程序代写代做代考 kernel go Resampling, part 1

Resampling, part 1
Jennifer Wilcock STAT221 2020 S2
We are going to look at several methods known as resampling methods. These methods include the bootstrap, the jack-knife, cross-validation, and permutation tests.
All resampling methods have the same basic idea, of using observed sample(s) of data and repeatedly sampling from the original observed sample(s) to obtain a large collection of re-samples, and then use the distribution of the resamples to tell us about the original sample.
Applications to classical statistical problems
Resampling methods are inherently computational, but directly complement classical approaches to the same problems.
We can use resampling methods for tasks where there are already parametric approaches, including: • estimating confidence intervals
• hypothesis tests.
Classical approaches are typically parametric, computational approaches are typically non-parametric (just as we have already considered the difference between methods in regression that are paramet-
ric/analytical/classical and which are non-parametric/computational/modern).
Classical/parametric approaches work well and are very quick to use when the assumptions they make are true (or close to being true). Often the assumptions are not true however, so we need methods we can use in these cases.
• For example, if we have one large sample of observations, all sampled from the same population, and the plot of these observed values looks roughly Normal, then we can assume a 95% confidence interval for the true population mean is the observed sample mean ± 1.96 standard errors.
Modern/non-parametric approaches work well when the assumptions in parametric approaches don’t hold, or when the maths is too hard/the problem is too complex to deal with in any other way.
• For example, we have a small sample of observations, all sampled from the same population, and the plot of these observed values doesn’t look roughly Normal, then we can use the bootstrap to find a 95% confidence interval for the population mean.
Although we can also use non-parametric methods even when there are parametric methods available, generally we wouldn’t because non-parametric solutions take longer to run and are less powerful.
Applications to inherently computational problems
Here, applications include:
• finding optimal values, for example of smoothing parameters
which we might do using, for example, cross-validation or the jack-knife.
1

The remainder of the course
In the remaining weeks we will look at:
• a review of hypothesis testing, including p-values and confidence intervals
• permutation tests (also known as randomisation tests)
• power of a test, and effect size
• bootstrapping
• if time permits, the jack-knife and cross-validation and applying these methods to finding optimal bandwidths in smoothing.
2

New Section: Hypothesis Testing
The basic idea
A statistical hypothesis is a conjecture about the distribution of some random variable.
For example, I might conjecture (or speculate, or believe but with no specific evidence to support the belief) that the mean weight of students at UC is 80kg.
I must make my conjecture before I see the evidence I use to test if the conjecture is true.
The idea with hypothesis testing is to formally examine two opposing hypotheses: • null hypothesis, H0; and
• alternative hypothesis, H1.
These two hypotheses are “usually” mutually exclusive and exhaustive. Only one of them can be true and the other false.
Setting up a hypothesis test
Statistical hypothesis testing is like a proof by contradiction, using data-based evidence.
There is an analogy with the criminal justice system:
• the null hypothesis states that the accused is innocent
• the burden of proof lies with the prosecution to provide enough evidence (“beyond reasonable doubt”) to convince the jury to reject the null hypothesis.
In a hypothesis test, evidence is accumulated by collecting and analyzing sample data for the purpose of determining which of the two hypotheses is true and which of the two hypotheses is false.
Null hypothesis, H0:
• default assumption that null hypothesis is true
• typically contains value for population parameters to be tested • must contain a “simple hypothesis”, so a single population value • see if there is evidence to contradict the null hypothesis
Alternative hypothesis, H1 (sometimes labelled Ha):
• opposite of null hypothesis
• often the hypothesis of interest that would like to be prove (the ‘research’ hypothesis) • typically a “composite hypothesis”, containing multiple values.
We then ask: Is there enough evidence in the data to refute the null hypothesis, in favour of the alternative?
They can be two-side or one-sided
Hypothesis tests can be two-sided or one-sided (also called two-tailed or one-tailed). Two tailed tests are typically of this form:
• H0 :μ=0 • H1 :μ̸=0
3

Here the null includes the “simple hypothesis” of equality μ = 0,
and the alternative is a “composite hypothesis”, which does not include a single specific value.
Two tailed tests are not directional, but one tailed tests are directional, e.g.: • H0 :μ≤0
• H1 :μ>0
or
• H0 :μ≥0 • H1 :μ<0 Again, in both cases the null includes the “simple hypothesis” of equality μ = 0. The need for scientific transparency It is possible to test hypotheses like • H0 :μ=0 • H1 :μ>0
which are not mutually exhaustive.
But it would be good scientific practice in this example to use H0 : μ ≤ 0 instead, to be transparent in
your study results.
This is because if one rejects the null of μ = 0 in favour of the alternative μ > 0 then one would also
reject the null for any case of μ < 0, as these possibilities are even more different to the alternative. In general, the null and alternative should be mutually exclusive and exhaustive, as this usually avoids the lack of transparency in interpreting study results. The testing procedure A typical testing process has four steps: 1. Assume null hypothesis H0 is true. 2. Use statistical theory to derive a statistic (function of the data) used to measure difference from the null hypothesis. Calculate this test statistic from the sample data. 3. Find the probability that the test statistic would take a value “as extreme or more extreme” than that actually observed. Referred to as the p-value for the test. 4. If probability from step 3 is: • high then sample test statistic is likely if null hypothesis is true, so we have “insufficient evidence to reject the null”; or • low the sample test statistic is unlikely if null hypothesis is true, so we have “sufficient evidence to reject the null” An example might be doing a t-test. p-values and rejecting the null hypothesis Test statistic essentially measures the (in)compatibility between the null hypothesis and the sample data. After calculating the test statistic, a p-value is obtained by comparing the observed test statistics to the distribution of the test statistic under the null hypothesis. The p-value is a measure of how unusual the test statistic is, under the null hypothesis. 4 The level of significance, α, is specified before the experiment, to define rejection region: • p-value ≤ α =⇒ Reject H0 at significance level α • p-value > α =⇒ Do not reject H0 at significance level α
The rejection region is the set of all test statistic values for which H0 will be rejected. Interpreting p-values
Smaller p-values indicate that the data provides stronger evidence against H0. We commonly use a ‘rule of thumb’ for the interpretation of “small” p-values:
• p-value > 0.1 provides insufficient evidence against H0;
• 0.05 < p-value ≤ 0.1 provides weak evidence against H0; • 0.01 < p-value ≤ 0.05 provides moderate evidence (or evidence) against H0; and • p-value ≤ 0.01 provides strong evidence against H0. The decision rule is made by comparing the p-value with the significance level α. As with confidence intervals, the significance level is chosen by the application, typical values are α = 0.1, 0.05 or 0.01. Rejection regions: Two-sided test Two−sided Rejection Region 1−α α 2 −4 −2 0 2 4 Test Statistic Red rejection region area is significance level α, i.e. probability of rejection if null is true. Blue non-rejection region area is remainder 1 − α, i.e. probability of non-rejection if null is true. Do not Reject α 2 Density 0.0 0.1 0.2 0.3 0.4 5 Reject Rejection regions: One-sided upper tail test One−sided Upper Rejection Region 1−α Do not Reject Reject α −4 −2 0 2 4 Rejection regions: One-sided lower tail test Reject α One−sided Lower Rejection Region 1−α Do not Reject Test Statistic Density Density 0.0 0.1 0.2 0.3 0.4 0.0 0.1 0.2 0.3 0.4 −4 −2 0 2 4 Test Statistic 6 Potential results from one-sided tests Let’s focus on the p-value defined by the upper tail probability, i.e. the probability of getting that value of test statistic or something more extreme under the null hypothesis. With p-value≤ α: Reject Null With p-value> α:
Test Statistic
Do Not Reject Null
−4 −2 0 2 4
X
Test Statistic
−4 −2 0 2 4
Test Statistic
If the test statistic is in the red rejection region area, the p-value is less than significance level α.
If the test statistic is in the blue non-rejection region area, the p-value is greater than significance level α.
7
X
Test Statistic
p−value
≤α
p−value

Density Density
0.0 0.1 0.2 0.3 0.4 0.0 0.1 0.2 0.3 0.4

Potential Results From Two-sided Test Results
In general for a two-sided test, the significance level cut-off is split between the two tails (α/2 per tail). Thus, the p-value is commonly defined by doubling the minimum of the upper and lower tail p-values.
With p-value≤ α:
Reject Null
With p-value> α:
Test Statistic
Do Not Reject Null
−4 −2 0 2 4
X
Test Statistic
−4 −2 0 2 4
Test Statistic
If the test statistic is in one of the two α/2 red rejection regions, the p-value is less than the significance level α.
If the test statistic is in the blue non-rejection region area, the p-value is greater than the significance level α.
8
Double this p−value
≤α
X
Test Statistic
Double this p−value

Density Density
0.0 0.1 0.2 0.3 0.4 0.0 0.1 0.2 0.3 0.4

Critical values & decision rule
The cut-off value(s) for the test statistic which define the rejection regions are defined as the critical value(s) of the test.
These are often denoted:
• Xα and X1−α for a one sided test for the lower and upper tail respectively • Xα/2 and X1−α/2 for a two-sided test
where the subscript represents the relevant upper or lower tail probability. Therefore, the decision rule can equivalently be based on the:
• p-value; or the
• critical value,
and both give the same conclusions.
Possible errors when testing
Consider the possible outcomes of hypothesis testing:
H0 Not Rejected
H0 true Type I Error
H0 Rejected (α)
Right Decision (1 − β)
Right Decision (1−α)
H1 true Type II Error (β)
Type I error: To wrongly reject null when it is true, often called false positive. Type II error: To wrongly not reject the null when it is false, false negative.
A way to remember these: the fable of the boy who called wolf: A shepherd boy looked after all the sheep of the village, watching them in the hills. For fun, he came down to the village many times and shouted ‘wolf, wolf’, and the villagers would go into the hills to protect the sheep, only to find the boy was lying (the villagers made a Type I error – the truth was that there was no wolf). One day a wolf really did arrive and so the boy ran down to the village and cried ‘wolf, wolf’, but the villagers ignored him and the
wolf ate all their sheep (the villagers made a Type II error).
9

Start of lecture 9
Significance and power
The power of the test is the probability of correctly rejecting the null, (1 − β).
The significance level is the probability of making a Type I error (α), and the confidence level is (1 − α). It’s important to realise that:
• rejecting H0 does not mean that we have proved H0 is false
• conversely, failing to reject H0 does not mean that we have proved H0 to be true. As a result, failure to reject H0 represents an inconclusive result.
Ideally, a statistical test should have small α and small β.
In practice, it is easy to control α but not β.
There is always a trade-off between the power of a test and the likelihood of false positives.
There is some statistical testing controversy …
There is some controversy over hypothesis testing (see Wikipedia on “Statistical Hypothesis Testing”), especially over interpretation and language.
In the Fisher school you either ‘Reject’ or ‘Do Not Reject’ the null, you never accept anything!
Under the Neyman-Pearson school you accept H1 in the rejection region and you accept H0 in the do
not reject region.
However, although you ‘accept’ one of the two hypotheses, this doesn’t mean it is in fact true, only that you carry on under the assumption that it is – so it effectively becomes your working hypothesis of the truth.
Parametric and nonparametric tests
• Parametric Tests: rely on theoretical distribution of the test statistic under the null hypothesis, typically based on asymptotic theory (e.g. Central Limit Theorem) or assumptions about the distribution of the sample data (e.g., normal population)
• Nonparametric Tests: do not necessarily assume that the sample data are drawn from any particular distribution
10

New Section: Permutation tests
The most common application of these nonparametric tests is for comparing two samples.
Two-sample permutation test
Suppose we have samples collected from two different populations, with values denoted by X = {X1,…,Xn}
and
Y = {Y1,…,Ym}. • the two samples are independent of each other; and
• the data in each individual sample are iid (independent and identically distributed), meaning that each set of data points are randomly sampled from a single population.
Hypotheses are framed in terms of population values, so if X is a random sample from population (distribution) FX , and Y from FY , then in general, we want to test the hypotheses:
H0 :FX =FY against H1 :FX ̸=FY
When samples are collected, they are ‘labelled’ with the name of the population they are sampled from.
For example: Suppose you buy 4 regular and 3 expensive premium lightbulbs, and you want to test whether the premium lightbulbs have a longer lifetime than the regular ones.
The lifetime measured in 10,000 hours are: • Regular bulbs: 90, 11, 94, 118
• Expensive bulbs: 197, 107, 752
Here the ‘labels’ are ‘Regular’ and ‘Expensive’, and we could instead have written the data as:
90 Regular
11 Regular 197 Expensive
94 Regular 107 Expensive 752 Expensive 118 Regular,
so that each observation is individually labelled.
Returning to the general case: under the null hypothesis, both samples X and Y come from the same population.
Therefore, under the null we can consider the combined sample (X, Y ) as coming from a single population.
Thus, under the null hypothesis, any labels are entirely arbitrary.
Therefore, we can consider randomly rearranging (or ‘permuting’) the labels in order to evaluate the distribution of the test statistic.
So it is the labels that are permuted, and we don’t touch the values of the observations. And after ‘permuting’ we have exactly the same number of observations as before, and the values are exactly the
same, but the labels have been rearranged.
We can visualise this using the software InzightVIT (downloadable from https://inzight.nz) using the data in the file packaging.csv.
We assume:
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Start of lecture 10
The permutation test distribution
Under the null hypothesis, any statistical function of the original data (X, Y ) will have same distribution as when the same function is applied to the permuted dataset (X∗, Y ∗).
For example, under the null the distribution of X ̄ −Y ̄ will be the same as the distribution of X ̄∗ −Y ̄∗.
The ‘sampling distribution’ of the test statistic under the null hypothesis is therefore found by calculating that statistic for a very large number of permuted datasets (X∗, Y ∗), where the labels have been randomly allocated.
The actual test statistic obtained from the original data (X,Y) is then compared to this sampling distribution.
If the value of the test statistic from (X, Y ) is ‘unusual’ in the permutation sampling distribution of (X∗, Y ∗), then this provides evidence against the null hypothesis.
Why use permutation tests?
Permutation tests are often used as a substitute for standard parametric tests (like those studied in STAT101) when the usual assumptions might fail, or when the sample size is sufficiently small that asymptotic theory (e.g. Central Limit Theorem) may not hold.
For example, a permutation test may be used to replace a two sample t-test when the normal population assumption doesn’t not hold.
You can still use the t-test statistic to compare the two groups when the population is not normal, but the problem is that without the normal assumption the distribution of the t test statistic is unknown – it is not likely to follow a Student-t distribution as was used in STAT101.
Permuting the original values many times gives us a distribution of the t-test statistic under the null hypothesis, without assuming the population is normal.
First example: Doing a two-sample permutation test in R
Suppose you buy 4 regular and 3 expensive lightbulbs.
You want to test whether the expensive lightbulbs have a longer lifetime than the regular ones:
• this would be a one sided (upper tail) test. We can’t use a parametric test because:
• such lifetimes are not normally distributed and can be highly skewed (the exponential distribution is commonly used for lifetimes), and further
• the samples are sufficiently small that it is not appropriate to assume any asymptotic models (because, for example, the Central Limit Theorem may not hold).
Therefore we’ll use a permutation test instead of a parametric t-test, but we still use the t-test statistic!!!
12

