1

Topic 3: Hypothesis TestingSection 10.1 Introduction

# Within Topics 1 and 2, we employed point and interval estimation to estimate some unknown population parameter. I.e. µ and σ.

# Within Topic 3, we now want to test the ________of some statement about the __________.

Example: Did the price of ________increase by an average of only 10¢ a litre last month?

Example: Was the average amount of _______ fees charged to Bank of Nova Scotia customers in January of 2000 higher than in January of 1999?

Example: Is the proportion of _____paid as tax the same in 1989 as in 1999?

# These assumptions about _______ values of population parameters are generally referred to as STATISTICAL HYPOTHESES.

# Determining the validity of an assumption of this nature is called “hypothesis _______.”

“ The primary goal of hypothesis testing is to choose between two __________ and ________ exclusive competing hypotheses about the value of a population parameter.

2

Types of Hypotheses:

pWhen specifying the competing hypothesis with respect to some statement about a population parameter, it is convenient to distinguish between ______ hypotheses and _________ hypotheses:

1) ______ Hypotheses: only one value of the population parameter is specified.

Example: Test whether the average price of gasoline last month µ=70¢/litre.

Example: The variability of the increase in gas prices last month was σ2=100 ¢.

2) _________ Hypotheses: Specifies a range of values that the population parameter may assume.

With _________ hypotheses, more than one value is specified in each case.

Examples: The average price of gasoline last month was µ …70¢/litre. The price of gasoline was µ> 70¢/litre. The variability of gas prices σ2…100 ¢2

%______ hypotheses are generally easier to test than _________ hypotheses.

,With ______ hypotheses we only have to determine whether or not the population parameter equals the specified value.

,With _________ hypotheses, we must determine whether or not the population parameter takes on any one of a set of values.

3

˜The two mutually exclusive hypotheses in a statistical test are referred to as the ____ hypothesis and the ___________ hypothesis:

(I) ____ Hypothesis: þ It is the statement being tested

þ Denoted by .H0

þ It is called “____’ because this hypothesis corresponds to a theory about a population parameter that is thought not to be true. Hence, the title: ‘null’ or ‘_______’.

Example: Θi~(µ, σ2) : µ=0 Testing to see if the population mean is ____. H0

is pronounced “H-nought.”H0

þWe can either reject or fail to reject the ____ hypothesis.

(II) ___________ Hypothesis: þ Is the situation which prevails if the null is _____.

þ Denoted .Haþ Generally, specifies the values of the parameter that theHa

researcher believes is ____.

Example: Θi~(µ, σ2)

(a) µ>0 ___ sided alternative.

(b)µ<0

(c) µ 0 ___-sided alternative.

4

Remarks:

(i) (the ___ hypothesis) and (the alternative hypothesis), H0 Ha must be ________ _________ ( i.e. no overlap). P .( )A BI = 0

þIf is ____, then is _____.H0 Ha

(ii) The null and alternative hypotheses can both be either simple or composite:

Examples:

HH Simple null and alternative

HH Composite and composite

HH Simple null composite alternative

a

a

a

0

0

0

01

00

00

::

.

::

.

:: , .

µµ

µµ

µµ

==

⇐

≤ ⇐

= ⇐

>

<

Regardless of the form of the two hypotheses, the ____ populationparameter under consideration must be either in the set specified byH0

or in the set specified by þ __ _______!!Ha

5

’To guarantee there is no _______, one convention is to create

_____________ hypotheses: HHa

0 00

::µµ=≠

(iii) The hypothesis is always specified about the __________ parameter, and not about the ______ statistic.

H correct formH X incorrect form

0

0

2525

:: !µ = ⇐

= ⇐

One and Two-Sided _____

’If the null hypothesis is ______, ( ), then the alternative H0 0: µ =

hypothesis may specify value(s) for the _________parameter that are

entirely above ( ), entirely below, ( ) or on Ha: µ > 0 Ha: µ < 0

other sides of the value specified by the ____ hypothesis,

( ). Ha: µ ≠ 0

6

Notation:

___-sided Test: A statistical test where specifies that the Ha population parameter lies either ________ ‘above’ or ‘below’ the value specified in the ____ hypothesis.

Example: ___ sided test, since specifies that µ lies on HHa

0 77

::µµ=

> Ha

one particular side of 7.

___-Sided Test: A statistical test where the (alternative Ha hypothesis) specifies that the parameter can lie on

______ side of the value indicated by (null H0

hypothesis).

Example: ___-sided test since specifies that µ lies on HHa

0 77

::µµ=≠

Ha

______ side of _.

7

Decision Problem

p We must use ______ information to determine whether to reject or not reject the _____hypothesis.

ö Hence, we must deal with the “uncertainty” associated with using _______ to test hypotheses pertaining to __________ parameters.

ö We again are faced with using ______ information to say something about the unknown __________ parameter

ö “_______ of _____”.

” In hypothesis testing, the usual procedure to solving these decisions is to: (1) initially assume that the ____ hypothesis is_____; (2) establish a probability _______ criteria; (3) take a ______; and (4) then employ ___________ based ________criteria.

ö We can then decide whether there is sufficient evidence to _____ the ____.

p The probability value which we base our conclusion, that the null is _____, is extremely important.

p Moreover, since the decisions to accept or reject the ____ are probability based, there are “chances of error” in these decisions.

8

Two Types of Potential Error

Based on the sample result:(1) Type _ Error: ______ the null hypothesis when the null is

actually ____.

(2) Type __ Error: _____t the null hypothesis when the null is actually _____.

TrueSituation: is True is FalseH0 H0

Action:

Accept H0

Reject H0

Notes: Type I and Type II errors are ___________ probabilities.

1)Type I Error:

ö Conditional on being ____.H0

ö The probability of Type I error is denoted by “α”.ö “α” is referred to as the level of significance or____ of the

test.

“α” = P(Type I error) = P(______ | is ____).H0 H0

öThe level of ____________ of a statistical test is comparable to the probability of an error, also referred to as ‘α’, as discussed in Topic 2.

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The value of “(1-α)” is referred to as the __________ level and represents the complement of P(Type I Error).

(1-α) = Confidence level = 1-P(Type _ Error) = P(______ | is ____).H0 H0

Example: Set α = 10%; We will ______ when it is ____, in 10% of H0

the samples.

2) Type II Error:

”The probability of a Type II error is denoted by ‘β’.

β = P(Type II error) = P(______ | is _____).H0 H0

p The complement of this probability is known as the ______of a statistical test.,It indicates the ability of the test to correctly recognize that the null hypothesis is _____ and should be ________. (1-β) = (______) = P(______ | is _____).H0 H0

=1- P(______ | is _____).H0 H0

” The researcher always wants to create a test that will yield a power

close to _. I.e., ‘β’ close to ____, when is false. H0

10

,Thus, we have the decision problem now, in terms of probability:

Decision Problem:

TrueSituation: is True is FalseH0 H0

Action:

Accept H0

Reject H0

Sum 1 1

Remarks:y We want a test for which both α and β are ____. (i.e. the probability of the two errors are low).y The probability of each decision outcome is a conditional probability.

y Elements in each column sum to one, because the column events are complements.yα and β need not add to 1, since their probabilities are not complementary.

y α and β are not ___________ of each other. öWhen α is lowered, β normally _________ if the sample size

remains the same.

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yα and β are not _________ of the sample size, n. öIf n increases, ____ α and β ________, since we use more

information about the population and potentially reduce the sampling error.

ö Researchers must decide between the higher cost of sampling to ________ the sample size, and the potential sampling error and size of α and β.

y In classical hypothesis testing, we set α and try to design tests such that β is as small as possible.

12

Section 10.2 The Standard Format of Hypothesis Testing

pThe general procedure for testing hypotheses follows 5 STEPS:

STEP 1: State the ____ and ___________ hypotheses. i.e. Formulate the and .H0 Ha“ The form of the test will depend on both and .H0 Ha

Example:

HH Clearly specify the two conflicting hypothesesa

0 100100

:: .µµ=≠

, The two hypotheses must be ________ _________.

, The___ value of the population parameter must be included in one of these hypotheses.

STEP 2: Determine the Test _________ Used To Test the _____Hypothesis.

Example: In testing:

HH We could use X to do the testa

0 100100

:: .µµ=≠

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In this example, we must decide whether is true or false byH0

determining if is “_____ enough” to 100.X, If is far above or below 100, this will lead us to _____t the X

____.

, If the value of is slightly below or above 100, we will fail X to reject the ____, and we conclude that is true.H0

So: “ How do we decide what is far away and what is slightly

above/below the null hypothesis?”

¸We need to use probability — specifically the _______ distribution of the test statistic — to determine whether to reject or not reject .H0

¸To obtain a sampling distribution which is known, we _________ X to the standard normal distribution form:

, ZX

n=

− µσ

0

where ö ~N(µ,σ2/n), X ö σ is known, and ö µ0 is the value of µ under the null. (Here µ0=100)

¸ The random variable, Z, is the test _________.

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¸ The ____ statistic is a random variable employed to determine whether a specific sample result falls in one of the hypotheses being tested.

¸ The test statistic (1) must have its _._._. known under the condition that is ____;H0

(2) must contain the _________ being tested; and

(3) all of its remaining terms musts be _____ or calculable from the sample.

So, , ZX

n=

− µσ

0

(1) contains µ

(2) we know the distribution of Z ~N(0,1)

(3) in our example n and σ are assumed known and would be specified, and

(4) can be determined from the sample.X

15

STEP 3: Determine the “________ Region(s)” of the Test

¸ Before the sample is taken, it is important to specify which values of the test statistic will lead to the _______of , (called the _______ H0

region,) and which values will lead to the acceptance of .H0

¸ The Critical Region is the region of test statistic values over which we believe to be _____, leading to a _________ of the null H0

hypothesis.

¸ The Acceptance Region is the complement of the rejection region. It is the region of test statistic values over which we believe the null to be ____: acceptance of the null.

¸ The ________ value is the value of the test statistic which separates the ________ region and __________ region.

The Acceptance and Critical Regions for H0

(Two-sided Alternative)

Acceptance Region

(Accept )H0 _______ Region

(Reject )H0

_______Region

(Reject )H0

Large Negative Z

Large Positive Z 0

Z values close to zero._______ Value _______

Value

16

Question: “How do we determine the exact location of the ________ Values?”

¸Since the sampling distribution gives values of statistics and associated probabilities, we can use the sampling distribution of the ____ statistic, assuming is ____, to evaluate the probability of H0

observing different values of the test statistic under the null hypothesis.

¸Then we can derive _______values which depend on the level of ____ of Type I errors (α) and Type II errors (β).

Note:¸__________ the size of α will ________ the size of β.

¸Traditionally, the value of α is set and then we choose the critical region that yields the ________ value of β.

¸The value of α is an indicator of the degree of importance that a researcher attaches to the consequences of __________ rejecting the null .H0

¸ Usually in Social Sciences, we use a level of significance of α=0.__. I.e. we are willing to accept a _% chance of being wrong when we reject . In contrast, hard sciences, like H0

pharmacology, α is set much lower , α=0.000_ or α=0.0000_, indicating a greater concern of incorrectly rejecting .H0

17

Example: Continuation:

HHa

0 100100

::µµ=≠

, Suppose the researcher desires a level of significance of α=0.05. I.e., no more than a 5% chance of committing a Type I error.

, The researcher wants the critical region to cut off 5% of the appropriate p.d.f. , Z- distribution.

, When the null hypothesis is ___-sided, the optimal critical region will cut off ___ of the area in each tail. (Same procedure as construction of a __________ interval.)

, If α=0.05, then α/2 =0.___. The Z-values that correspond to α/2=0.025 in each tail are ±____.

Figure 3.1

0

Acceptance Region

(Accept )H0

Critical Region

(Reject )H0

Critical Region

(Reject )H0

-1.96 1.96

18

¢ The above illustration expresses the critical value (±____) in terms of the Z-distribution. i.e. We standardize , so we could determine the _______values for X a known probability distribution.

¢ We can also express the critical values in terms of __.

¢ It is simple to transform a Z ________ value into its__ critical value format by solving for :X

ZX

nOur test statistic

rearranging

X Zn

*

*

( )

:

=−

⇐

=

+

µσ

σµ

0

0

µ0 +

+Z

nσ

µ0

These 2 illustrations are equivalent means of depicting the critical regions.

−

+Z

nσ

µ 0

19

STEP 4: Take Your Sample and Calculate the Value of the ____ Statistic

Recap: 1) Specified ____hypotheses: and .H0 Ha 2) Determined the appropriate ____ _________. 3) Find the ________ region(s).

Now we must determine if the sample result lands in the acceptance orrejection regions.

¸ If we use _-values to define the ________values, we must standardize the sample results into _-values.

,This is called determining the “________-Z” value: Zc

ZX

n

Equationc =−

⇐( )

.µ

σ0 101

Notation:Z* Ô ________ Z-value

Zc Ô ________ Z value

t* Ô Critical t-value

tc Ô Computed t-value

20

STEP 5: Compare the Calculated Test Statistic With the Critical Value and Make the Statistical Decision

Decision Rule:

p If Zc value of the test statistic falls ______ the critical region, thenthe belief that is ____, is ________. H0

p If Zc value of the test statistic falls ______ the acceptanceregion, is ________.H0

Remarks:

The final decision depends on:

% the particular ______% α, the level of significance/size of the test% form of the alternative hypothesis: Ha

Example: Two-Sided TestLet Xi ~N(µ,σ2)We want to test the belief that µ=$50.

HHa

0 5050

::µµ=≠

We decide that we will test at the _% significance level α=0.05. Suppose that n=___; σ2=__; and =48.5.XQuestion: Is the value of (=48.5) “close enough” to the value of µXunder the null?

21

The sampling distribution ~ N(µ, σ2/n) =X ~ N(50, __/___=0.25) if is true.H0

P( | is true)X H0

Low 50 UpperX X

α=5%P(Reject | is true) = 0.05.H0 H0

P( < *Low or > *

upper | is true)=0.05X X X X H0

, 2P( > *upper | is true)=0.05X X H0

, 2P( > *upper |µ=50)= 0.__X X

, P( > *upper |µ=50)= 0.___X X

, P(Z > Z*)=0.___

Where Z* is the critical value for Z.

22

From the standard normal table we get:Z*upper =_.__Z*Low =-_.__

We reject if either: Zc >____H0

Zc<-____

ZX

nc =

−=

−= = −

µσ

0 485 506

122 15 3

( . )( . )

Since Zc =-3 which is less than -____, we _____t at 5% level ofH0

significance.

-3 -1.96 0 1.96 3

Notice we split the significance level into 2 parts — one relating toeach condition under which we would reject H0

23

One-Sided Tests

For these tests, the only difference from the two-sided tests is that theone-sided tests have a one-sided ___________ hypothesis, such that wehave a single ________ rejection region.

α=probability in upper tail.

α=probability in lower tail.

24

Example:

Let Xi ~N(µ,σ2 =36)Test to determine if the population mean is equal to 50 (µ=50)

HH One sided null hypothesisa

0 5050

::

.µµ=

⇐ −

<

Suppose n=___; = 48.5; Test at 5% significance α =0.05.X“Is close enough to the value of µ under the null?” X ~ N(µ, σ2/n) =X ~ N(50, 36/___=0.25) if is true.H0

50X *

Rejection region Acceptance region

is the critical value for .X * X

25

Standardize :X

ZX

nc =

−=

−= − = −

µσ

0 485 506

122 15 3

( . )( . )

Zc ~N(0,1)

× ________ valueZα = = −0 05 1645.* .

Since, , ______ the null: The population mean does not Zc < Z =0.05*α

equal 50.

Z-3 -1.645 0

26

In terms of the original units:

Critical value

X Zn

X

X

*

*

*

.

. .

= −

+

= −

+

= − + =

α

σµ

16456

1250

08225 50 491775

Calculated value

X Zn

X

X

c c

c

c

= −

+

= −

+

= − + =

σµ

306

1250

150 50 485

.

. .Since 48.5 is less than and left of 49.1775, we ______ the null.

X

48.5 49.177 50

27

What if we change the level of ____________?Test at 1% significance α =0.05.

“Is close enough to the value of µ under the null?”X

ZX

nc =

−=

−= − = −

µσ

0 485 506

122 15 3

( . )( . )

Zc ~N(0,1)

− = −=Zα 0 01 2 326.* .

We still ______ the null.

******************************************************“What if we change the level of significance again?”:

Test at 0.1% significance: α =0.001.

“Is close enough to the value of µ under the null?”X

ZX

nc =

−=

−= − = −

µσ

0 485 506

122 15 3

( . )( . )

Zc ~N(0,1)

− = −=Zα 0 001 308.* .

We cannot ______ the null at the α =0.001 level.

28

The P-Value

˜ Test results can vary depending on α, the level of significance.˜ It is not ________ for researchers to avoid specifying α before seeing the data, and instead reporting a ___________ that depends on the computed value of the test statistic, say Zc.

“Changing the previous example such that =__.X Let Xi ~N(µ,σ2 =__)HH One sided null hypothesisa

0 5050

:: .µµ=

⇐ −<

Suppose n=___; =___.; Test at 5% significance α =0.05.XThe computed value is now:

ZX

nc =

−=

−= − = −

µσ

0 49 506

122 1 2

( )( )

We usually compare Zc with the ________ values to decide whether to rejector not reject the ____.

− = −=Zα 0 01 2 326.* .

Zα = = −0 05 1645.* .

Zα = = −0 10 128.* .

Zc is outside the 5% and 10% acceptance region but inside the 1% acceptanceregion.

29

The _-value equals the ___________ that the random variable Z wouldtake on a value as _______ as Zc, given that the ___ hypothesis is ____.

( )( )

P value P Z Z

P ZP Z

c− = ≤

= ≤ −= − ≥= −=

(( )

..

21 21 0 97720 0228

“A sample mean of __ or _____ will occur only _.__% of the time when the true mean is __. Hence, the true mean may not actually be 50.”

Decision Rule:

¸ If α is ______ than _-value, ______ H0.¸ If α is _____ than _-value, ______ H0.

The above example relates to a one-sided test about µ, when Ha is one-sided on the _____ side:

( )P value P Z Zc− = ≤

When Ha is one-sided on the upper side:

( )P value P Z Zc− = ≥

30

¸ If the __________ hypothesis is ___-sided, the one-sided probability must be _______ to obtain the p-value:

( )( )

P value P Z Z

P ZP Z

c− = ≤

= ≤ −= − ≥= −= =

2

2 22 1 22 1 0 97722 0 0228 0 0456

*

* (*( ( ))( . )*( . ) .

So, if µ=50, we would observe a sample mean with the difference fromthe mean higher than ___, (50-49), either above or below µ, in 4.56% ofthe samples.

Relating back to the example above, _____ H0 at the α=0.05, but _____H0 at α =0.01.

Given that we have a 2-sided alternative:

-2.575 -2 -1.96 -1.645 0 1.645 1.96 2 2.575

At α=5% , Z*=____At α= 10%, Z*=_____At α=1% , Z*=_____

31

For any test, the p-values can be obtained (at least approximately orwithin a range) from interpolation within a probability table such as Z-table, t-table, etc. Exact p-values can be found using EViews.

Example: Your research group is hired as consultants to a major financial institution in _______ who plan to market a new financial product, targeting people living in major urban areas. In a previous census 5 years ago, the mean income of people 25-40 years old was ________ distributed with a mean of $______ per year, and a standard deviation of $_,___. Within wage gaps closing due to union and pay equity, it is believed that the ____ income has _________. Your team takes a random sample of __ incomes and finds =$43,500. X Report a p-value, assuming the standard deviation has not changed. Would you accept or reject the null hypothesis at the 5% significance level?

(1) Establish the Hypotheses:

HHa

0 42: (in thousands):µµ=> 42 (one - sided)

(2) Test Statistic:

ZX

nc =

−

( )µ

σ0

(3) Determine critical regions: Since α=0.05 and Ha is one sided on the high side, the critical value is Zα=Z0.05=_____.

32

(4)Calculate the test statistic value and report a p-value:

ZX

nc =

−=

−= =

µσ

0 435 428

75

150 924

1623( . ) .

..

Zc =1.623P(Z >1.623) =1-P(Z<1.623)=1-0.9474=0.0526P-value= ______

(5) Decision rule:

Since the p-value is _______ than 0.05, we _____ reject the ___. P-value falls ______ the acceptance region. The ____ income has ___ changed.

0 1.623 1.645

33

Section 10.3 Testing Hypotheses About µ When σ2 is _________

˜In general, there are two types of tests involving __.

,_ assumed to be known: use of the _-distribution is appropriate.

,_ assumed to be unknown: _-distribution appropriate test statistic (although Z-distribution is OK if n > 30.

˜If we need to solve problems involving when σ is _______, use X the t-distribution if the parent population is ______.

˜ So, if µ0 is the value of µ specified by the ____hypothesis, then when σ is _______ and the population is ______, the appropriate test statistic for tests on µ is:

tX

Standardize X and form t statisticn( ) __

− =−

⇐ −10µ

34

Example: In ____, the average _____-weight was ___ lbs. We believe that newborn birth-weights are _____now than they were 25 years ago. We sample of __ such weights and find that

=____lbs. with s2=4 lbs2. XUsing a significance level of _%: ,[ ]α = =P Reject H H is true0 0 0 05.

test to see if birth-weights have _________.

(1)Determine the hypotheses:

Test: HHa

0: _ _ _:µµ

=>_ _ _

(2) Test Statistic: tXsn

n( )− =−

10µ

(3) Determine Critical Region:

t t

X

critical = =

=

+ =

0 05 24

25

7 5

. ,*

*

_ _ _ _ _

_ _ _ _ . _ _ _ _ _

(4) Calculate the test statistic and p-value:

tXsn

t valuec =−

=−

= ⇐ −µ0 8 2 7 5

25

. ._ _ _ _ _ _ _ _ _ _ _ _ _ _ _

35

t

X

calculated

c

=

=

+ =

_ _ _ _

. . .17525

7 5 8 2

(5) Since t* is ____ than tc and the calculated value is _______ the acceptance region, we ______ the null to be true, and conclude that birth-weights have _________.

Or since > , we ______ the null and conclude that birth-Xc X *

weights have _________.

t24

X

Notice at the 1% level, t*=2.___, and at the 2.5% level, t*=2.___We would ______ the null that birth-weights have not changed intwenty-five years.

P-Value Using EViews

36

Tests: ( )H versus HP value P t

a0 7 5 7 5175

: . : ..

µ µ=− = ≥

>

EViews Command Code:scalar t=1.75show 1-@ctdist(t,24)

************************************

*******************Use the @ctdist(t, n-1) command to get p-value;*Note the p-value depends on the alternative hypothesis*that is specified.*******************************************************

|_*t is the calculated value.

37

|_distrib t/type=t df=24 T DISTRIBUTION DF= 24.000 VARIANCE= 1.0909 H= 1.0000 DATA PDF CDF 1-CDF T ROW 1 1.7500 0.87989E-01 0.95355 0.0464481

( )P value P t CDF− = ≥ = − =175 1 0 04644. .

If we had a two sided ___________ hypothesis:

HHa

0 7 57 5

: .: .µµ

=≠

( )

( )

P value P tP t P tP t CDF

− == ≥ + ≤ −= ≥ = −= =

> 175175 175

2 175 2 12 0 04644 0 09288

.( . ) ( . )( . ) *( )*( . ) .

p Cannot ________ the null at the _% level, but ______ at the __% significance level.

38

Section 10.4Measuring β and the _______of A Test

p Up to this point, we have not calculated the value of β: P(Type __ error),

primarily because the alternative hypothesis has been _________ in format, and hence there is no one value specified for µ that makes H0 false.Example: H0: µ=100 Ha: µ…100

p A Type I error, incorrectly _________ H0, is α, since it will occur when µ=100.

p A Type II error, incorrectly _________ H0, can occur for any value of µ not equal to 100.

¸The probability of incorrectly accepting H0:µ=100, is much higher when the true value of µ=99.9 than when the true value is µ=125.

, And, the value of β is _________ for these 2 situations.

p So, one can calculate the values of β for other values of µ which may be possible under the alternative hypothesis.

p These different probabilities of β that occur when Ha is _________ can be summarized in either _____ format, by graphical illustration, or described by a functional relationship.

p Often a _____ curve illustrates the value of (1-β) — the _____ function — which depicts the ability or power of the test to correctly reject a _____ null hypothesis.

39

Example: Illustration of β Error Calculation:

Suppose Xi~N(µ=__,σ2= __); n=144 ; α=0.05 Zα/2=____.

Step 1: Determine your __________:

H0: µ = __Ha: µ …__

Step 2: Test _________:

ZX

n=

− µσ

0

A Type I error occurs when we believe µ… __, when the null is actually____: ,H0: µ=___.

A Type II error occurs when we believe µ=__, when the null is actually_____: , Ha: µ…__.

Step 3: ________ ______:

µσ

α0 2 1966

120 98± = ± = ±Z

n*

/ _ _ ( . ) _ _ .

, Acceptance region is from __.__ to __.__

40

How Do We Calculate β?

¸ Since β is a conditional probability that depends on the ____ value of µ, we will assume that µ= __._.

β=P( Accept H0:µ0=__*µ=__._)

We know that H0 will be accepted whenever X lies between __.__ and__.__. So,

β=P(49.02 # # 50.98 *µ=__._).X¸ Using the same Zα/2 and test statistic, we transform this problem into the standardized normal format, but allow µ = __._, σ= _, and n= 144:

[ ]

P X PX

n

P Z

P ZF F

( . _ _ ._ _ ). _ _ ._ _ _ ._ _ _ _ ._

.

...

. .( ._ _ ) ( _ ._ _ )

. . ._ _

49 0249 02

612

612

14805

0 4805

2 96 0 960

08315 0015 0

≤ ≤ =−

≤−

≤−

=−

≤ ≤

= − ≤ ≤= − −= − =

µσ

' P(Type II error) =0.__Interpretation: When µ=__._, we will _________ accept H0:µ0=__ as being true 83% of the time using our test procedure. + The power of this test is (1- β)= 1-0.__= 0.17.

41

This test will ________ recognize this _____ null hypothesis 17% of thetime when µ= __._.

Remark: This is only one possible result using µ= ____.

What about when µ=__._?

[ ]

P X PX

n

P Z

P ZF F

( . . ). _ _ ._ . _ _ _

.

...

. .(_ _ _ _ ) ( . )

. ( . ) ._ _

49 02 50 9849 02

612

50 986

12

0 4805

14805

0 96 2 960 96

0 9985 1 08315 0

≤ ≤ =−

≤−

≤−

=−

≤ ≤

= − ≤ ≤= − −= − − =

µσ

Symmetry! ˜The power of this test is (1- β)= 1-0.__= 0.__. This test will correctly recognize this ____e null hypothesis __% of the time when µ= 49.5.What if µ= __._

42

[ ]

P X PX

n

P Z

P ZF F

( . . ). _ _ ._ . _ _ ._

.

...

. .( . ) ( _ _ _ _ )

._ _ _ _ . ._ _ _ _

49 02 50 9849 02

612

50 986

12

05805

13805

116 2 762 76

0 0123 0

≤ ≤ =−

≤−

≤−

=−

≤ ≤

= − ≤ ≤= − −= − =

µσ

P(Type II error) = 0.____

˜When µ= __._, we will ___________ accept H0:µ= __ as being true __.__% of the time.,The power 1-β =0.____.¸We will correctly recognize this false null hypothesis 12.5% of the time when µ=____.

What about when µ=_____?

[ ]

P X PX

nP ZF F

( . . ). _ _ _ _ _ . _ _ _ _ _

. .( . ) ( . )

. ( . )

._ _ _ _ . ._ _ _ _

49 02 50 9849 02

612

50 986

12

0 46 346346 0 46

0 9997 1 0 67720 0 3228 0

≤ ≤ =−

≤−

≤−

= − ≤ ≤= − −= − −= − =

µσ

P(Type II error)= 0.6769

43

What about when µ=49.1?

[ ]

P X PX

nP ZF F

( . . ). . . .

. .( . ) ( . )( . ). .

49 02 50 9849 02 4910

612

50 98 49106

12

016 376376 016

1 1 056361 0 4264 05635

≤ ≤ =−

≤−

≤−

= − ≤ ≤= − −= − −= − =

µσ

P(Type II error)= 0.5635

What about when µ=48.9?

[ ]

P X PX

nP ZF F

( . . ). . . .

. .( . ) ( . )

. ).

49 02 50 9849 02 48 9

612

50 98 48 96

12

0 24 416416 0 24

1 059480 4052

≤ ≤ =−

≤−

≤−

= − ≤ ≤= −= −=

µσ

P(Type II error)= 0.4052

44

What about when µ=48?

[ ]

P X PX

nP ZF F

( . . ). .

. .( . ) ( . )

. .

49 02 50 9849 02 48

612

50 98 486

12

2 04 596596 2 04

1 0 9793 0 0207

≤ ≤ =−

≤−

≤−

= ≤ ≤= − −= − =

µσ

P(Type II error)= 0.0207 Table of Results:

µ β Power=(1-β)

0.9793

0.5948

0.4365

0.3231

0.1700

0.1259

0.0500

0.1259

0.1700

0.3231

0.4365

0.5948

0.9793

45

¸The most powerful tests are ones with the _______ ascending power. ¸These are tests that quickly recognize a ____ null hypothesis for ____ differences between µ0, the hypothesized value of the parameter and µ, the true value.

The Trade-Off Between α and β

˜ When the sample size, n, is _____, α and β have an _______ relationship. I.e. Cannot decrease α without __________ β.

Illustration: repeat the example from above, but this time α= 10% instead of α = 5%.

Since α has increased, the _________ region will be _______ and β willbe _____ than before. Hence the _____ of the test will increase.

Z0.10 = ± 1.___ Ô Critical values Hypotheses: H0: µ0=__ Ha: µ0…__

Critical and Acceptance Regions:

X Zn

* */ _ _ ( . ) _ _ .= ± = ± = ±µ

σα0 2 1645

612

08225

Critical Values are __.__ and __.__, so the acceptance region is from49.18 to __.__.

So,

46

β= P(Accept H0:µ0=__| H0 is _____) = P(Accept H0:µ0=__| µ=__._)

Since H0 is accepted whenever lies between 49.18 and 50.82:X

β = P(49.18 # # 50.83 | µ=__._)X

[ ]

P X PX

nP ZF F

( . . ). _ _ ._ . _ _ ._

. .( . ( . )

._ _ _ _ ( ._ _ _ _ )

._ _ _ _

4918 50824918

612

50826

12

2 64 0 640 64 2 64

0 1 00

≤ ≤ =−

≤−

≤−

= ≤ ≤= − −= − −=

µσ

β=0.____ (which is less than 0.____).

This means that when µ=__._, we will ___________ accept H0:µ0=__ asbeing true about 73% of the time. The power of this test is (1-β)=__%.

¸This test will correctly reject the false null 27% of the time when µ is actually __._.

Note: Trade-off is not one-to-one.

( Increasing α resulted in a _____ β and a higher power.

¸Recall, in classical testing we fix the P(Type I Error) and attempt to

47

construct a test that minimizes P(Type II error) to maximize _____:

Definition: The “Power Curve” associated with a test plots power against the value of the _________ under the test.

A) H0:µ=µ0 versus Ha:µ>µ0

Power=(1- β)

1P reject H( ?)0 µ =

µ0 µ* H0 true H0 is false

P reject H P reject H H is true( ) ( )0 0 0 0µ µ α= = =

48

B) H0:µ=µ0 versus Ha:µ<µ0

Power=(1- β) 1

P reject H( ?)0 µ =

µ* µ0 H0 is false H0 trueP(Reject H0|µ=µ*)=Power at µ*=µ.

C) H0:µ=µ0 versus Ha: µ…µ0 Power

1.0

(1-α)

0.05

}P(reject H0 when H0 is true)

P reject H P reject H H is true( ) ( )0 0 0 0µ µ α= = =

49

Definition: An ________test is one such that Power $ Size everywhere: (1-β) $ _.Definition: A __________ test is one whose power approaches 1 as the sample size goes to ________.

Power as n⇒ ⇒ ∞1 .

Meaning, we will always reject the ____when the null is _____, assample size is large.

¸We are using unbiased and __________ test here.

Decreasing α and β by Increasing the Sample Size (n)

¸The former examples fixed the sample size in advance.

¸If _ changes, the size of α and β may be changed, since the size of n affects the __________ of the underlying probability distribution and the location of the acceptance and critical regions.

¸When n is _________, the test usually becomes more sensitive in distinguishing H0 and Ha since the acceptance region is _________.

, α and β ________.

50

Example: H0: µ=__ Ha: µ…__

n=625σ=6α= 0.05Zα/2=±____

Critical and Acceptance Regions:

X Zn

* */ _ _ (_ _ _ _ ) _ _ .= ± = ± = ±µ

σα0 2

625

0 4704

Acceptance region is between 49.53 to 50.47.

What if µ=__._?

[ ]

P X PX

n

P Z

P ZF F

( . . ). _ _ ._ . _ _ ._

..

..

. .( ._ _ _ _ ) ( )

.

49 53 50 4749 53

625

50 476

25

0 970 24

0 030 24

4 04 0131 0 1 1

0 4483

≤ ≤ =−

≤−

≤−

=−

≤ ≤−

= − ≤ ≤ −= − − −=

µσ

β=0.4483(1-β)=0.5517 Ô_____ has increased.

51

Section 10.7 One Sample Test On σ2

Test On A Population Variance σ2

¸Recall ~ , is a χ2 random variable with:( )n s− 1 2

2σχ ( )n−1

2

n = sample sizeσ2 = population variances2 = sample varianceParent population is _______ and (n-1)=ν= degrees of freedom.

”We can use this random variable to test hypotheses about the _______ population variance σ2 based on the sample estimator s2.

”The appropriate ___statistic for hypotheses about σ2 for (n-1) degrees of freedom is:

Equation 10.5χ ( )

( )__n

n− =

−1

22

02

1

where is the ___________value of σ2 when H0 is ___ and s2 is theσ 02

sample variance:

.sn

X Xii

n2 2

1

11

=−

−=∑( )

52

Example: Suppose Xi ~ N(__, σ2)α =0.__n=__s2=__

Step 1: H0: σ2=__ Ha: σ2<__ Ô(One-sided alternative in the lower direction)

”We will _____ H0 for small values of the test statistic, and all ‘___’ is in the ____-hand tail.

Step 2: Test statistic: χσ( )

( )n

n s− =

−1

22

02

1

Step 3: ________ Values: : P( < )= 0.05.χ 2* χ242 χ 2*

Using the Chi-square table: = __._.χ 2*

Step 4: Calculate the test statistic: for this sample:χ c2

χ c2 600

2142=

= =

(_ _ )(_ _ )_ _ _ _

. .

53

Step 5: Decision Rule:

(i) Reject H0 if is less than __._: < ______ the null.χ c2 χ c

2 χ242*

(ii) ______ H0 if is greater than __._: > do not reject χ c2 χ c

2 χ242*

the ____.

oSince > , we ______ H0. χ c2 χ24

2*

We cannot reject H0 that σ2=__ at the _% significance level.

P( |H0 true)χ242

χ242

12 13.8 21.42

54

Example: Two Sided Test

Suppose Xi ~N(__, σ2)α =0.__n=__s2=__

Step 1: H0: σ2=__ Ha: σ2 …__

” So ______ H0 if the test statistic is “too _____” or “too _____.”

Step 2: Test statistic: χ σ( )

( )n

n s− =

−1

22

02

1

Step 3: Critical Values: Must split _% rejection region into 2 tails: __% in each ____.

( )( )

χ

χ

χ χ χ χ

χ χ χ χ

Upper

Lower

n Upper n Lower

Upper Lower

critical upper tail value

critical lower tail valueThen

P or H is true

P or H is true

2

2

12 2

12 2

0

242 2

242 2

0

0

0

=

=

=

=

− −

.

