Download Inference for 1 Sample

Purposes of These Notes
• Describe point estimation, interval estimation, and hypothesis testing.
• Describe a random sample.
• Define a confidence interval and its level.
• Derive some confidence intervals in 1 sample problems: means and proportions.
• Discuss difference between Fisher and NeymanPearson.
• Describe ingredients of Neyman-Pearson hypothesis testing.
1
Purposes Continued
• Define null and alternative hypotheses.
• Define a test statistic, rejection region, level.
• Define a Type I and Type II errors.
• Differentiate between one-tailed and twotailed problems.
• Specific formulas for hypotheses about means
and proportions.
• Define a P -value.
• Understand technical meaning of statistically significant.
2
List of Statistical Problems
• Name most likely value of parameters: point
estimation.
• Name range of likely values: confidence interval.
• Assess evidence against hypothesis about
parameters: hypothesis testing.
• Make forecasts, do interpolation.
• And more.
3
Point Estimation
• Estimate: number which is our best guess
for parameter value.
• Estimator : rule for computing estimate from
data.
• An estimator is a random variable which is
a function of the data.
• Example. Newcomb & Michelson measured
speed of light in 1880s.
• Made 66 measurements of time taken by
light to travel 7.44373 km.
4
• Measured values are X1, X2, . . . , Xn with n =
66.
• Use lower case letters for observed values.
• First measurement was 24.828 millionths
of a second.
• Convert measurement to speed of light.
• x1 = 109·7.44373/24.828 = 2.998119×108
m/s.
• x2 = 2.998361 × 108 m/s.
• Point estimate of speed of light is 2.998336×108
m/s.
5
Estimators
• We were using the rule: average the data.
• So our estimator was
X + · · · + Xn
.
X̄ = 1
n
• Model for measurement error.
• Several parts: X1, . . . , Xn independent and
identically distributed.
6
• Let µ = E(Xi) be the population mean.
• Long run average measurement.
• Population SD is σ.
• Speed of light is c — standard notation.
• Relate µ to c:
µ = c + bias
• Often assume bias is 0.
7
Newcomb data
8
Point Estimation
• Have data and model for population.
• Model describes population in terms of some
parameters.
• Binomial(n, α) model: α is a parameter.
• Sample from a N (µ, σ 2) model. Parameters are µ and σ.
• Sample from the Gamma density
1
x
f (x; α, β) =
βΓ(α) β
!α−1
exp(−x/β)
x > 0.
• Parameters are α and β.
• Generic notation: θ.
9
Standard Errors
• Estimates should always be accompanied
by some assessment of their likely accuracy.
• For unbiased estimators with approximately
normal sampling distributions we use the
Standard Error.
• The SE of an estimator θ̂ of θ is
SE =
q
Var(θ̂).
10
• That is: Standard Error of an estimator is
another name for its SD.
• The standard error of α̂ in the Binomial(n, α)
problem is
q
• The SE of X̄ is
p(1 − p)
√
n
σ
√ .
n
11
Estimated Standard Errors
• What accompanies our point estimate is a
number, not a formula.
• The SE is usually a formula with unknown
parameters in it.
• We estimate the SE by plugging in estimates of the parameters.
12
• The SE for α̂ is
SE is
q
α(1 − α)/n so Estimated
q
α̂(1 − α̂)
.
√
n
• And you plug in data to get a number to
put in your report.
• We use Standard Errors in Confidence Intervals.
13
Confidence Interval Definition
• A level β confidence interval for a parameter θ is the interval [L, U ] between two
statistics L and U such that
P (L ≤ θ ≤ U ) ≥ β
for all possible parameter values.
• We prefer to replace ≥ by = or ≈.
14
• We use CIs by:
– Deciding how to do data analysis before
gathering data (decide on formulas for
L and U before getting data).
– Get data; compute observed values of L
and U , say l and u.
– Say ‘I am 100β% confident that θ is the
interval [l, u]’.
• 1 − β is the error rate or non-coverage rate.
15
Populations and Samples
• Meaning of a sample from a population.
• Population is group we want to find out
about.
• Can be real: all Canadian adults of working
age.
• Can be ‘conceptual’: all possible outcomes
of some experiment.
• Populations often thought of as populations of numbers.
• Conceptual populations often described by
probability density or pmf.
16
• Examples: heights of adults. Think of
population as being normally distributed with
mean µ and sd σ.