The observed lifetimes measured in units of 10,000 hours are:
x = c(90, 11, 94, 118) y = c(197, 107, 752)
The steps in our permutation test are:
# Sample size
nx = length(x) ny = length(y)
# Combined dataset and corresponding labels
xy = c(x, y)
xylabels = c(rep(“x”, nx), rep(“y”, ny))
# t-test statistic for original sample
teststat = (mean(y) – mean(x)) / sqrt(var(x)/nx + var(y)/ny) teststat
## [1] 1.348412
# Number of permutation samples to take
N = 10000 set.seed(1)
# Create vector to store permutation t-statistics
tstat = numeric(N) for (i in 1:N) {
newlabels = sample(xylabels, replace = F) # sample labels without replacement
# permute the data and obtain the t-test statistic
newx = xy[newlabels == “x”]
newy = xy[newlabels == “y”]
tstat[i] = (mean(newy) – mean(newx))/sqrt(var(newx)/nx + var(newy)/ny)
}
# Calculate upper tail one-sided p-value
# (proportion of permuted test statistics above observed value) pvalue = mean(tstat >= teststat)
pvalue
## [1] 0.0562
## [1] 273.750000 1.348412 0.056200
# Output mean difference, t-test statistic and p-value
print(c(mean(y) – mean(x), teststat, pvalue))
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Now we can plot the 1000 test statistics (to get the t-distribution) and compare the original test statistic against this distribution:
# Plot permutation sampling distribution using kernel density estimate
plot(density(tstat, bw=0.1), xlab=”Test Statistics”, main=”Permutation Distribution and Actual Test Statistic”)
abline(v = teststat, lwd = 2, col = “red”) rug(tstat)
Permutation Distribution and Actual Test Statistic
Density
0.0 0.2 0.4 0.6 0.8
−1.5 −1.0 −0.5 0.0 0.5 1.0 1.5
Test Statistics
Notice that the sampling distribution of the test statistic is nothing like the usual t distribution.
The observed test statistic (vertical line) is out in the upper tail, so is unusual under the null hypothesis.
The conclusions of this test are:
The above is a one-sided upper tail test with hypotheses:
• H0 :μY ≤μX orequivalentlyμY −μX ≤0
• H1 : μY − μX > 0
The observed difference in means is μY − μX = 273.5.
The test statistic (teststat) is 1.348412.
The p-value is 0.056200.
There is therefore insufficient evidence to reject the null hypothesis at the α = 5% level.
The critical value for a permutation test can also be obtained using the following code:
alpha = 0.05
# Sort permutation t-statistics and extract upper (1-alpha)% quantile
tstat = sort(tstat)
tcrit = tstat[ceiling((1 – alpha) * N)] tcrit
## [1] 1.348412
14

Second example: Two-sample, two-sided permutation test
Consider making the above example a two-sided test with hypotheses: • H0 :μY =μX orequivalentlyμY −μX =0
• H1 : μY − μX ̸= 0
The observed difference in means and the test statistic are the same.
But the p-value will be doubled to give 0.1126, so there is insufficient evidence to reject the null hypothesis
at α = 5% (or even at the α = 10% level). Permutation test p-values revisited
The above formula for calculating the p-value is not quite correct, since when carrying out the hypothesis test, you are also assuming the observed test statistic is actually from the null distribution.
Thus this datum should be included in calculating the p-value, and we need to add one to the numerator and denominator:
## [1] 0.05629437
However, for most practical purposes this usually makes no difference to the conclusion from the test, except in ‘marginal cases’ or where the sample size is small.
‘Marginal significance’ is terminology used when the p-value is only a little smaller than the significance level.
Third example: Two-sample, one-sided, permutation test with more data
The first example used a very small dataset, with 7 observations in total, to do a one-sided test. Here we will repeat this but using a larger dataset (now with a total of 30 observations, but unequal sample sizes).
The earlier examples used real data for lifetimes where lifetimes are known to follow an exponential distribution. This time we will simulate some data from two different exponential distributions.
# Calculate correct upper tail one-sided p-value
pvalue = (sum(tstat >= teststat) + 1) / (N + 1) pvalue
# Simulate larger samples from Exponential populations with lambda=1 and lamba=0.5
nx = 10
ny = 20
set.seed(1)
x = rexp(nx, rate = 1)
y = rexp(ny, rate = 0.5)
# Combined dataset and corresponding labels
xy = c(x,y)
xylabels = c(rep(“x”, nx), rep(“y”, ny))
# t-test statistic for original sample
teststat = (mean(y) – mean(x)) / sqrt(var(x)/nx + var(y)/ny) # Number of permutation samples to take
N = 1000
# Create vector to store permutation t-statistics
tstat = numeric(N) for (i in 1:N) {
newlabels = sample(xylabels, replace = F) # sample labels without replacement
15

# permute data and obtain t-test statistic
newx = xy[newlabels == “x”]
newy = xy[newlabels == “y”]
tstat[i] = (mean(newy) – mean(newx)) / sqrt(var(newx)/nx + var(newy)/ny)
}
# Calculate upper tail one-sided p-value
pvalue = (sum(tstat >= teststat) + 1) / (N + 1)
# Sort permutation t-statistics and extract upper (1-alpha)% quantile
alpha = 0.05
tstat = sort(tstat)
tcrit = tstat[ceiling((1 – alpha) * N)]
# Output mean different, test statistic, critical value and p-value
print(c(mean(y) – mean(x), teststat, tcrit, pvalue))
## [1] 1.56737386 2.67828573 2.04469990 0.00999001
Conclusions:
The p-value of 0.01 is less than the significance level (0.05), so there is evidence that the mean of the
Y ∼ Exponential(λ = 0.5) population is greater than for X ∼ Exponential(λ = 1). The sample difference in means is 1.57.
The test statistic is 2.67 which is well above the critical value of 2.04.
# Plot permutation sampling distribution using kernel density estimate
hist(tstat, breaks = 100, freq = FALSE, xlab = “Test Statistics”, ylim = c(0, 0.8), xlim = c(-4,4),
main = “Permutation Distribution and Actual Test Statistic”)
lines(density(tstat, bw = 0.25)) abline(v = teststat, col = “red”) abline(v = tcrit, col = “blue”) legend(“topleft”,
legend = c(“Kernel Density Estimate”, “Test Statistic”, “Critical Value”), lty = 1, col = c(“black”, “red”, “blue”))
16

Permutation Distribution and Actual Test Statistic
Kernel Density Estimate Test Statistic
Critical Value
−4 −2 0 2 4
Test Statistics
Notice that with larger samples from each population that the sampling distribution for the test statistic is now much smoother and close to symmetric.
In fact, it is now close to the Student-t distribution, which is assumed in the classical parametric testing approach demonstrated in STAT101.
Classical t-test in R, using the same data as the third example It is easy to carry out a classical t-test in R:
##
## Welch Two Sample t-test
##
## data: y and x
## t = 2.6783, df = 26.274, p-value = 0.006297
## alternative hypothesis: true difference in means is greater than 0
## 95 percent confidence interval:
## 0.5696045 Inf
## sample estimates:
## mean of x mean of y
## 2.4099961 0.8426222
In this example, we have unequal sample sizes and unequal variances. Under these criteria it is called ‘Welch’s two sample t-test’.
# This is a one sided t-test with mean(y) > mean(x) as alternative hypothesis # (hence “greater” option)
t.test(y, x, alternative = “greater”)
The t-test statistic for evaluating whether the population means are different is:
t=Y−X (1)
sY −X where the standard deviation of the difference in the means:
􏰃 s 2X s 2Y
sY−X= n+n. (2)
XY
17
Density
0.0 0.2 0.4 0.6 0.8

Note that s2∗ are the unbiased estimators of the variance of the two samples and n∗ are the sample sizes. Under the null hypothesis, the distribution of the test statistic is approximately a Student-t distribution
with the degrees of freedom calculated using the ‘Welch-Satterthwaite equation’: (s2X /nX + s2Y /nY )2
df = (s2X/nX)2/(nX − 1) + (s2Y /nY )2/(nY − 1). (3) Let’s check out the classical t-test in R to make sure we understand how all the output is calculated:
# One sided t-test with mean(y) > mean(x) as alternative hypothesis
t.test(y, x, alternative = “greater”)
##
## Welch Two Sample t-test
##
## data: y and x
## t = 2.6783, df = 26.274, p-value = 0.006297
## alternative hypothesis: true difference in means is greater than 0
## 95 percent confidence interval:
## 0.5696045 Inf
## sample estimates:
## mean of x mean of y
## 2.4099961 0.8426222
The test statistic t = 2.1528 is the same as that calculated above using equation (1). The degrees of freedom in output is calculated using equation (3):
## [1] 26.2741
The standard deviation is calculated using equation (2):
## [1] 0.5852153
The t-test statistic is calculated using equation (1):
## [1] 2.678286
The p-value for one-sided upper tail test is given by upper tail probability from Student-t distribution:
## [1] 0.00629742
It is also worth noting that the permutation test p-value of 0.00999001 and that from Welch’s test 0.00629742 are similar, demonstrating that the Student-t approximation is fairly good even for such small samples from non-normal populations (ie the t-test is robust against non-normality).
# degrees of freedom for Welch’s test
ttestdf = (var(x)/nx+var(y)/ny)^2 / ((var(x)/nx)^2/(nx-1)+(var(y)/ny)^2/(ny-1)) ttestdf
# standard deviation for Welch’s test
ttestsd = sqrt((var(x)/nx + var(y)/ny)) ttestsd
# $t$-test statistic for Welch’s test
tteststat = (mean(y)-mean(x)) / ttestsd tteststat
# p-value for Welch’s test
pt(tteststat, df = ttestdf, lower.tail = FALSE)
18

