| title | author | output | ||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
Statistical Inference Course Notes |
Andrey G |
|
w1
################## center of mass
##Ex
library(manipulate)
data(galton)
myHist <- function(mu){
g <- ggplot(galton, aes(x = child))
g <- g + geom_histogram(fill = "salmon",
binwidth=1, aes(y = ..density..), colour = "black")
g <- g + geom_density(size = 2)
g <- g + geom_vline(xintercept = mu, size = 2)
mse <- round(mean((galton$child - mu)^2), 3)
g <- g + labs(title = paste('mu = ', mu, ' MSE = ', mse))
g
}
manipulate(myHist(mu), mu = slider(62, 74, step = 0.5))
w2
################## binomial distr
choose(8, 7) * .5 ^ 8 + choose(8,8) * .5 ^ 8
#or
pbinom(6, size = 8, prob = .5, lower.tail = FALSE)
##################
################## normal distr
#Ex1: Assume the number of daily ad clicks for companies is approximately
#normally distributed with a mean of 1,020 clicks per day,
#and a standard deviation of 50 clicks per day.
#What is a probability of 1,160 clicks?
pnorm(1160, mean = 1020, sd = 50, lower.tail = FALSE)
# or
pnorm(2.8, lower.tail = FALSE) # (1160-1020)/50 = 2.8
#Ex2: Assume that the number of daily ad clicks for this company is
#approximately normally distributed with a mean of 1020 and a standard
#deviation of 50. What number of daily ad clicks would represent the one
#where 75% have fewer clicks?
# 1020 + 50 = 1sigma (68 %)
qnorm(0.75, mean = 1020, sd = 50)
##################
################## poisson distr
#Ex: if the number of people that show up to a bus stop is Poisson with
#a mean of 2.5 people per hour. We watch the bus stop for four hours.
#What's the probability that three or four, three or fewer people show up
#the whole time?
ppois(3, lambda = 2.5 * 4)
################## law of large numbers
#Ex: coin flip
n <- 1000
means <- cumsum(sample(0:1, n, replace = TRUE))/(1:n)
plot(means) #mean of the large population is close to M(X)
##################
################## confidence intervals
#Ex: give confidence intervals for the average height of sons
#Galtons data
library(UsingR)
data(father.son)
x <- father.son$sheight
(mean(x) + c(-1,1) * qnorm(0.975) * sd(x)/sqrt(length(x))) / 12
#######
#Ex2: you were running for political office and your campaign advisor told you
#that in a random sample of 100 likely voters, 56 intended to vote for you.
#Can you relax?
# 1. quick (intuitive) calculation
#1/sqrt(100)= 0.1 - 95% interval of (0.46, 0.66)
# 2. binomial interval
0.56 + c(-1,1) * qnorm(0.975) * sqrt(0.56 * 0.44/100)
#or
binom.test(56, 100)$conf.int
# answer: nope, more adv needed
#########
#Ex3: Simulation
# 20 coin flips
# 1000 simulations
n <- 20 # 20 coin flips
pvals <- seq(0.1, 0.9, by = 0.05)
nosim <- 1000 # 1000 simulations
coverage <- sapply(pvals, function(p) { #loop
phats <- rbinom(nosim, prob = p, size = n)/n #foreach success prob
#generate 1000 sets of 10 coin flips
ll <- phats - qnorm(0.975) * sqrt(phats * (1 - phats)/n) #lower limit of confidence
ul <- phats + qnorm(0.975) * sqrt(phats * (1 - phats)/n) #upper -//- for each case
mean(ll < p & ul > p) #calculate the proportion of times that they can
#cover that true value of p that I used to simulate the data
})
ggplot(data.frame(pvals, coverage), aes(x = pvals, y = coverage)) +
geom_line(size = 2) + geom_hline(yintercept = 0.95) + ylim(.75, 1.0)
#result: not good. Need more coin flips in a row: n <- 20 change to n <- 100
#########
#Ex4:a nuclear pump that failed 5 times out of 94.32 days,
#given 95% confidence interval for the failure rate per day
x <- 5
t <- 94.32
lambda <- x/t
round(lambda + c(-1,1) * qnorm(0.975) * sqrt(lambda/t), 3)
#or
poisson.test(x, T = 94.32)$conf
#result: variance estimation: confidence interval
#########
#Ex5: Small lambda simulations
lambdavals <- seq(0.005, 0.10, by = .01);
nosim <- 1000;
t <- 100
# calculate coverage using Poisson intervals
coverage <- sapply(lambdavals, function(lambda){
# calculate Poisson rates
lhats <- rpois(nosim, lambda = lambda * t) / t
# lower bound of 95% CI
ll <- lhats - qnorm(.975) * sqrt(lhats / t)
# upper bound of 95% CI
ul <- lhats + qnorm(.975) * sqrt(lhats / t)
# calculate percent of intervals that contain lambda
mean(ll < lambda & ul > lambda)
})
# plot CI results vs 95%
ggplot(data.frame(lambdavals, coverage), aes(x = lambdavals, y = coverage)) + geom_line(size = 2) + geom_hline(yintercept = 0.95)+ylim(0, 1.0)
#result: bad on small lambda values: should not use asymtotic interval for small
# values of lambda (not enough coverage).
#todo: extend monitoring time to t <- 100 to t <- 1000 (number of days)