• Example: repeatedly measure speed of light
in a vacuum. Each measurement is ‘truth’
plus ‘measurement error’. Population of
errors describe by density: N (0, σ 2) perhaps.
17
Populations and Samples
• Sample is part of the group for which data
is obtained.
• Use n for number of items sampled.
• Call it a “single sample” problem if we
measure 1 number for each item sampled.
• Call measurements X1, . . . , Xn .
• Random sampling: fixed number of members of group selected by random mechanism playing no favourites.
18
• With replacment: pick one at a time. On
ith selection each member of population
has same chance of being drawn, even if
that member has been picked before.
• Usual model for conceptual populations.
• Without replacment: pick one at a time.
On ith selection each member of population who has not been drawn yet has same
chance of being drawn.
• Common model (sampling method) for real
populations.
• Neither is usual selection method in real
surveys.
19
Simplest Derivation of Confidence
Interval
• Mathematical model for a single sample:
X1, . . . , Xn are independent and identically
distributed. Write ‘iid’.
• Simplest populations to describe – approximately normal, like heights.
• Suppose X1, . . . , Xn are independent N (µ, σ 2).
• Suppose (quite unrealistically) that σ is known.
• I now show you a 95% confidence interval
for µ, based on the data.
20
• Consider the random variable
Z=
X̄ − µ
√ .
σ/ n
• Then, regardless of what µ is, Z has a standard normal distribution.
• So
P (a ≤ Z ≤ b)
does not depend on µ.
• No matter what µ is
P (−1.96 ≤ Z ≤ 1.96) =
Z 1.96
−1.96
φ(z) dz = 0.95.
21
The Confidence Interval
• The event −1.96 ≤ Z ≤ 1.96 can be rewritten in a number of ways.
• It is the event
−1.96 ≤
X̄ − µ
√ ≤ 1.96.
σ/ n
√
• Multiply by σ/ n (which is positive):
σ
σ
−1.96 √ ≤ X̄ − µ ≤ 1.96 √
n
n
Notice this is still the event.
22
• Rearrange second inequality:
σ
L ≡ X̄ − 1.96 √ ≤ µ
n
• Rearrange first inequality:
σ
µ ≤ X̄ + 1.96 √ ≡ U
n
• So no matter what µ is:
P (L ≤ µ ≤ U ) = 0.95
• The range L to U is a 95% confidence interval.
23
An example with data
• Simon Newcomb made 66 measurements
of time taken by light to travel 7.44373
km.
• I round off a bit from real data.
• Convert to list of 66 speeds.
• Sample mean is 299,833,533 m/s.
• Temporarily assume σ = 130, 000 m/s is
known.
24
• 95% confidence interval is
299, 833, 553 − 1.96 ×
130, 000
√
66
to
130, 000
299, 833, 553 + 1.96 × √
m/s.
66
• We say we are 95% confident that the
speed of light is between 299,802,189 and
299,864,917 m/s.
25
Caveats and improvements
• More digits than is wise but 6 leading digits
worth reporting.
• The quantity
130, 000
√
m/s
66
is called the standard error of the sample
mean.
26
• Pretty well everything is an approximation
so many data analysts round 1.96 to 2.
• We are only pretending we know σ.
• Usually we have to use the data to tell us
about σ as well as about µ.
• Notation: define upper α critical point of
normal by:
P (N (0, 1) > zα) = α.
• So z0.025 = 1.96.
27
The role of normality
• We assumed initially that the population
we are sampling is, itself normally distributed.
• But our basic probability was:
P
−1.96 ≤
!
X̄ − µ
√ ≤ 1.96
σ/ n
=
Z 1.96
−1.96
φ(z) dz = 0.95.
• Accuracy depends on sampling distribution
of X̄.
28
• Central limit theorem says: if n large enough
this is normal for (nearly) any population
distribution.
• More skewness means larger n needed.
• Heavy tails mean larger n needed.
• We often use rule of thumb: n ≥ 30.
• Message: use same formula if n large:
σ
X̄ ± zα/2 √ .
n
29
Unknown SD, lots of data
• Actually Newcomb did not know σ at all.
• He measured s, the SD of his 66 measurements.
• In fact s = 130, 026 m/s.
• When n is large s will be close to σ so
X̄ − µ
X̄ − µ
≈
√
√ .
σ/ n
s/ n
30
• So just replace σ by s in confidence interval.
• We are 90% confident that the speed of
light is in the range
130, 026
299, 833, 553 − 1.645 × √
66
to
299, 833, 553 + 1.645 ×
130, 026
√
.
66
• The Estimated Standard Error is
130, 026
√
66
• It estimates the Standard Deviation of X̄.
• Notice use of z0.05 = 1.645 not z0.025 =
1.96.
31
Small samples – Student’s t distribution
• How good is the approximation?
• Estimated SD of X̄ using same data from
which we computed the mean.
• So we should use something a bit bigger
than 1.645.
• For 66 observations that ‘bit bigger’ is 1.669.
• Correct critical point comes from Student’s
t distribution.
32
More probability – small samples
• When sampling from a normally distributed
population we have:
P
!
x
X̄ − µ
fT,n−1(u)du
√ ≤x =
S/ n
−∞
Z
where fT,n−1 is Student’s t-density “with
n − 1 degrees of freedom”.
• To be precise – but this density is not part
of this course:
Γ((ν + 1)/2)
fT,ν (u) = √
(1 + u2/ν)−(ν+1)/2
πνΓ(ν/2)
• As ν → ∞ this converges to the standard
normal density.
• Curve looks a lot like normal but heavier
tails.
33
Specific scientific settings
• Specific settings have specific formulas for
Estimated SE.
• Scenario 1: sample from normal population, σ (population SD) known, CI for population mean, µ.
• Interval (already done) is
√
X̄ ± zα/2σ/ n
• Scenario 2: sample from general population, σ (population SD) unknown, sample
size n large, CI for population mean, µ.
• Interval (already done) is
√
X̄ ± zα/2s/ n
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• Scenario 3: sample from normal population, σ (population SD) unknown, sample
size n anything, CI for population mean, µ.
• Interval is
√
X̄ ± tα/2,n−1s/ n
• Multipliers tα/2,n−1 from other table at back
of text.
• Statistical packages always do Scenario 3
arithmetic.
35
Confidence intervals for proportions
• Common scientific framework
• Sequence of Bernoulli trials.
• Number n fixed, p is “Success Probability”
on each trial.
• X is the number of successes.
• Goal: confidence interval for proportions.
• Based on Central Limit Theorem.
36
Using the CLT
• Recall p̂ = X/n and X = X1 + · · · + Xn;
each Xi is Bernoulli.
• So p̂ is a sample mean of the Xi.
• Population mean is µ = E(Xi) = p.
• Population variance is
σ 2 = Var(Xi ) = p(1 − p).
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√
• So SE of p̂ is σ/ n =
q
√
p(1 − p)/ n.
• Estimated SE is usually taken to be
q
√
p̂(1 − p̂)/ n.
• CLT says
q
p̂ − p
p(1 − p)/n
⇒ N (0, 1).
38
Using the CLT 2
• Law of large numbers says:
lim p̂ = p
n→∞
• So it is also true that
p̂ − p
q
p̂(1 − p̂)/n
⇒ N (0, 1).
• Result is:


p̂ − p


lim P −zα/2 ≤ q
≤ zα/2
n→∞
p̂(1 − p̂)/n
=
Z z
α/2
−zα/2
φ(z)dz = 1 − α.
• Leads to approximate level 1−α confidence
interval.
39
Solving inequalities to get limits
• Temporary notation A is the event
−zα/2 ≤ q
p̂ − p
p̂(1 − p̂)/n
≤ zα/2.
• Solve inequalities in A to isolate p: multiply
through by SE:


s
p̂(1 − p̂)
A = −zα/2
≤

n
p̂ − p ≤ zα/2
s

p̂(1 − p̂) 
n
.

• Rearrange each inequality: right hand gives
q
p̂ − zα/2 p̂(1 − p̂)/n ≤ p.
• Similarly for left inequality get
q
p ≤ p̂ + zα/2 p̂(1 − p̂)/n.
40
General points
• Most essential: the meaning of confidence:
• If we analyze 100 data sets and compute
100 (exact) confidence intervals at the 95%
level we expect that some of the 100 intervals will contain the truth and some won’t.
• The expected number which contain the
truth is 95.
41
• The number which contain the truth is random.
• Rule of thumb: if np > 10 and n(1−p) > 10
then normal approx is fine.