Start of lecture 11
Permutation test exact p-values
The sampling distribution from the original small samples on page 14 is very different from that for the larger samples on page 17.
In fact, for the small samples it is not even close to being symmetric. Why is so oddly behaved? The key reason is that:
• it is a discrete distribution, so that
• only certain permutations of the sample are possible, so that
• there are limited values that the test statistic and p-value can take.
This motivates the idea of an ‘exact test’ where only the limited number of permutations of the actual sample data (from each group) are considered.
We could do this instead of doing many randomly chosen permutations for which many will simply be repeated (as happened in the first example).
The hypothesis test could use combinations only
The test statistic in the first example will be the same for many permutations, e.g. • X = {90,11,94,197} and Y = {118,107,752}
• X = {90,11,197,94} and Y = {118,107,752}
• X = {90,197,11,94} and Y = {118,107,752}
• etc.
will all give same test statistic, since we are comparing X ̄ to Y ̄ .
Therefore, only the unique permutations (in terms of which observations are allocated to each group) are relevant.
These unique permutation are called the combinations. For two groups the number of combinations is given by:
􏰀nx+ny 􏰁 (nx+ny)! n=n!n!.
‘nx+ny choosenx’.
However, these can still be referred to as ‘permutation tests’.
Note that if the test statistic does depend on the orderings, then this simplification will not hold. So the simplification doesn’t hold in general.
􏰀7􏰁
In our small example there are 4 = 35 combinations.
xxy
19

Fourth example: An ‘Exact’ two-sample test
The choose function gives the total number of combinations:
x = c(90, 11, 94, 118) y = c(197, 107, 752)
# Sample size
nx = length(x) ny = length(y)
# Combined dataset and corresponding labels
xy = c(x, y)
xylabels = c(rep(“x”, nx), rep(“y”, ny))
# t-test statistic for original sample
teststat = (mean(y) – mean(x)) / sqrt(var(x)/nx + var(y)/ny)
# Number of unique permutation samples to take
N = choose(nx + ny, nx) N
## [1] 35
So there are 35 combinations for this exact test. R has a function combn which will produce a list of these
unique combinations (selecting nx from nx + ny possibilities): check the help file using ?combn. An aside on how this function works:
letters[1:4]
## [1] “a” “b” “c” “d”
combn(letters[1:4], 2) # all combinations of a,b,c,d taken 2 at a time
## [,1] [,2] [,3] [,4] [,5] [,6]
## [1,] “a” “a” “a” “b” “b” “c”
## [2,] “b” “c” “d” “c” “d” “d”
which gives 6 unique combinations.
1:4
## [1] 1 2 3 4
combn(1:4, 2) # all combinations of 1,2,3,4 taken 2 at a time
## [,1] [,2] [,3] [,4] [,5] [,6] ##[1,] 1 1 1 2 2 3 ##[2,] 2 3 4 3 4 4
combn(4, 2) # all combinations of 1,2,3,4 taken 2 at a time (the same as before)
## [,1] [,2] [,3] [,4] [,5] [,6] ##[1,] 1 1 1 2 2 3 ##[2,] 2 3 4 3 4 4
20

Returning to our test, we can use combn(7, 4) to tell us the observations to label as x, with the remainder labelled as y:
## [1] 7
nx
## [1] 4
## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
# Obtain all unique permutations
nx + ny
allpermutations = combn(nx + ny, nx) # all combinations of 1:7 taken 4 at a time allpermutations
##[1,]111111111 ##[2,]222222222 ##[3,]333344455 ##[4,]456756767
## [,14] [,15] [,16] [,17] [,18] [,19] [,20] [,21] [,22] [,23] [,24] ##[1,] 1 1 1 1 1 1 1 2 2 2 2 ##[2,] 3 3 3 4 4 4 5 3 3 3 3 ##[3,] 5 5 6 5 5 6 6 4 4 4 5 ##[4,] 6 7 7 6 7 7 7 5 6 7 6 ## [,25] [,26] [,27] [,28] [,29] [,30] [,31] [,32] [,33] [,34] [,35]
##[1,] 2 2 2 ##[2,] 3 3 4 ##[3,] 5 6 5 ##[4,] 7 7 6
End of the aside
Recall that the observations are:
xy
2 2 2 3 3 3 3 4
4 4 5 4 4 4 5 5
5 6 6 5 5 6 6 6
7 7 7 6 7 7 7 7
1 1 1 1
2 3 3 3
6 4 4 4
7 5 6 7
##[1] 90 11 94118197107752
We can now do the ‘exact’ test by calculating the test statistics for these 35 unique permutations only:
## [1] 35
# Number of unique permutation samples to take
N = choose(nx + ny, nx) N
# Create vector to store permutation t-statistics
tstat = numeric(N)
for (i in 1:N) {
# permute data and obtain t-test statistic
newx = xy[allpermutations[, i]]
newy = xy[-allpermutations[, i]]
tstat[i] = (mean(newy) – mean(newx)) / sqrt(var(newx)/nx + var(newy)/ny)
}
# Calculate upper tail one-sided p-value
# (proportion of permuted test statistics above observed value) pvalue = mean(tstat >= teststat)
# Sort permutation t-statistics and extract upper (1-alpha)% quantile
alpha = 0.05
21

