Chapter 5: Inference for a single population Outline The Central

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Outline
Chapter 5: Inference for a single population
5.1 Central Limit Theorem
5.2 A confidence interval for µ
Introductory Statistics for Engineering Experimentation
Peter R. Nelson, Marie Coffin and Karen A.F. Copeland
Slides by Douglas Bates
5.3 Prediction and tolerance intervals
5.4 Hypothesis tests
5.5 Inference for Binomial Populations
The Central Limit Theorem
Other properties of the distribution of the sample mean
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The central limit theorem is one of the most important results
in mathematical statistics. It says that the sample means from
a random sample (meaning independent samples from a stable
process) will be normally distributed, regardless of what the
original distribution was, when n is sufficiently large.
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Formally, if Y1 , Y2 , . . . , Yn is a random sample from a
distribution with σ 2 < ∞ then for large samples, Ȳ is
approximately normally distributed.
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This is a remarkably powerful result; first, because it is very
general and secondly because it is a description of the
asymptotic or “limiting” distribution but it holds for quite
small values of n.
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If the random variables Y1 , Y2 , . . . , Yn are a random sample
(sometimes also described as a “independent and identically
distributed” or i.i.d. sample) from a distribution with mean µ
and variance σ 2 then E(Ȳ) = µ and Var(Ȳ) = σ 2 /n.
So the central limit theorem states that, for large n,
σ2
Ȳ ∼ N µ,
n
Exactly how large n must be depends on the form of the
original distribution. If it is continuous and reasonably
symmetric then n = 15 may be large enough. If it is skewed
but continuous we may need n = 30 or more. For discrete and
skewed we may need as much as n = 100.
Although in practice we only have one sample and one
average, ȳ we can use computer simulation to consider the
sorts of samples we could have gotten and the distribution of
the statistic.
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Suppose we wish to simulate the value of a statistic (e.g.
mean or median or variance or standard deviation) from
samples of size n drawn from a certain distribution. Let K be
the number of replicates we want to obtain.
The sample size, n, is typically small. The number of
replicates, K, can be very large. The larger the value of K,
the more accurately we can determine the distribution of the
statistic. With modern computers we can afford to use values
of K in the hundreds of thousands or more.
Mean of samples of size 5 from U(-1,1)
What is the shape of the distribution of the mean of a sample of
size n = 5 from a U (−1, 1) distribution?
> mns5 <- replicate(50000, mean(runif(5, min = -1, max = 1)))
> histogram(~mns5,breaks = seq(-1, 1, len = 40))
8
6
Percent of Total
Conducting a simulation study (not part of the course)
First determine how to evaluate the statistic from a single
sample of size n then use the replicate function to repeat
the process K times.
Sampling densities of statistics
4
2
0
−1.0
The idiom
replicate(K, <statfn>(r<distab>(n, <pars>)))
produces K replicates of the statistic calculated by <statfn>
(examples are mean, median, var and sd) on samples of size n
from distribution <distab> with parameter(s) <pars>.
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Typically K is large and n is small. Values of 10,000 or 100,000
are used for K on modern computers. The larger the value of K
the smoother the approximation to the sampling density. n is
the size of the actual sample you can afford to collect.
0.0
0.5
1.0
Effect of changing the sample size, n
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−0.5
>
>
>
>
Performing multiple simulations allows us to see how
characteristics of the distribution of Ȳ depends on n.
mns1 <- runif(50000, -1, 1)
mns10 <- replicate(50000, mean(runif(10, -1, 1)))
mns20 <- replicate(50000, mean(runif(20, -1, 1)))
sapply(list(mns1, mns5, mns10, mns20), mean)
[1]
0.0007864216 -0.0006131859 -0.0004411292
0.0004612862
> sapply(list(mns1, mns5, mns10, mns20), var)
[1] 0.33107065 0.06678985 0.03341942 0.01646286
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As n increases the expected value of the sample mean stays
near 0.
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As n increases the variance of the sample mean decreases.
Roughly, V (X̄n ) = 13 · n1
Shape of distribution of X̄n
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More detail on the shape of the distribution of Ȳ
As n increases, the shape of the distribution of X̄n tends to
the “bell-curve” or Gaussian shape and it has less variability.
That is, it tends to a “central limit”.
−1.0
−0.5
mns1
0.0
0.5
1.0
−1.0
mns5
mns10
−0.5
0.0
0.5
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In addition to the histogram we can use normal probability
plots to evaluate the deviations of the distribution of Ȳ from
normality.
1.0
mns20
−3
15
1.0
Percent of Total
0.5
10
0.0
−0.5
5
−1.0
−2
−1
0
1
2
3
mns1
mns5
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3
Means of samples of size n from U(−1,1)
0
−1.0
−0.5
0.0
0.5
1.0
−1.0
−0.5
0.0
0.5
1.0
Means of samples of size n from U(−1,1)
Overlaid normal probability plots for Ȳn
mns1
mns5
mns10
Sample means from an exponential distribution
mns20
1.0
>
>
>
>
emns01
emns05
emns15
emns50
<<<<-
replicate(50000,
replicate(50000,
replicate(50000,
replicate(50000,
mean(rexp(1, rate = 1/7)))
mean(rexp(5, rate = 1/7)))
mean(rexp(15, rate = 1/7)))
mean(rexp(50, rate = 1/7)))
0.5
−4
0
2
4
−4
emns05
emns15
2
−1.0
10
8
0
0
4
5
5
0
6
10
10
15
40
20
−2
emns50
20
60
25
0.0
−0.5
−2
15
80
emns01
−4
−4
−2
0
2
4
−2
0
2
4
−4
−2
0
2
4
Standard normal quantiles
Standard normal quantiles
The conclusion is that the distribution of means from an i.i.d.
sample of a uniform distribution is very close to a normal, even for
n = 5.
Even for n = 50 there is noticeable skewness in the distribution
(althought we would not be far wrong in assuming normality at
n = 50).
4
Elementary uses of the C.L.T.
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If we have plausible values of the variance of our process,
perhaps from a pilot study, we can use the normal distribution
and the Central Limit Theorem (C.L.T.) to evaluate
probabilities regarding the sample mean.
Example 5.1.3 discusses product lifetimes that have an
unknown mean and a variance of approximately 8 years. The
number of products to sample so that we are 95% certain that
ȳ will be within 1 year of the true mean is derived from
Approximations for binomial or Poisson distributons
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The text describes approximations of the probabilities for a
binomial or Poisson distribution based on the normal
distributon.
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These are interesting from the point of view of understanding
that these distributions will tend to have a “bell-curve” shape
when n is large and p is moderate for the binomial or λ t is
large for the Poisson.
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In practice, though, you can evaluate probabilities for such
distributions exactly so there is no need to use approximations.
0.95 = P (|Ȳ − µ| < 1)
The distribution of Ȳ will be approximately normal with mean
√
µ and standard deviation σ/ n. For a standard normal, 95%
of the probability is within “2” standard deviations of the
mean (the actual multiple is qnorm(0.025)= -1.95996) so we
want 1 = qnorm(0.025)2 n8 . That is, n >
> 8 * qnorm(0.025)^2
[1] 30.73167
Confidence intervals
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Our “best guess” at a parameter is called a point estimate.
For example, we usually use the sample mean, ȳ, as the point
estimate of µ.
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An interval estimate or confidence interval is an interval of
plausible values for the parameter. Values outside the interval
are “unreasonable” and values inside are “not unreasonable”.
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To calibrate the meaning of “unreasonable” we assign a value
α to the probability of getting data like we did or even more
extreme when the parameter is outside. This corresponds to
the “p-value” in a hypothesis test.
The coverage probability or confidence level is 1 − α.
Typically we set α = 0.05 or α = 0.01 resulting in 95% or
99% confidence intervals.
Formally, the coverage probability is the probability that an
interval constructed in this way will cover the true parameter
value.
A confidence interval on µ
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In the unlikely event that someone were to tell us what the
standard deviation, σ, of the population was but somehow not
know much about the mean, µ, we could create a (1 − α)
confidence interval as
ȳ ± z(α/2)
σ
n
where z(α/2) is the upper α/2 quantile of the standard
normal distribution.
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For example, the upper 0.025 quantile of the standard normal
is
> qnorm(0.025, low = FALSE)
[1] 1.959964
so a 95% confidence interval on µ for this artificial, “known
sigma” case is
σ
ȳ ± 1.959964 √
n
Use of Student’s T distribution
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In the real world no one tells us what σ is and we must
estimate it as s. A statistician named William Gossett, who
wrote under the pseudonym “A Student”, derived the
distribution of the shifted, scaled sample mean when the scale
is based on the estimate, s, not the theoretical value σ.
This distribution is called the “Student’s t distribution”. It is
similar to the standard normal distribution but a bit more
spread out. The spreading depends on the number of “degrees
of freedom” in the estimate of σ 2 . The degrees of freedom are
written as ν. For a single sample ν = n − 1.
As ν increases the T distribution approaches the standard
normal. If we were using tables we would call anything with
ν > 30 a standard normal. When using a computer we don’t
bother.
Notation: the t distribution with ν degrees of freedom is
written t(ν). The corresponding R functions are dt, pt, qt
and rt. The upper α quantile is written t(α; ν).
General form of the confidence interval
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The general form of the confidence interval on µ is
α
s
ȳ ± t
,n − 1 √
2
n
We can use this formula for any values of n. If n is large we
don’t need strong assumptions on the shape of the original
distribution. If n is small we must assume that the original
distribution is close to the normal (but, of course, we can’t
check this with a small sample - a “Catch 22” situation).
The R function to create this interval is t.test. The name
comes from the corresponding hypothesis test, which we will
discuss later.
Graphical comparison of t(ν) and Z
T25
Z
T10
T5
0.4
0.3
0.2
0.1
0.0
−4
−2
0
2
4
Example 5.2.2
The example provides (probably fictitious) discharge times for a
particular electric vehicle
> sd(charge <- c(5.11,2.1,4.27,5.04,4.47,3.73,5.96,6.21))
[1] 1.3108
> summary(charge)
Min. 1st Qu.
2.10
4.14
Median
4.76
Mean 3rd Qu.
4.61
5.32
Max.
6.21
> t.test(charge)
One Sample t-test
data: charge
t = 9.9502, df = 7, p-value = 2.211e-05
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
3.5154 5.7071
sample estimates:
mean of x
4.6113
Example 5.2.2 (cont’d)
Another R evaluation of confidence intervals
Because the degrees of freedom, ν = 7, are quite small we should
check for normality.
0.3
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5
0.2
●
x
Density
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3
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2
0.0
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2
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4
−1.5 −1.0 −0.5 0.0
6
0.5
1.0
1.5
8
Clear-coat thickness (example 5.2.4)
0.15
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● ●●
60
−2
> summary(fm2)
0.05
−1
0
1
2
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0.00
●
55
60
Standard normal quantiles
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65
thickness
> with(ccthickn, summary(thickness))
Min. 1st Qu.
58.2
62.3
Median
64.4
Mean 3rd Qu.
64.3
66.2
●
70
●
75
Estimate Std. Error t value Pr(>|t|)
(Intercept) 64.2600
0.4297
149.5
<2e-16
Residual standard error: 2.718 on 39 degrees of freedom
The values in this summary include
ȳ = 64.26 The parameter estimate, µ̂.
Max.
71.3
> sd(ccthickn$thickness)
[1] 2.7176
> confint(fm2 <- lm(thickness ~ 1, ccthickn))
2.5 % 97.5 %
(Intercept) 63.391 65.129
> confint(fm1 <- lm(charge ~ 1))
The summary of the fit of the “trivial” model includes many of the
statistics from the data.
0.10
Density
thickness
65
To use confint we fit what we sometimes call the “trivial”
model
Yi = µ + i , i = 1, . . . , n
Clear-coat thickness (cont’d)
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2.5 % 97.5 %
(Intercept) 3.5154 5.7071
Discharge times
70
Another way of evaluating a confidence intervals on µ is with
the confint function, which provides confidence intervals on
the parameters in a fitted model.
The estimate of µ, µ
b = ȳ will be called (Intercept) in the
output. The formula for the model contains the constant
term, 1, as the only predictor.
●
●
0.1
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s = 2.718 The sample standard deviation, σ̂
n − 1 = 39 The degrees of freedom, ν, for the variance estimate,
s2 .
p
√s = 0.4297 The standard error of the mean,
Var(Ȳ)
n
Sample sizes
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The half-width of a confidence interval, also called the margin
of error depends on
The confidence level Higher confidence levels require wider
intervals
The standard deviation More variability in the original
distribution results in wider intervals.
The sample size Larger samples produce narrower intervals.
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Given a working value for σ we can determine the sample size
needed to attain a given margin of error.
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If we are willing to assume that n is large we can use z(α/2)
in the calculation. For small n it gets tricky because
ν = n − 1 determines the multiplier which, in turn, affects the
sample size. We must solve a nonlinear equation but
computers are good at that.