• You don’t know p but you use p̂ in the rule
of thumb.
• Text uses 5 instead of 10. That is ok, too.
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A catalogue of confidence intervals
• Intervals for population proportions; done
earlier.
• Intervals for population means.
– Samples from Normal populations with
known σ.
– Samples from Normal populations with
unknown σ.
– Large samples from more general populations.
43
Confidence statements, normal
populations
• Normal sample, σ known:
X̄ − µ
P (−z ≤
√ ≤ z) = Φ(z) − Φ(−z)
σ/ n
so if we find z so that Φ(z)−Φ(−z) = 1−α
then
√
√
X̄ − zσ/ n to X̄ + zσ/ n
is an exact level 1 − α confidence interval
for µ.
• Value of z is denoted zα/2 because
P (N (0, 1) > z) = α/2 = P (N (0, 1) < −z)
in this case. We call zγ the upper tail γ
critical point.
44
Confidence statements, normal
populations
• Normal sample, σ unknown:
t
X̄ − µ
fT,n−1(u)du
P (−t ≤
√ ≤ t) =
S/ n
−t
Z
Rt
so if we find t so that −t fT,n−1(u)du =
1 − α then
√
√
X̄ − tS/ n to X̄ + tS/ n
is an exact level 1 − α confidence interval
for µ.
• Value of t is denoted tα/2,n−1 because
P (T > t) = α/2 = P (T < −t)
Again tγ,ν is notation for the upper γ critical point of a Student’s t-distribution on
n − 1 degrees of freedom.
45
Confidence statements, large samples,
general populations
• Sample from population mean µ and unknown SD σ:
t
X̄ − µ
P (−t ≤
fT,n−1(u)du
√ ≤ t) ≈
S/ n
−t
≈ Φ(t) − Φ(−t)
Z
so
√
√
X̄ − tα/2,n−1S/ n to X̄ + tα/2,n−1S/ n
and
√
√
X̄ − zα/2S/ n to X̄ + zα/2S/ n
both approximate large sample level 1 − α
confidence intervals for µ.
• Very rarely: σ is known so replace S by σ
and use zα/2.
46
Confidence statements, large samples,
general populations
• Books traditionally recommend z for n ≥
30 or n ≥ 40 or some such rule of thumb.
• BUT I say just use t; software always does
and the t approximation is generally better.
• Rule of thumb comes from DARK AGES
before computers when people used the tables in the book.
• Those are for statistics exams, nothing else.
47
Typical hypothesis testing science
questions
• New drug for blood pressure. Get 200 patients. Pick 100 at random to get new
drug; others get old.
• Choose between two possibilities: drug reduces BP or doesn’t.
• Speed of light in vacuum is known. Measure speed of neutrinos. Is speed equal to
speed of light or not?
48
• Are far away galaxies moving away from
earth faster than nearby ones or not?
• Is speed of light same in north south and
east west directions?
• Does some intervention program in prison
reduce recidivism or not?
• Common feature: choose between two scientific alternatives.
49
Methodology
• Conduct experiment in which response (BP,
speed of neutrinos, two light speeds, recidivism) is measured.
• Formulate statistical models: data are like
a sample from a normal population; number of patients surviving has binomial distribution; north south speeds and east west
speeds like samples from 2 populations.
• Phrase the scientific alternatives as alternatives about the parameter values in the
model: mean north south speed equals mean
east west speed OR not; probability of reoffense in treatment group equals probability of re-offense in control group OR not
...
50
• Develop a rule to make a choice between
two alternatives.
• Understand error rates.
• Apply rule to data.
• Details follow.
51
Example 1: Measurement bias
• Newcomb makes n = 66 measurements of
time for light to travel 7.44373 km.
• Modern value for that time is 24.82961 microseconds.
• Is Newcomb biased?
• Model: each measurement is like draw from
a population of possible measurements. Data
is X1, . . . , Xn sample from population with
mean µ and SD σ.
52
• No bias translates to µ = 24.82961 microseconds.
• We say our null hypothesis, H0, is µ =
24.82961.
• Our alternative hypothesis, Ha, becomes
µ 6= 24.82961.
• H0 is pronounced “H nought” (“H not”).
53
The test statistic
• To make the decision we find a test statistic, T , which is function of data.
• It will depend on the number 24.82961 as
well.