tstat = sort(tstat)
tcrit = tstat[ceiling((1 – alpha) * N)]
# Output mean different, test statistic, critical value and p-value
print(c(mean(y) – mean(x), teststat, tcrit, pvalue))
## [1] 273.75000000 1.34841208 1.34841208 0.05714286
Notice that the p-value 0.05714286 is close to the 0.056200 obtained by the repeated permutations approach, on page 14.
Monte Carlo and exact sampling
The two sampling approaches for getting the test statistics under the permutation scheme are referred to as:
• Monte Carlo sampling – random samples from possible permutations
• Exact test – only unique permutations used.
The former is an approximation to the latter, but is more practical for large samples which lead to too many permutations.
Exact p-values frequently come up in other areas of statistics that have discrete data, such as binomial distributions (testing for a proportion) and contingency tables.
Back to the example: An exact two-sample test
The unique values of the test statistic are given in the following table:
table(tstat)
## tstat
## -1.46012848312594 -1.39582759622129 -1.37666316011324
## 1 1 1
## -1.34261724030786 -1.32401252310949 -1.26435428150407
## 1 1 1
## -1.04313743925751 -0.998231311458993 -0.989246691069474
## 1 1 1
## -0.973759375119742 -0.946744011035543 -0.931216941089718
## 1 1 1
## -0.923810953579016 -0.88200187821866 -0.691211289507945
## 1 1 1
## -0.645502648506479 -0.631610122091557 -0.607035509594958
## 1 1 1
## -0.593276520708821 -0.548674435835161 0.657956193884955
## 1 1 1
## 0.669239975188824 0.706253147387712 0.737987095868991
## 1 1 1
## 0.912531923383827 0.955279265069289 0.968717372641221
## 1 1 1
## 0.977926215782566 0.992123649633063 1.00587512834482
## 1 1 1
## 1.05121904971437 1.27696549603357 1.29352200921875
## 1 1 1
## 1.3484120784389 1.39620685221911
## 1 1
and these are the values of the 35 tick-marks in the rug on the plot on page 14. Here’s the plot from the full permutation test again (from page 14):
22

# Plot permutation sampling distribution using kernel density estimate
plot(density(tstat, bw=0.1), xlab=”Test Statistics”, main=”Permutation Distribution and Actual Test Statistic”)
abline(v = teststat, lwd = 2, col = “red”) rug(tstat)
Permutation Distribution and Actual Test Statistic
Density
0.0 0.2 0.4 0.6 0.8
−1.5 −1.0 −0.5 0.0 0.5 1.0 1.5
Test Statistics
Note that the critical value of the test statistic is one of these unique values.
What p-values are possible for an exact test?
Only a limited set of p-values are possible due to the discrete nature of the exact test. There are 35 combinations:
length(table(tstat))
## [1] 35
so a maximum of 35 p-values are possible.
Under the null hypothesis, p-values are uniformly distributed1. Therefore we know that the 35 p-values
in this exact test take the values {i/35 : i = 1 . . . , 35} at each of the test statistic values: (1:35)/35
## [1] 0.02857143 0.05714286 0.08571429 0.11428571 0.14285714 0.17142857
## [7] 0.20000000 0.22857143 0.25714286 0.28571429 0.31428571 0.34285714
## [13] 0.37142857 0.40000000 0.42857143 0.45714286 0.48571429 0.51428571
## [19] 0.54285714 0.57142857 0.60000000 0.62857143 0.65714286 0.68571429
## [25] 0.71428571 0.74285714 0.77142857 0.80000000 0.82857143 0.85714286
## [31] 0.88571429 0.91428571 0.94285714 0.97142857 1.00000000
Notice that the p-value of 0.05714286 for this example is second in the list.
However, you should be aware this is the maximum possible number of p-values. If the test statistic is not unique for each combination then a reduced set of p-values is possible.
1 The reason for this is really the definition of alpha as the probability of a type I error. We want the probability of rejecting a true null hypothesis to be alpha, we reject when the observed p-value < α, the only way this happens for any value of alpha is when the p-value comes from a uniform distribution. The whole point of using the correct distribution (normal, t, f, chisq, etc.) is to transform from the test statistic to a uniform p-value. If the null hypothesis is false then the distribution of the p-value will (hopefully) be more weighted towards 0. The Pvalue.norm.sim and Pvalue.binom.sim functions in the TeachingDemos package for R will simulate several data sets, compute the p-values and plot them to demonstrate this idea, if you would like to investigate this further. 23 For example, if the dataset is X = {90, 90, 94, 118} and Y = {197, 107, 752} then the duplicated data values will lead to non-unique values for the test statistics. p-values of 1! How? What does a p-value of 1.0 mean? When doing exact testing the p-values can be 1, meaning that your data is as consistent as possible with the null hypothesis. This could happen, for example, if the null hypothesis is that the mean of X is at least the mean of Y (as it was in our small sample, one-sided test), and you observe xi > yj for every combination of i and j.
Fifth example: p-value of 1
For example, if the dataset were X = {197, 107, 752, 118} and Y = {90, 11, 94}.
So, as stated in the first example on page 14, which was a one-sided upper tail test with hypotheses:
• H0 :μY ≤μX orequivalentlyμY −μX ≤0 • H1 : μY − μX > 0.
x = c(197, 107, 752, 118) y = c(90, 11, 94)
# Sample size
nx = length(x) ny = length(y)
# Combined dataset and corresponding labels
xy = c(x, y)
xylabels = c(rep(“x”, nx), rep(“y”, ny))
# t-test statistic for original sample
teststat = (mean(y) – mean(x)) / sqrt(var(x)/nx + var(y)/ny) # Number of unique permutation samples to take
N = choose(nx + ny, nx)
# Obtain all unique permutations
allpermutations = combn(nx + ny, nx)
# Create vector to store permutation t-statistics
tstat = numeric(N)
for (i in 1:N) {
# permute data and obtain t-test statistic
newx = xy[allpermutations[, i]]
newy = xy[-allpermutations[, i]]
tstat[i] = (mean(newy) – mean(newx)) / sqrt(var(newx)/nx + var(newy)/ny)
}
# Calculate upper tail one-sided p-value
# (proportion of permuted test statistics above observed value) pvalue = mean(tstat >= teststat)
pvalue
## [1] 1
24

