1

Power of the Test & Likelihood Ratio Test

<Today’s Topic>

Power Calculation (Inference on one population mean)

Likelihood Ratio Test (one population mean, normal

population)

You can watch this and other related ‘funny’ videos

about statistics (power of the test etc.) at YouTube:

https://www.youtube.com/watch?v=kMYxd6QeAss

Truth

0H aH

Decision 0H Type II error

aH Type I error

β = P(Type II error) = P(Fail to reject 0H | aH )

Power = 1- β = P(Reject 0H | aH )

2

Relationship between α and β (and of course, 1-β) The following diagrams illustrate the relationship between the Type I error and the Type II error (and thus power of the test). You can see that for the given sample, α and β have a competitive relationship. When you decrease one, the other will increase.

Power Calculation (Inference on one population mean)

1.

)(:

:

00

00

aaH

H

1st scenario, Normal population, 2 is known.

Test statistic : )1,0(~0

00 N

n

XZ

H

At the significance level , we reject 0H if ZZ 0

3

Power of the test = 1-

= P(reject 0H | aH ) = P( )|0 aZZ

= )|( 0

aZn

XP

,

If a , )1,(~ 000

nN

n

XZ a

Therefore,

1- = )|( 0a

aa Znn

XP

= )|( 0a

a

nZZP

, )1,0(~ NZ

2nd scenario, Normal population, 2 is unknown.

Test statistic : 10

0 ~0

n

H

tnS

XT

At the significance level , we reject 0H if ,10 ntT

Power = 1- = P(reject 0H | aH ) = P( )|,10 antT

4

= )|( ,10

antnS

XP

= )|( ,10

anaa t

nSnS

XP

= )|( 0,1 a

an

nStTP

, Here 1~ ntT

Recall that the Shapiro-Wilk test: can be used to determine

whether the population is normal.

3rd scenario, Any population (*usually we use this test when the

population is found not normal), large sample (n30 )

Test statistic : )1,0(~0

00 N

nS

XZ

H

At the significance level , we reject 0H if ZZ 0

Power = 1- = P(reject 0H | aH ) = P( )|0 aZZ

5

= )|( 0

aZnS

XP

= )|( 0a

aa ZnSnS

XP

= )|( 0a

a

nSZZP

, )1,0(~ NZ

2.

)(:

:

00

00

aaH

H

1st scenario, Normal population, 2 is known.

Test statistic : )1,0(~0

00 N

n

XZ

H

At the significance level , we reject 0H if ZZ 0

Power = P(reject 0H | aH ) = P( )|0 aZZ

6

= )|( 0aZ

n

XP

=

)|( 0a

aa Znn

XP

= )|( 0a

a Zn

ZP

, )1,0(~ NZ

The derivation of the remaining two scenarios is very similar

to that of the first pair of hypotheses too and is thus omitted.

3.

0

00

:

:

aaH

H

1st scenario, Normal population, 2 is known.

Test statistic : )1,0(~0

00 N

n

XZ

H

At the significance level , we reject 0H if 20 ZZ

Power = P(reject 0H | aH ) = P( )|( 20 aZZ

7

= )|()|( 2020 aa ZZPZZP

=

)|()|( 20

20

aaa

aaa Z

nn

XPZ

nn

XP

=

)|()|( 02

02 a

aa

a

nZZP

nZZP

(*Without loss of generality, for the plot below, we assume

0 a )

The derivation of the remaining two scenarios is very similar

to that of the first pair of hypotheses too and is thus omitted.

8

Ex1) Jerry is planning to purchase a sports goods store. He

calculated that in order to make profit, the average daily sales must

be 525$ . He randomly sampled 36 days and found 565$X

and 50$S

If the true average daily sales is $550, what is the power of Jerry’s

test at the significance level of 0.05?

Solution:

𝐻0: 𝜇 = 𝜇0 = 525

𝐻𝑎: 𝜇 = 𝜇𝑎 = 550 > 𝜇0

Power = P(Reject aHH |0 )

9123.02

9131.09115.0

)355.1()3650

525550645.1(

)|(

)|(

)|(

0

0

0

ZPZP

nSZ

nS

XP

ZnS

XP

ZZP

aaa

a

a

Ex2) John Pauzke, president of Cereal’s Unlimited Inc, wants to

be very certain that the mean weight of packages satisfies the

package label weight of 16 ounces. The packages are filled by a

machine that is set to fill each package to a specified weight.

However, the machine has random variability measured by 2 .

John would like to have strong evidence that the mean package

weight is about 16 oz. George Williams, quality control manager,

advises him to examine a random sample of 25 packages of cereal.

From his past experience, George knew that the weight of the

9

packages follows a normal distribution with standard deviation 0.4

oz. At the significance level 05.0 ,

(a) What is the decision rule (rejection region) in terms of the

sample mean X ?

(b) What is the power of the test when 13.16 oz?

Sol) Let X denote the weight of a randomly selected package of

cereal, then )4.0,16(~ NX

16:

16:0

aH

H

16:

16:0

aH

H

(a) Test Statistic : )1,0(~0

00 N

n

XZ

H

if 160

ZcHcZP )|( 00

We reject 0H at 05.0 if

)(1316.16

25

4.0645.1160

00

oz

nZXZ

n

XZ

(b)

0

00

13.16:

16:

aaH

H

(n=25)

Power = P(Reject aHH |0 )

49.0)02.0()254.0

1613.16645.1(

)|(

)|(

)|(

0

0

0

ZPZP

nZ

n

XP

Zn

XP

ZZP

aaa

a

a

10

Sample size determination in the hypothesis

test scenario

1.

)(:

:

00

00

aaH

H

1st scenario, Normal population, 2 is known.

1 = )|( 0a

a Zn

ZP

, )1,0(~ NZ

Z

nZ

nZ aa

00

a

ZZn

0

)(

2

0

22

)(

)(

a

ZZn

2.

)(:

:

00

00

aaH

H

11

1st scenario, Normal population, 2 is known.

2

0

22

)(

)(

a

ZZn

(*the same result as in Scenario 1 – in

summary, the formula is identical for the one-sided tests.)

3.

0

00

:

:

aaH

H

1st scenario, Normal population, 2 is known.

1 =

)|()|( 02

02 a

aa

a

nZZP

nZZP

(Assume 0 a . Then, 0)|( 02

a

a

nZZP

.

So, we can neglect it.)

12

ZZn

nZZ

nZZP

a

a

aa

2

0

0

2

0

2 )|(1

2

0

22

2

)(

)(

a

ZZn (* in summary, this hand-calculated formula

for the two sided test differs from that for the one-sided tests in

two aspects:

(1) α is replaced by α/2;

(2) it is an approximate formula, not an exact formula.)

13

Likelihood Ratio Test (one population mean, normal population, two-sided)

1. Please derive the likelihood ratio test for H0: μ = μ0 versus Ha: μ ≠ μ0, when the population is normal and population variance σ2 is known. Solution: For a 2-sided test of H0: μ = μ0 versus Ha: μ ≠ μ0, when the population is normal and population variance σ2 is known, we have:

0: and :

The likelihoods are:

n

i i

n

in

ix

x

LL

1

2

02

2

22

2

0

1 2

0

2

1exp

2

1

2exp

2

1

There is no free parameter in L , thus LL ˆ .

n

i i

n

in

ix

x

LL

1

2

2

2

22

2

1 2 2

1exp

2

1

2exp

2

1

There is only one free parameter μ in L . Now we shall find

the value of μ that maximizes the log likelihood

n

i ixn

L1

2

2

2

2

12ln

2ln

.

By solving

n

i ixd

Ld12

01ln

, we have x̂

It is easy to verify that x̂ indeed maximizes the

loglikelihood, and thus the likelihood function. Therefore the likelihood ratio is:

14

2

0

2

0

1

22

02

1

2

2

2

2

1

2

02

2

20

0

2

1exp

/2

1exp

2

1exp

2

1exp

2

1

2

1exp

2

1

ˆ

maxˆ

ˆ

zn

x

xxx

xx

x

L

L

L

L

L

L

n

i ii

n

i i

n

n

i i

n

Therefore, the likelihood ratio test that will reject H0 when

* is equivalent to the z-test that will reject H0 when

cZ 0 , where c can be determined by the significance level α

as / 2c z .

You see that from the beauty contest example that to set which theory as the null hypothesis may not always follow the same rule. That is why for the normality test, we put the hypothesis that “the population from which the sample was drawn is a normal population” as the null hypothesis. Similarly, in the beauty contest problem, it is easier to interpret when we set the null hypothesis as “Stony Brook is the most beautiful University”, versus the alternative that we are not.

15

2. Please derive the likelihood ratio test for H0: μ = μ0 versus Ha: μ ≠ μ0, when the population is normal and population variance σ2 is unknown. Solution: For a 2-sided test of H0: μ = μ0 versus Ha: μ ≠ μ0, when the population is normal and population variance σ2 is unknown, we have:

2 2

0, : ,0 and

2 2, : ,0

The likelihood under the null hypothesis is:

n

i i

n

in

ix

x

LL

1

2

02

2

22

2

0

1 2

2

0

2

1exp

2

1

2exp

2

1

,

There is one free parameter, σ2, in L . Now we shall find the

value of σ2 that maximizes the log likelihood

22

02 1

1ln ln 2

2 2

n

ii

nL x

. By solving

2

02 2 4 1

ln 10

2 2

n

ii

d L nx

d

, we have

22

01

1ˆ

n

iix

n

It is easy to verify that this solution indeed maximizes the loglikelihood, and thus the likelihood function. The likelihood under the alternative hypothesis is:

n

i i

n

in

ix

x

LL

1

2

2

2

22

2

1 2

2

2

1exp

2

1

2exp

2

1

,

There are two free parameter μ and σ2 in L . Now we shall

find the value of μ and σ2 that maximizes the log likelihood

n

i ixn

L1

2

2

2

2

12ln

2ln

.

By solving the equation system:

2 1

ln 10

n

ii

Lx

and

2

2 2 4 1

ln 10

2 2

n

ii

L nx

16

we have x̂ and 22

1

1ˆ

n

iix x

n

It is easy to verify that this solution indeed maximizes the loglikelihood, and thus the likelihood function. Therefore the likelihood ratio is:

2

2

2

22 2

00 0 1

2 2

2,

2

1

2 2 22 2

0 0 01 1

2 2

1 1

exp2ˆ 2max , ,ˆ

ˆ ˆ ˆmax , ,

exp22

1

n

n

ii

n

n

ii

n nn n

i ii i

n n

i ii i

n n

xL LL

L LLn n

x x

x x x n x n x

x x x x

2 2

2

1

2 2

01

1

n

n

ii

n

x x

t

n

Therefore, the likelihood ratio test that will reject H0 when

* is equivalent to the t-test that will reject H0 when 0t c ,

where c can be determined by the significance level α as

1, / 2nc t .

17

Likelihood Ratio Test (one population mean, normal population, one-sided)

1. Let 𝑓(𝑥𝑖|𝜇) =1

√2𝜋𝜎2𝑒−(𝑥𝑖−𝜇)

2

2𝜎2 , 𝑓𝑜𝑟 𝑖 = 1,⋯ , 𝑛, where 𝜎2 is

known. Derive the likelihood ratio test for the hypothesis

𝐻0: 𝜇 = 𝜇0 𝑣𝑠. 𝐻𝑎: 𝜇 > 𝜇0.

Solution:

𝝎 = {𝜇: 𝜇 = 𝜇0} 𝛀 = {𝜇: 𝜇 ≥ 𝜇0}

Note: Now that we are not just dealing with the two-sided

hypothesis, it is important to know that the more general

definition of 𝝎 is the set of all unknown parameter values

under 𝐻0, while 𝛀 is the set of all unknown parameter values

under the union of 𝐻0 and 𝐻𝑎.

Therefore we have:

{

𝐿𝜔 = 𝑓(𝑥1, 𝑥2, ⋯ , 𝑥𝑛|𝜇 = 𝜇0) =∏𝑓(𝑥𝑖|𝜇 = 𝜇0)

𝑛

𝑖=1

=∏1

√2𝜋𝜎2𝑒−(𝑥𝑖−𝜇0)

2

2𝜎2

𝑛

𝑖=1

= (2𝜋𝜎2)−𝑛2𝑒

−12𝜎2

∑ (𝑥𝑖−𝜇0)2𝑛

𝑖=1

𝐿Ω = 𝑓(𝑥1, 𝑥2, ⋯ , 𝑥𝑛|𝜇 ≥ 𝜇0) =∏𝑓(𝑥𝑖|𝜇 ≥ 𝜇0)

𝑛

𝑖=1

=∏1

√2𝜋𝜎2𝑒−(𝑥𝑖−𝜇)

2

2𝜎2

𝑛

𝑖=1

= (2𝜋𝜎2)−𝑛2𝑒

−12𝜎2

∑ (𝑥𝑖−𝜇)2𝑛

𝑖=1

Since there is no free parameter in 𝐿𝜔,

sup(𝐿𝜔) = (2𝜋𝜎2)−

𝑛

2𝑒−

1

2𝜎2∑ (𝑥𝑖−𝜇0)

2𝑛𝑖=1 .

∵

{

𝜕𝑙𝑜𝑔𝐿Ω

𝜕𝜇=𝑛

𝜎2(�̅� − 𝜇)

𝜕2𝑙𝑜𝑔𝐿Ω𝜕𝜇2

= −𝑛

𝜎2< 0

,

∴ sup(𝐿Ω) = {(2𝜋𝜎2)−

𝑛2𝑒

−12𝜎2

∑ (𝑥𝑖−�̅�)2𝑛

𝑖=1 , 𝑖𝑓 �̅� > 𝜇0

(2𝜋𝜎2)−𝑛2𝑒

−12𝜎2

∑ (𝑥𝑖−𝜇0)2𝑛

𝑖=1 , 𝑖𝑓 �̅� ≤ 𝜇0

.

18

∴ 𝜆 =sup(𝐿𝜔)

sup(𝐿Ω)= {𝑒

−𝑛2𝜎2

(�̅�−𝜇0)2

, 𝑖𝑓 �̅� > 𝜇01, 𝑖𝑓 �̅� ≤ 𝜇0

.

∴ 𝑃(𝜆 ≤ 𝜆∗|𝐻0) = 𝛼 ⇔ 𝑃 (𝑍0 =�̅� − 𝜇0𝜎 √𝑛⁄

≥ √−2𝑙𝑛𝜆∗|𝐻0) = 𝛼.

∴ Reject 𝐻0 in favor of 𝐻𝑎 if 𝑍0 =�̅�−𝜇0

𝜎 √𝑛⁄≥ 𝑍𝛼

2. Let 𝑋1, 𝑋2, …𝑋𝑛 ~𝑁(𝜇, 𝜎2), where 𝜎2 is known

Find the LRT for 𝐻0: 𝜇 ≤ 𝜇0 𝑣𝑠 𝐻𝑎: 𝜇 > 𝜇0

Solution:

𝝎 = {𝜇: 𝜇 ≤ 𝜇0} 𝛀 = {𝜇: − ∞ < 𝜇 < ∞}

Note: Now that we are not just dealing with the two-sided

hypothesis, it is important to know that the more general

definition of 𝝎 is the set of all unknown parameter values

under 𝐻0, while 𝛀 is the set of all unknown parameter values

under the union of 𝐻0 and 𝐻𝑎.

(1) The ratio of the likelihood is shown as the following,

𝜆 = 𝐿(�̂�)

𝐿(Ω̂)

𝐿(�̂�) is the maximum likelihood under 𝐻0: 𝜇 ≤ 𝜇0

𝐿(�̂�) =

{

(

1

2𝜋𝜎2)𝑛/2

exp (−∑(𝑥𝑖 − �̅�)

2

2𝜎2) , �̅� < 𝜇0

(1

2𝜋𝜎2)𝑛/2

exp(−∑(𝑥𝑖 − 𝜇0)

2

2𝜎2) , �̅� ≥ 𝜇0

𝐿(Ω̂) is the maximum likelihood under 𝐻0 𝑢𝑛𝑖𝑜𝑛 𝐻𝑎

𝐿(Ω̂) = (1

2𝜋𝜎2)𝑛/2

exp (−∑(𝑥𝑖 − �̅�)

2

2𝜎2)

The ratio of the maximized likelihood is,

𝜆 = {

1, �̅� < 𝜇0

𝑒𝑥𝑝 (−(∑(𝑥𝑖 − 𝜇0)

2 − ∑(𝑥𝑖 − �̅�)2)

2𝜎2) , �̅� ≥ 𝜇0

⇒

𝜆 = {1, �̅� < 𝜇0

𝑒−𝑛(�̅�−𝜇0)

2

2𝜎2 , �̅� ≥ 𝜇0

19

By LRT, we reject null hypothesis if 𝜆 ≤ 𝜆∗. Generally, 𝜆∗ is

chosen to be less than 1. The rejection region is,

𝑅 = {𝑋 ∶ 𝑒−𝑛(�̅�−𝜇0)

2

2𝜎2 ≤ 𝜆∗ } = {𝑋: �̅� ≥ 𝜇0 +√−2𝜎2

𝑛ln 𝜆∗}

∴ 𝑃(𝜆 ≤ 𝜆∗|𝐻0) = 𝛼 ⇔ 𝑃 (𝑍0 =�̅� − 𝜇0𝜎 √𝑛⁄

≥ √−2𝑙𝑛𝜆∗|𝐻0) = 𝛼.

∴ Reject 𝐻0 in favor of 𝐻𝑎 if 𝑍0 =�̅�−𝜇0

𝜎 √𝑛⁄≥ 𝑍𝛼

Now we have shown that the one-sided hypothesis tests for the

following pairs of hypotheses are indeed, the same.

0 0 0 0

0 0

: :

: :a a

H H

H H