# Examples and Exercises from Think Stats, 2nd Edition

http://thinkstats2.com



In [1]:

from __future__ import print_function, division

%matplotlib inline

import numpy as np

import brfss

import thinkstats2
import thinkplot



## The estimation game

Root mean squared error is one of several ways to summarize the average error of an estimation process.



In [2]:

def RMSE(estimates, actual):
"""Computes the root mean squared error of a sequence of estimates.

estimate: sequence of numbers
actual: actual value

returns: float RMSE
"""
e2 = [(estimate-actual)**2 for estimate in estimates]
mse = np.mean(e2)
return np.sqrt(mse)



The following function simulates experiments where we try to estimate the mean of a population based on a sample with size n=7. We run iters=1000 experiments and collect the mean and median of each sample.



In [3]:

import random

def Estimate1(n=7, iters=1000):
"""Evaluates RMSE of sample mean and median as estimators.

n: sample size
iters: number of iterations
"""
mu = 0
sigma = 1

means = []
medians = []
for _ in range(iters):
xs = [random.gauss(mu, sigma) for _ in range(n)]
xbar = np.mean(xs)
median = np.median(xs)
means.append(xbar)
medians.append(median)

print('Experiment 1')
print('rmse xbar', RMSE(means, mu))
print('rmse median', RMSE(medians, mu))

Estimate1()




Experiment 1
rmse xbar 0.3666916796469523
rmse median 0.4440447654801178



Using $\bar{x}$ to estimate the mean works a little better than using the median; in the long run, it minimizes RMSE. But using the median is more robust in the presence of outliers or large errors.

## Estimating variance

The obvious way to estimate the variance of a population is to compute the variance of the sample, $S^2$, but that turns out to be a biased estimator; that is, in the long run, the average error doesn't converge to 0.

The following function computes the mean error for a collection of estimates.



In [4]:

def MeanError(estimates, actual):
"""Computes the mean error of a sequence of estimates.

estimate: sequence of numbers
actual: actual value

returns: float mean error
"""
errors = [estimate-actual for estimate in estimates]
return np.mean(errors)



The following function simulates experiments where we try to estimate the variance of a population based on a sample with size n=7. We run iters=1000 experiments and two estimates for each sample, $S^2$ and $S_{n-1}^2$.



In [5]:

def Estimate2(n=7, iters=1000):
mu = 0
sigma = 1

estimates1 = []
estimates2 = []
for _ in range(iters):
xs = [random.gauss(mu, sigma) for i in range(n)]
biased = np.var(xs)
unbiased = np.var(xs, ddof=1)
estimates1.append(biased)
estimates2.append(unbiased)

print('mean error biased', MeanError(estimates1, sigma**2))
print('mean error unbiased', MeanError(estimates2, sigma**2))

Estimate2()




mean error biased -0.11357007322862302
mean error unbiased 0.03416824789993983



The mean error for $S^2$ is non-zero, which suggests that it is biased. The mean error for $S_{n-1}^2$ is close to zero, and gets even smaller if we increase iters.

## The sampling distribution

The following function simulates experiments where we estimate the mean of a population using $\bar{x}$, and returns a list of estimates, one from each experiment.



In [6]:

def SimulateSample(mu=90, sigma=7.5, n=9, iters=1000):
xbars = []
for j in range(iters):
xs = np.random.normal(mu, sigma, n)
xbar = np.mean(xs)
xbars.append(xbar)
return xbars

xbars = SimulateSample()



Here's the "sampling distribution of the mean" which shows how much we should expect $\bar{x}$ to vary from one experiment to the next.



In [7]:

cdf = thinkstats2.Cdf(xbars)
thinkplot.Cdf(cdf)
thinkplot.Config(xlabel='Sample mean',
ylabel='CDF')






The mean of the sample means is close to the actual value of $\mu$.



In [8]:

np.mean(xbars)




Out[8]:

89.9861531227758



An interval that contains 90% of the values in the sampling disrtribution is called a 90% confidence interval.



In [9]:

ci = cdf.Percentile(5), cdf.Percentile(95)
ci




Out[9]:

(85.67721360886706, 94.03550577193234)



And the RMSE of the sample means is called the standard error.



In [10]:

stderr = RMSE(xbars, 90)
stderr




Out[10]:

2.4985255047938177



Confidence intervals and standard errors quantify the variability in the estimate due to random sampling.

## Estimating rates

The following function simulates experiments where we try to estimate the mean of an exponential distribution using the mean and median of a sample.



In [11]:

def Estimate3(n=7, iters=1000):
lam = 2

means = []
medians = []
for _ in range(iters):
xs = np.random.exponential(1.0/lam, n)
L = 1 / np.mean(xs)
Lm = np.log(2) / thinkstats2.Median(xs)
means.append(L)
medians.append(Lm)

print('rmse L', RMSE(means, lam))
print('rmse Lm', RMSE(medians, lam))
print('mean error L', MeanError(means, lam))
print('mean error Lm', MeanError(medians, lam))

Estimate3()




rmse L 1.0991868744984798
rmse Lm 1.4874648694733599
mean error L 0.32089631663544593
mean error Lm 0.35349315821339045



The RMSE is smaller for the sample mean than for the sample median.

But neither estimator is unbiased.

## Exercises

Exercise: In this chapter we used $\bar{x}$ and median to estimate µ, and found that $\bar{x}$ yields lower MSE. Also, we used $S^2$ and $S_{n-1}^2$ to estimate σ, and found that $S^2$ is biased and $S_{n-1}^2$ unbiased. Run similar experiments to see if $\bar{x}$ and median are biased estimates of µ. Also check whether $S^2$ or $S_{n-1}^2$ yields a lower MSE.