There is no evidence against the null.
Permutation test for correlation
In addition to two independent samples being tested with permutation tests, we might wish to test paired data (X1, Y1), (X2, Y2), . . . , (Xn, Yn).
If the data can be assumed to come from a bivariate normal distribution (i.e., X and Y random variables are both normal, but might be correlated), then the classical parametric test is another t-test where
• H0 :ρ=0
• Test statistic t = r􏰂 (n−2) which follows a Student-t distribution on n − 2 degrees of freedom.
(1−r2)
This test is implemented in R with the test.cor function.
If the Student-t distribution is inappropriate (i.e., the populations are not normal), can we do a permutation test?
The null hypothesis is that there is no correlation, which typically means no relationship between X and Y.
As usaual, the sampling distribution of the test statistics under the null is found by calculating that statistic for a large number of permuted datasets (X∗,Y∗) where the labels have been randomly allocated.
We could permute the indices on both the X and Y values, but it is sufficient to permute just one of the datasets, since this permutation will destroy any relationship/correlation between X and Y .
This can be done using the basic steps we have already followed in other examples:
1. use the sample function to permute the data;
2. recompute the correlation from the permuted data; and
3. see what proportion of the correlations are at least as large as the actual value in the original dataset.
Finally: Matched pairs t-test (or paired t-test) for the mean
The same permutation approach can be used for paired data, and is useful if you don’t wish to assume normal populations.
In a matched pairs test, the difference Di = Xi − Yi, i = 1, . . . , n, between the two paired observations is calculated. The null hypothesis is that D = 0, and the alternative is that the mean of these differences D is significantly different to 0.
• H0 :D=0
• H1 :D̸=0
In correlation testing we could permute individual datapoints to destroy any relationship between X and Y . With paired data, however, we must retain the relationship between X and Y , so we have to permute each of the n pairs together to obtain the sampling distribution of the test statistic.
Sixth example: Matched pairs permutation test
# Simulate correlated samples from standard normals
n = 20
set.seed(2)
x = rnorm(n)
y = rnorm(n, 0.5*x + 0.5)
25

# t-test statistic for original sample extracted from t.test() function
teststat = t.test(y, x, alternative = “greater”, paired = TRUE)$statistic # Number of permutation samples to take
N = 1000
# Create vector to store permutation t-statistics
tstat = numeric(N) for (i in 1:N) {
# decide which ones to interchange
newlabels = sample(c(TRUE, FALSE), size = n, replace = TRUE)
# permute the pairs where newlabels = TRUE, and obtain t-test statistic
newx = x
newy = y
newx[newlabels] = y[newlabels]
newy[newlabels] = x[newlabels]
tstat[i] = t.test(newy, newx, alternative = “greater”, paired = TRUE)$statistic
}
# Calculate upper tail one-sided p-value and critical value
pvalue = (sum(tstat >= teststat) + 1) / (N + 1) alpha = 0.05
tstat = sort(tstat)
tcrit = tstat[ceiling((1 – alpha) * N)]
print(c(mean(y) – mean(x), teststat, tcrit, pvalue))
## t
## 0.4019877 1.3513186 1.7299853 0.0989011
# Plot the permutation sampling distribution using kernel density estimate
hist(tstat, breaks = 100, freq = FALSE, xlab = “Test Statistics”, ylim = c(0, 0.8), xlim = c(-4,4),
main = “Permutation Distribution and Actual Test Statistic”)
lines(density(tstat, bw = 0.25)) abline(v = teststat, col = “red”) abline(v = tcrit, col = “blue”) legend(“topleft”,
legend = c(“Kernel Density Estimate”, “Test Statistic”, “Critical Value”), lty = 1, col = c(“black”, “red”, “blue”))
26

Permutation Distribution and Actual Test Statistic
Kernel Density Estimate Test Statistic
Critical Value
−4 −2 0 2 4
Test Statistics
Notice in the R code that the t-test statistic is extracted from the t.test function, rather than manually calculated.
The permutation test p-value, 0.0989011, is similar to that of the parametric test, 0.09623:
t.test(y, x, alternative = “greater”, paired = TRUE)
##
## Paired t-test
##
## data: y and x
## t = 1.3513, df = 19, p-value = 0.09623
## alternative hypothesis: true difference in means is greater than 0
## 95 percent confidence interval:
## -0.1123914 Inf
## sample estimates:
## mean of the differences
## 0.4019877
End of notes for Permutation tests. Next is ‘Power’.
27
Density
0.0 0.2 0.4 0.6 0.8