.:

._ _

._ _

> <

> <

From the Table VII: χχLower

Upper

2

2

=

=

_ _ _ __ _ _ _

55

Step 4: Compute the test statistic:

χ c2 24

2142=

= =

( )(_ _ )_ _

_ _ __ _

. .

Step 5: Decision rule:

(i) _____ H0 if calculated value of test statistic, , for this sample is χ c2

less than ____ or greater than ____.

(ii) ______ H0 if calculated value of the test statistic, , is between χ c2

____ and ____.

ª Since =_____, which falls into the __________ region, we χ c2

cannot ______ the hypothesis that σ2 =__ against the 2-sided alternative at the __% level of significance.

12.4 39.4 χν2

χ lower2 χ c

2 χupper2

56

Chapter 11: Hypothesis Testing: Multi-Sample TestsSection 11.1-3Introduction

pIn Chapter 10 we tested hypotheses about the population parameters from a _______ sample.pThis chapter will examine hypothesis testing when two or more _______ are drawn and we want to determine if these samples originate from the ____ population.

Example: Are their population _____ or variances the same? If they are, then the same population generated the various samples.

Typical Examples:

1) Testing the different treatment of two groups for ____ development: p _______ Group p_________ Group

2) Tests on income ________: Is the mean _____of university graduates in Victoria and _______ the same?

3) Is the mean salary of ______ employees the same as ____ employees in the __________ industry?

4) Is the male unemployment rate in _________ the same as the female unemployment rate?

pWe will explore two primary cases:

57

I) Testing equality of two means when are both _____.( )σ σ12

22and

II) Testing ________ of two means when are both ( )σ σ12

22and

_______.

Case I: Testing Equality of Two _____

(Population variances are _____)( )σ σ12

22and

Let µ1 = mean of population 1.Let µ2 = mean of population 2.

If we restrict the null hypothesis to be _______in format, the commonform of H0 is:

H0:µ1 - µ2 =___ or

H0:µ1 = µ2

These two means are _____.

The alternative hypothesis could then be either:

Ha: µ1 - µ2 >0 Ôµ1 _______ µ2or

Ha: µ1 - µ2< 0 Ôµ2 _______ µ1or

Ha: µ1 … µ2 Ôµ1 not equal to µ2

pSuppose a simple random sample of size n1 is taken from population

58

1 and . Also, suppose a simple random sample of size X i1 12~ N( 1µ σ, )

n2 is drawn from population 2 and .X i2 22~ N( 2µ σ, )

pDetermine the Test Statistic:

Recall, when we standardize the Z-variable is:

(11.1)Z =−Point _ _ _ _ _ _ _ _ Hypothesized _ _ _ _

Standard _ _ _ _ _ of point estimator

The “best” ________ point estimator of (µ1 - µ2) is (_______), where:

= ____ of sample 1( )X1

= ____ of sample 2( )X2

The standard error of (______), denoted is:( )σ X X1 2−

V X Xn nX X( )1 2

12

1

22

21 2

− = = +−σσ σ

Note: Assuming independence:V V V

V V

n n

V Xn

(_ _ ) (_ ) ( _ )(_ ) (_ _ ) (_ )

( ) .

1 2 1 2

12

2

12

1

22

22

− = + −

= +

= +

=

σ σ

σsince

pIf and are normal, is ______. (It is a linear X1 X2 ( )X X1 2−

59

function of normal random variables.)

Hence,(________)~ N n n( ), .µ µ

σ σ1 2

12

1

22

2− +

Thus, we can derive the test _______ for finding the 2-sample computedZ-value for H0:µ1 - µ2 =__:

~ N(0, 1).Z =(_ _ ) ( )1 2 1 2

12

1

22

2

− − −

+

µ µσ σn n

pThe ________ values are determined the same way as in the former sections:

If α=0.__, then α/2=0.___Z*=±____ will be the critical values.

pCompute the test statistic:

Z

X X

n n

C =− − −

+

( ) (_ _ ).1 2 1 2

12

1

22

2

σ σ

pDecision Rule :Reject H0 if: the calculated values of Z are _______ than the ‘+’

60

critical value or ____ than the ‘-’ _______ value.Reject H0: Z* < __ or -Z* > __.

Cannot reject otherwise.

Example: Determine if the length of _____to find employment in their field of study of university _________ in Ottawa (µ1) and ________ (µ2) is the same:

pTest Hypotheses: H0: µ1 - µ2 = __ }| __ = __

Ha: µ1 - µ2 … __ }| __ … __

α=0.__=_% | Z*=±_____

Let: n X weeks weeksn X weeks weeks

1 1 12 2

2 2 22 2

122 3

= = =

= = =

_ _; _ _ _ ; . ._ _ ; _ _ _ ; . .

σσ

pCompute Z Value:

ZC =− −

+

=−+

=−

=−

= −

(_ ._ _ ._ ) ( ). .

.. .

..

..

_ ._ _

01250

2 375

190 024 0 03067

190 0546

190 2338

Since -Z* >Zc, we ______ H0.

pInterpretation:

61

We believe the _______ amount of _____these graduates search for jobsin their field of study in each city is statistically _________.

pUsing the P-value:

p-value =2P(Z#-____) = 2(1-P(Z # ____)) =(2)(1-_) =2(0.____) =_____

Since ____ is ____ than 0.__, we ______ the null hypothesis.

62

Case II:Difference Between Two Means

(Variances Are _______ But Assumed _____)( )σ σ12

22and

p The two-sample Z test is appropriate only when both

are _____.( )σ σ12

22and

p When the _________ are unknown, the _-distribution can be used to test the two means if both parent populations are ______.

p If the parent populations are not normal, but n1 and n2 are _____ (ni $15), the _-distribution can be used.

p The other difference from the last section is the assumption that

are the ____ population variance; Both ( )σ σ12

22and

must represent two estimates of the ____ population s s12

22and

variance.