In [12]:

# Solution

def Estimate4(n=7, iters=100000):
"""Mean error for xbar and median as estimators of population mean.

n: sample size
iters: number of iterations
"""
mu = 0
sigma = 1

means = []
medians = []
for _ in range(iters):
xs = [random.gauss(mu, sigma) for i in range(n)]
xbar = np.mean(xs)
median = np.median(xs)
means.append(xbar)
medians.append(median)

print('Experiment 1')
print('mean error xbar', MeanError(means, mu))
print('mean error median', MeanError(medians, mu))

Estimate4()




Experiment 1
mean error xbar 0.0005094416922616375
mean error median 0.0008580279056516486




In [13]:

# Solution

def Estimate5(n=7, iters=100000):
"""RMSE for biased and unbiased estimators of population variance.

n: sample size
iters: number of iterations
"""
mu = 0
sigma = 1

estimates1 = []
estimates2 = []
for _ in range(iters):
xs = [random.gauss(mu, sigma) for i in range(n)]
biased = np.var(xs)
unbiased = np.var(xs, ddof=1)
estimates1.append(biased)
estimates2.append(unbiased)

print('Experiment 2')
print('RMSE biased', RMSE(estimates1, sigma**2))
print('RMSE unbiased', RMSE(estimates2, sigma**2))

Estimate5()




Experiment 2
RMSE biased 0.5135938607123939
RMSE unbiased 0.5763054869567986




In [14]:

# Solution

# My conclusions:

# 1) xbar and median yield lower mean error as m increases, so neither
# one is obviously biased, as far as we can tell from the experiment.

# 2) The biased estimator of variance yields lower RMSE than the unbiased
# estimator, by about 10%.  And the difference holds up as m increases.



Exercise: Suppose you draw a sample with size n=10 from an exponential distribution with λ=2. Simulate this experiment 1000 times and plot the sampling distribution of the estimate L. Compute the standard error of the estimate and the 90% confidence interval.

Repeat the experiment with a few different values of n and make a plot of standard error versus n.



In [15]:

# Solution

def SimulateSample(lam=2, n=10, iters=1000):
"""Sampling distribution of L as an estimator of exponential parameter.

lam: parameter of an exponential distribution
n: sample size
iters: number of iterations
"""
def VertLine(x, y=1):
thinkplot.Plot([x, x], [0, y], color='0.8', linewidth=3)

estimates = []
for _ in range(iters):
xs = np.random.exponential(1.0/lam, n)
lamhat = 1.0 / np.mean(xs)
estimates.append(lamhat)

stderr = RMSE(estimates, lam)
print('standard error', stderr)

cdf = thinkstats2.Cdf(estimates)
ci = cdf.Percentile(5), cdf.Percentile(95)
print('confidence interval', ci)
VertLine(ci[0])
VertLine(ci[1])

# plot the CDF
thinkplot.Cdf(cdf)
thinkplot.Config(xlabel='estimate',
ylabel='CDF',
title='Sampling distribution')

return stderr

SimulateSample()




standard error 0.825791904537049
confidence interval (1.311259184670713, 3.7054533946140884)

Out[15]:

0.825791904537049




In [16]:

# Solution

# My conclusions:

# 1) With sample size 10:

# standard error 0.762510819389
# confidence interval (1.2674054394352277, 3.5377353792673705)

# 2) As sample size increases, standard error and the width of
#    the CI decrease:

# 10      0.90    (1.3, 3.9)
# 100     0.21    (1.7, 2.4)
# 1000    0.06    (1.9, 2.1)

# All three confidence intervals contain the actual value, 2.



Exercise: In games like hockey and soccer, the time between goals is roughly exponential. So you could estimate a team’s goal-scoring rate by observing the number of goals they score in a game. This estimation process is a little different from sampling the time between goals, so let’s see how it works.

Write a function that takes a goal-scoring rate, lam, in goals per game, and simulates a game by generating the time between goals until the total time exceeds 1 game, then returns the number of goals scored.

Write another function that simulates many games, stores the estimates of lam, then computes their mean error and RMSE.

Is this way of making an estimate biased?



In [17]:

def SimulateGame(lam):
"""Simulates a game and returns the estimated goal-scoring rate.

lam: actual goal scoring rate in goals per game
"""
goals = 0
t = 0
while True:
time_between_goals = random.expovariate(lam)
t += time_between_goals
if t > 1:
break
goals += 1

# estimated goal-scoring rate is the actual number of goals scored
L = goals
return L




In [18]:

# Solution

# The following function simulates many games, then uses the
# number of goals scored as an estimate of the true long-term
# goal-scoring rate.

def Estimate6(lam=2, m=1000000):

estimates = []
for i in range(m):
L = SimulateGame(lam)
estimates.append(L)

print('Experiment 4')
print('rmse L', RMSE(estimates, lam))
print('mean error L', MeanError(estimates, lam))

pmf = thinkstats2.Pmf(estimates)
thinkplot.Hist(pmf)
thinkplot.Config(xlabel='Goals scored', ylabel='PMF')

Estimate6()




Experiment 4
rmse L 1.4126517617587144
mean error L -0.001049




In [19]:

# Solution

# My conclusions:

# 1) RMSE for this way of estimating lambda is 1.4

# 2) The mean error is small and decreases with m, so this estimator
#    appears to be unbiased.

# One note: If the time between goals is exponential, the distribution
# of goals scored in a game is Poisson.

# See https://en.wikipedia.org/wiki/Poisson_distribution




In [ ]: