribution of streak lengths for the
# simulated shooter? Using this comparison, do you have
# evidence that the hot hand model fits Kobe's shooting
# patterns? Explain.
# Answer - Kobe Bryant's distribution is also rightly skewed
# just like the simulation. The simulated model fits Kobe
# Bryant's pattern however the max hot hand differs - 4 for
# Kobe vs 6 for Simulation