• It should tend to be big if the alternative
hypothesis is right.
• It should NOT tend to be big if the null
hypothesis is right.
• We will calculate T and choose alternative
if it is “too big”.
54
• First obvious suggestion: T = |X̄−24.82961|.
• How big is too big? Compare T to variability of X̄ − 24.82961.
• Estimate that variability using Estimated
√
Standard Error s/ n of X̄.
• So change to
X̄ − µ 0
T =
√
s/ n 55
How big is too big?
• Two big approaches – assess evidence versus make firm decision.
• Fisher: summarize size of T by a P -value
and interpret this P value as strength of
evidence against null hypothesis.
• Formal decision making: select rejection
region. If T lands in rejection region we
reject the null hypothesis and behave as if
alternative hypothesis is true.
• Two approaches very closely connected.
• Neyman-Pearson approach first — formal
decision making.
56
• Recognize two kinds of errors.
• Type I error : Newcomb has no bias but we
say he did. Null hypothesis is true but we
say it is false.
• Type II error : Newcomb was biased but we
miss that fact. Null hypothesis is false but
we decide it is true.
• Language used in book: reject null hypothesis or fail to reject null hypothesis.
• Other places: “fail to reject” null hypothesis is called “accept null hypothesis”. You
behave as if null is true.
57
Making a decision
• For Newcomb our rejection region is
X̄ − µ 0
T =
>c
√
s/ n • c is critical point.
• How do we select c?
• Neyman Pearson method.
58
• Choose c to control Type I error rate.
• Select a pre-specified tolerable error rate:
usually 5%. Call this rate α.
• Find c so that
PHo (T > c) = α.
• PHo is notation to show that we compute
this chance assuming that the null hypothesis is true.
59
Specific scientific settings
• Scenario 1: sample from normal population, σ (population SD) known, hypothesis
tests for population mean, µ.
• Two sided alternative: H0:µ = µ0, Ha:µ 6=
µ0
and
X̄ − µ 0
T =
√
σ/ n c = zα/2
60
• One sided alternative. H0:µ = µ0, Ha:µ >
µ0 or H0:µ ≤ µ0, Ha:µ > µ0.
X̄ − µ0
T =
√
σ/ n
and
c = zα
• I expect you to know what to do if inequalities reversed.
61
Scenario 2, σ unknown
• Scenario 2: sample from general population, σ (population SD) unknown, sample
size n large, hypothesis tests for population
mean, µ.
• Two sided alternative: H0:µ = µ0, Ha:µ 6=
µ0
and
X̄ − µ 0
T =
√
s/ n c = tα/2,n−1
62
• One sided alternative.
H0:µ = µ0, Ha:µ > µ0
or
H0:µ ≤ µ0, Ha:µ > µ0.
X̄ − µ0
T =
√
s/ n
and
c = tα,n−1
63
Small samples
• Scenario 3: sample from normal population, σ (population SD) unknown, sample
size n anything, CI for population mean, µ.
• Use same method as Scenario 2.
• But now the method is exact.
• Without the normal population assumption
we are relying on the CLT and LLN and
Slutsky’s theorem.
64
Hypothesis tests for proportions
• Common scientific framework
• Sequence of Bernoulli trials.
• Number n fixed, p is “Success Probability”
on each trial.
• X is the number of successes.
65
• Goal is a hypothesis test for proportions.
• Method based on application of the Central
Limit Theorem.
• Same list of null / alternative choices: H0:p =
p0 or H0:p ≤ p0
• H0:p = p0 allows either 1 or 2 sided alternatives.
66
Using the CLT (repeat from CI notes!)
• Recall p̂ = X/n and X = X1 + · · · + Xn;
each Xi is Bernoulli.
• So p̂ is a sample mean of the Xi.
• Population mean is µ = E(Xi) = p.
• Population variance is σ 2 = Var(Xi ) = p(1−
p).
√
• So SE of p̂ is σ/ n =
q
√
p(1 − p)/ n.
• CLT says: if p = p0 then
q
p̂ − p0
p0(1 − p0)/n
⇒ N (0, 1).
67
Using the CLT 2
• Our test statistic is either
p̂ − p0
T =q
p0(1 − p0)/n
for Ha:p > p0 or
p̂ − p0
T = q
p0(1 − p0)/n for Ha:p 6= p0
• Critical value c is
zα/2
for two-sided alternative or
zα
for one-sided alternative.