Section 5.3: Prediction and tolerance intervals
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A confidence interval on µ provides a measure of the precision
of the information regarding the unknown population
parameter. It does not directly tell us about bounds on where
we expect a future observation to fall.
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A prediction interval indicates where a single future
observation is likely to be.
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A tolerance interval indicates where a large proportion of the
population is likely to be.
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Unlike the confidence interval on µ, prediction intervals and
tolerance intervals depend strongly on the shape of the
distribution of the data.
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In theory one can make a confidence interval arbitrarily narrow
by taking a sufficiently large sample. You can’t do this for a
prediction interval.
Sample size calculations
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Example 5.2.5 shows calculations for the sample size from the
h
i2
when the desired margin of error, d,
formula n = t(α/2;∞)s
d
is 0.2, the working value of s is 0.4 and α is 5% and we round
the answer to the next largest integer.
> ceiling((qnorm(0.025)*0.4/0.2)^2)
[1] 16
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Because this
h is a small
i2 value of n we should instead solve for
t(α/2;n−1)s
n in n =
d
> ceiling(uniroot(function(x) x-(qt(.025,x-1)*0.4/0.2)^2,
+
c(2,100))$root)
[1] 18
Prediction intervals on a future observation
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If it is reasonable to assume that the data (i.e.
Y1 , Y2 , . . . , Yn ) are from normal distribution then we could
say that a model for the data is
Yi = µ + i ,
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i ∼ N (0, σ 2 )
Our estimate µ
b = Ȳn is independent of n+1 . The variability
in the difference between Yn+1 and Ȳn is the sum of the
2
variability in Ȳn − µ ( σn ) and the variability in n+1 (σ 2 ).
Because we estimate σ 2 the (1 − α) prediction interval
becomes
r
1
ȳ ± t(α/2; n − 1)s 1 +
n
Evaluating a prediction interval
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The prediction interval could be evaluated according to the
formula. For the clear-coat thickness data the 95% prediction
interval on a future thickness measurement is
> with(ccthickn, mean(thickness) + c(-1,1) * qt(0.975, 39) *
+
sd(thickness) * sqrt(1 + 1/40))
[1] 58.69478 69.82522
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Tolerance intervals
An alternative is to use the predict function applied to the
trivial model and with the optional argument interval =
"pred". This produces a matrix with n rows that are identical
so I just look at the first row.
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A tolerance interval is more difficult to describe and to
calculate than is a prediction interval.
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Methods for tolerance intervals are given in the text but we
will not cover this topic in this course.
> predict(fm2, int = "pred")[1,]
fit
lwr
upr
64.26000 58.69478 69.82522
Section 5.4 Hypothesis tests
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A hypothesis test is a procedure for deciding if a particular
value of a parameter is reasonable, given the observed data.
We have a probability model (e.g. our sample is a random
sample from a normal distribution with mean µ and variance
σ 2 ), the observed data, y1 , y2 , . . . , yn , and a particular value
of the parameter in mind (e.g. the mean clear-coat thickness,
µ, should be 65 microns).
We consider two competing claims called the null hypothesis,
written H0 , and the alternative hypothesis, written Ha .
H0 is the “no change” assumption. For our example, it is
µ = 65. Ha is the result we are trying to establish. It is also
the result indicated by the data. In our example ȳ = 64.26
microns. If we are interested only in whether we are “off
target” then Ha : µ 6= 65. If we are interested in whether the
clear coats are systematically too thin then Ha : µ < 65.
These are called “two-tailed” and “one-tailed” alternatives,
respectively.
The p-value for a hypothesis test
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We would like to establish Ha directly but, because of the
variability in the data, we can’t.
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Instead we try to “rule out” H0 . Again, because of the
variability we can’t rule it out completely. What we do is to
calculate “the probability of seeing the data that we did, or
something even more unusual, assuming that H0 is true”.
This is called the p-value for the test.
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Because the p-value is a probability it must be between 0 and
1. If the p-value is small (i.e. close to 0) we reject H0 in favor
of Ha . If the p-value is not small we fail to reject H0 .
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Note that we never confirm H0 . We either reject it (i.e. rule
it out) or fail to reject it (i.e. are unable to rule it out). The
latter conclusion represents “no decision”.
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The p-value is evaluated assuming H0 is true. The form of
the evaluation depends on Ha .
Performing a hypothesis test
Examples 5.4.5, 5.4.7 and 5.4.10
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To set up a hypothesis test you must first decide on H0 and
Ha . See pages 190–191 in the text for examples. Determine if
you want a one-sided or two-sided alternative. If two-sided
you are done. If one-sided then the only claim that makes
sense as Ha is the one indicated by the data.
The text describes methods based on rejection regions and
critical values of test statistics. These are used when you can’t
calculate probabilities for distributions like the T distribution.
We can do that so we use the more direct approach.
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H0 : µ = 230
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The t.test function in R is used for one- or two-sample t
tests on the population mean, µ.
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In the one-sample form you must specify the variable name as
the first argument. Use with or the $ operator to access a
variable in a data frame.
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You should specify the nominal value µ0 of the population
mean as the mu argument.
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The default alternative is "two.sided". Specify alt =
"greater" or alt = "less" for one-sided alternatives.
vs.
Ha : µ < 230
The observed t statistic, assuming that H0 is true, is
tobs =
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Use of the R function t.test
The situation described in examples 5.4.5 and 5.4.7 involves
drums of material that should have a mass of 230 kg. A
random sample of size 25 yields ȳ = 229 and s = 4.
The sample mean is less than the nominal value of 230 kg.
We wish to determine if it is “significantly less” than 230 kg.
Our test is of the form
229 − 230
√
= −1.25
4/ 25
The probability of seeing values like this, or even more
extreme, when H0 is true, is P (T24 ≤ −1.25) which we
evaluate as pt(-1.25, df = 24) = 0.1117. This value is not
sufficiently small for us to “rule out” or reject H0 .
Clear-coat thickness data
The ccthickn data are from a process with a target thickness of
65 microns but the sample mean, ȳ = 64.24 microns. Is this
“significant evidence” the µ < 65?
> with(ccthickn, t.test(thickness, mu = 65, alt = "less"))
One Sample t-test
data: thickness
t = -1.7221, df = 39, p-value = 0.04648
alternative hypothesis: true mean is less than 65
95 percent confidence interval:
-Inf 64.98398
sample estimates:
mean of x
64.26
Sample sizes
Use of power.t.test in R
I
Our conclusion in a hypothesis test is either to “rule out” or
reject H0 in favor of Ha or to fail to rule H0 out.
I
We don’t know if H0 is true or not. If it is true and we reject
it, we have made one type of “error” or mis-classification. If it
is false and we fail to reject it we have made another type of
“error”. These are called Type I and Type II, respectively.
I
For a test on a particular set of data the probability of a Type
I error in rejecting H0 is the p-value. For planning purposes
we control for this type of error by saying it should not exceed
some value, α. Typically α = 0.05 or α = 0.01.
I
To evaluate the probability of a Type II error we must specify
a specific alternative, such as µ1 , instead of a general
alternative like µ 6= µ0 . We also need a working value of σ.
I
The function power.t.test can be used to evaluate the power
or the sample size for one- and two-sample t-tests.
I
Two out of the three arguments n, delta (= µ1 − µ0 ) and
power must be specified and the third is calculated.
I
By default delta is in standard deviation units (i.e.
0
δ = µ1 −µ
σ ). If you want to specify delta in the units of the
response you should also give a value for the optional
argument sd (= working value of σ).
I
The default test type is "two.sample" (the next chapter). For
this chapter specify type = "one.sample".
I
The default alternative is "two.sided". For one-sided
alternatives specify alt = "one".
The power of a test is a function of µ1 . It is the probability of
rejecting H0 : µ = µ0 when µ = µ1 . Sometimes it is written
as 1 − β. In the text it is γ.
Example 5.4.13
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I
Example 5.4.14
This example involves a test of H0 : µ = 5 versus Ha : µ > 5
(i.e. a one-sample test with a one-sided alternative) with
σ = 1, α = 0.05, µ1 = 6 and a power of 80% or 0.8.
> power.t.test(delta = 1, sd = 1, sig = 0.05, power = 0.8,
+
type = "one", alt = "one")
One-sample
n
delta
sd
sig.level
power
alternative
t
=
=
=
=
=
=
test power calculation
7.727622
1
1
0.05
0.8
one.sided
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We round fractional sample sizes up so we would use a sample
of size n = 8. Note that the sample size calculated in this
example in the text, assuming a “known” value of σ and not
an estimated value, is n = 7.
I
The values sig = 0.05 and sd = 1 are the defaults and could
be omitted.
I
This example is like the previous one except that the
magnitude of δ = µ1 − µ0 is now 1.5 mg and the desired
power is 90%.
I
The larger tolerance decreases n and the higher power
increases n. The net effect is to decrease n.
> power.t.test(del = 1.5, pow = 0.9, type = "one", alt = "one")
One-sample
n
delta
sd
sig.level
power
alternative
t
=
=
=
=
=
=
test power calculation
5.471726
1.5
1
0.05
0.9
one.sided
Example 5.4.16
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I
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Section 5.5, Inference for Binomial Populations
As explained in the text, in practice we should use the T
distribution and not the standard normal distribution for
calculating sample sizes. This is what power.t.test does.
The text provides an alternative based on tables but they are
very coarse.
The important quantity is written in the text as d =
the R function it is delta
µ−µ0
σ .
In
Results from a binomial simply consist of the number of
“successes”, y, and the number of trials, n. This is already an
i.i.d. sample, Y1 , Y2 , . . . , Yn , if we consider each Yi as giving
a binary (0/1) response.
I
The parameter estimate is p̂ = ny .
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For this example d = 0.4 with 90% power on a one-tailed,
one-sample test.
> power.t.test(del = 0.4, pow = 0.9, typ = "one", alt = "one")
One-sample
n
delta
sd
sig.level
power
alternative
t
=
=
=
=
=
=
I
test power calculation
54.90553
0.4
1
0.05
0.9
one.sided
Example 5.5.4
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We observe y = 18 defectives in a sample of n = 1000 from a
process where the nominal defect rate is 1%. Test
H0 : p = 0.01 versus H0 : p > 0.01
> binom.test(18, 1000, p = 0.01, alt = "greater")
Exact binomial test
data: 18 and 1000
number of successes = 18, number of trials = 1000,
p-value = 0.01383
alternative hypothesis: true probability of success is greater than 0.01
95 percent confidence interval:
0.01166572 1.00000000
sample estimates:
probability of success
0.018
Often approximate confidence intervals and hypothesis tests
are formulated based on a normal distribution for P̂ using a
mean of p and a variance of p(1−p)
n .
However, we can do better. In particular, the p-value for a
hypothesis test can be calculated exactly. We use binom.test
for a one-sample binomial test and prop.test for comparing
results from multiple samples. Sample sizes are calculated
with power.prop.test.
Comparison to results in the text
I
In the text the p-values for tests on a binomial distribution are
calculated through a normal approximation and not the
binomial distribution.
I
This results in p-values that are lower than they should be,
which is dangerous. E.g. the text’s p-value for Example 5.5.5
is 0.0055 versus the exact calculation of 0.01383. That could
be important.
I
Example 5.5.6 in the text gives a p-value of 0.0089 for a test
of H0 : p = 0.05 versus Ha : p > 0.05 for y = 8 and n = 72.
The more accurate value is
> binom.test(8, 72, p = 0.05, alt = "greater")$p.value
[1] 0.02715970
> sum(dbinom(18:1000, 1000, 0.01)) # explicit P(Y >= 18) for H0 true > sum(dbinom(8:72, size = 72, prob = 0.05))
[1] 0.01383258
[1] 0.02715970
Confidence intervals on p
I
The binom.test function produces a confidence interval based
on “inverting” the exact hypothesis test. Usually we want to
two-sided alternative when forming a confidence interval.
I
The conf.level argument can be used to set the desired
confidence level (default is 95%).
I
These intervals are more reliable than those given in the book.
I
Example 5.5.1 has y = 27 out of n = 200. The 95%
confidence interval should be
> binom.test(27, 200)$conf.int
[1] 0.0908871 0.1903092
attr(,"conf.level")
[1] 0.95
Notice that it is wider on the side towards p = 0.5 than on
the other side. This is a characteristic of the binomial.
Sample sizes for tests on the binomial
I
Formulas (5.5.4), (5.5.7) and (5.5.8) in the text allow for
calculating sample sizes for a confidence intervals with a given
margin of error, d, or for tests with a null hypothesis of the
form H0 : p = p0 .
I
All of these formulas are based on the normal approximation
and will give values that are a bit too small.
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In the two-sample case (next chapter) we can use the function
power.prop.test to determine the sample size for the two
samples.
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