68
Some scientific examples
• Cadmium in a lake example.
• n = 17 measurements of cadmium concentration. x̄ = 211, s = 15, units are parts
per million or some such. (Important but
these numbers are made up.)
• Scientific question: decide between two possibilities – concentration below 200 vs above
200.
• Typical one-sided situation.
69
• Need to connect data to scientific question
of interest.
• Introduce notation: X1, . . . , Xn are the 17
measurements.
• Must assume that they are gathered and
measured in such a way that they are a
sample of size 17 from a population whose
mean µ is “concentration of cadmium in
the lake”
• Definition of that last is scientific problem.
• Issues to consider: is the whole lake sampled? are the measurements biased? are
the measurement errors independent?
• Assume issues dealt with.
70
Cadmium
• For first pass I consider BOTH possible
H0s.
• For H0:µ ≤ 200 use
T =
X̄ − 200
√
s/ n
and reject if T > t0.95,n−1 = 1.75. (Notice
rejection region.)
• Notice use of borderline value, 200, in T .
71
• Plug in values and find
211 − 200
√
= 3.02
T =
15/ 17
• Since 3.02 > 1.75 we reject the hypothesis
that µ ≤ 200.
72
P -values
• BUT: in fact we can say a bit more.
• 3.02 is quite a bit bigger than 1.75.
• If we had used α = 0.01 instead of 0.05
our rejection region would be
T > t0.01,16 = 2.58
and we would still have rejected.
• In fact we would reject for any α for which
tα,16 < 3.02
73
• Smallest possible α is when
tα,16 = 3.02.
Or
P (T16 ≤ 3.02) = 1 − α = 1 − P (T16 ≥ 3.02)
• This α is Fisher’s P -value.
• Compute P by finding area to right of observed statistic under null density of statistic.
74
P -values
• Reject H0 at level α if P < α.
• If H0 is right then P has a Uniform[0,1]
distribution.
• Interpret P as measure of evidence strength
– smaller P , stronger evidence against H0.
75
• Call evidence statistically significant if P <
0.05.
• Highly statistically significant and very highly
statistically significant are often used for
smaller thresholds like 0.01 or 0.001.
• Some statistics packages label P -values with
1 star for P < 0.05, 2 stars for P < 0.01
and 3 stars for P < 0.001.
• These are all simply conventions.
• For two tailed problems: P is twice the
area in the small tail.
76
Example from Devore, Page 342 Q 65
• Sample of n = 50 lens thicknesses. Given
x̄ = 3.05 and s = 0.34 (all in mm).
• Desired mean thickness 3.20 mm.
• Do “the data strongly suggest that the
true average thickness of such lenses is
something other than what is desired”?
• Clear two sided alternative. Null must be
H0:µ = 3.20.
77
• Test statistic is
3.05 − 3.2 √ = 3.12
T =
0.34/ 50 • P value? Twice area to right of 3.12 under
t on 49 df.
• P = 0.003 which is very significant. (Table
A.8 gives P in range of 0.002 to 0.004.)
• So we see very strong evidence against the
assertion that the true average thickness is
3.2mm.
• We would reject null at α = 0.05 or even
α = 0.01.
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Error rates and sample size calculations
• Type I error: incorrectly reject H0.
• Type II error: incorrectly fail to reject H0.
• Type I error rate is α; determined in advance.
• Type II error rate is β – depends on what
true parameter value is.
• Can sometimes compute β = P (don’t reject)
as a function.
79
• Answer will depend on n.
• Can then sometimes choose n to give suitable sample size.
• But often n depends on unknown parameters like σ.
• So we design for some hoped for value of
σ.
80
Sample size, Z test, 1 sided
• Imagine testing µ ≤ µ0 against µ > µ0.
• Assume that σ is known.
• Fix some α like 0.05.
• So reject if
Z=
X̄ − µ0
√ > zα .
σ/ n
• Compute β:
β=P
!
X̄ − µ0
√ < zα .
σ/ n
81
• For β > β0 we make a type II error is Z <
zα .
• Centre on correct µ:
β=P
µ − µ0
X̄ − µ
√ +
√ < zα
σ/ n
σ/ n
!
√
• Area to left of zα − (µ − µ0)/(σ/ n):
β = Φ zα −
µ − µ0
√
σ/ n
!
82