Example: Suppose X ~1i N n sample size( , ); .µ σ1 12

1 =

X N n1 112

1~ ( , )µ σ

X ~2i N n sample size( , ); .µ σ2 22

2 =

X N n2 222

2

~ ( , )µ σ

63

We want to test:H0: µ1 - µ2 = __ }| __ = __

Ha: µ1 - µ2 … __ }| __ … __

Test Statistic:

Use the t-distribution since X1 and X2 are normal and variances are_______ but _____:

Form:

( )

( )

sn

X X

sn

X X

ii

n

ii

n

12

11 1

1

2

22

22 2

1

2

11

11

1

2

=−

−

=−

−

=

=

∑

∑

r If are not _____, which estimate of s s12

22and

the population ________ of σ2 do we use?

Use a ________ average of the two estimators.

,This will produce a more ________ and reliable estimate of the population variance than using one or the other.

pUse ‘________’ degrees of freedom as the ______ relative to the total number of degrees of freedom.

64

Let: s2= weighted average.

s s s

sn s n s

2 1

1 212 2

1 222

2 1 12

2 22

1 2

12

12

1 12

=−

+ −∗ +

−+ −

=− + −

+ −

(_ )(_ _ )

(_ )(_ _ )

*

( ) ( )(_ _ )

p Now: ~ NormalX X1 2−

E X XandV X X V V by

n n

n n

( )

( ) (_ _ ) (_ _ ) _ _ _ _ _ _ _ _ _ _ _ _

1 2 1 2

1 2 1 2

12

1

22

2

2

1 2

1 1

− = −

− = + ⇐

= +

= +

µ µ

σ σ

σ

since σ2 denotes the two equal variances: .σ σ12

22 2= =_

Hence, the test statistic is: .( ) ( )

t

n n

=− − −

+

_ _

_

1 2 1 2

2

1 2

1 1

µ µ

p Using s2 as the estimator of σ2, we get the test statistic:

65

( ) ( )( )t

n n

=− − −

+

+ −

_ _

_

1 2 1 2

2

1 2

21 1

µ µ~ t

n n1 2

which is the two sample t-test statistic.

Example: Test the difference in average ________ paid to _____ in _______ and Vancouver, assuming salaries are normally distributed. Test at the α=0.__ level of significance.

_ _ _ _ _ _ _nXs

1

1

12

10000000

==

=

$_ _ ,$_ _ _ ,

VancouvernXs

2

2

22

9500

000

==

=

$39,$_ _ _ ,

Test:

H0: µ1 - µ2 = __ }| __ = __ Average ________ are the same.

Ha: µ1 - µ2 …__ }| __ … __ Average ______ are not the same.

,The degrees of freedom = __ + 9 - 2= __.

66

,t__ ~ under the null hypothesis. ,The ________ value from a two-sided alternative is: t*=±____

Computed Value:

( ) ( )

( ) ( )

tX X

n n

s

C =− − −

+

− −

+

1 2 1 2

2

1 2

2

1 1

000 39 500 0

110

19

µ µ

_

_ _ , ,=

Must determine s2 :

sn s n s2 1 1

22 2

2

1 2

1 12

1 000 9 1 0009 2

000 8 00017

1 350 000 1 240 000 000

=− + −

+ −

=− + −

+ −

=+

=+

= =

( ) ( )(_ _ )

(_ _ )(_ _ _ , ) ( )(_ _ _ , )(_ _ )

(_ )(_ _ _ , ) ( )(_ _ _ , )( )

, , , ,_ _

_ ,_ _ _ ,_ _

_ _ _ ,_ _ _ ._ _

So, calculating tc:

67

( ) ( )

( ) ( )

( )

tX X

sn n

s

C =− − −

+

− −

+

=−

+

= =

1 2 1 2

2

1 2

2

1 1

000 39 500 0

110

19

000 39 500

352 941

1019

1500179 29

µ µ

=_ _ , ,

_ _ , ,

_ _ _ , .

._ ._ _ _

t*= ± _.__ and tc= _.___

,Reject H0 if tc is _______ than _.__ or less than -_.___

p We clearly ______ the null. p The average salaries for women are statistically _______in each city.p

68

Section 11.6 Two Sample Tests for Population _________

˜Suppose we have random variables from two ___________:

X NX N

i

i

1 1 12

2 2 22

~~

( , )( , )µ σµ σ

%If the two populations have _____ variances , then ( )σ σ12

22=

the populations are _____________ ( “homo” = same, “scedastic” = variability) as opposed to _______________ (different variability).

Why would we be interested in whether populations are____scedastic or ______scedastic?

%To answer questions like:

(1) Is the variability in ____ worked different between the United Statesand ______?(2) Is the variability of the ____________ rate in Canada over the last year the _____as in the United States?(3) Is the variability of ___ prices in Victoria and ______ the same?

%To test if the ___________ is the same across populations, the null hypothesis is usually in the form:

Ô the variability is the ____ across these two H0 12

22:_ _=

populations.

69

%The alternative hypothesis can be any of the following:

Ô the variability is ___ the ____ across these two Ha:σ σ12

22≠

populations.

Ô the variability in population 2 is _______ than the Ha:σ σ12

22<

variability in population 1.

Ô the __________in population 1 is greater than the Ha:σ σ12

22>

variability in population 2.

%The form of the alternative hypothesis will affect the _______ values and p-values, but ___ your approach to testing hypotheses.

Illustration: ___-sided Alternative Hypothesis

Suppose or H0 12

22:σ σ= H0 1

222: / _ _σ σ =

or Ha:σ σ12

22≠ Ha : / _ _σ σ1

222 ≠

%Since the population variance is estimated using the sample variance, we can base the test of H0 on the _____ of ______ variances

:( )S and S12

22

70

Recall

Sn

X X

Sn

X X

ii

n

ii

n

12

11 1

2

1

22

22 2

2

1

11

11

1

2

=−

−

=−

−

=

=

∑

∑

( )

( )

The sampling distribution for testing hypotheses about two ________ iscalled the _-distribution.

Hence the test statistic is: F = __ .1

2

22

The _-distribution tests whether , by taking the _____ ofσ σ12

22=

the two sample variances, .( )S and S12

22

%If H0 is ____, we would expect ‘_’ to be close to _.

%Thus, the more _ differs from _, the more likely we will _____ H0 for the alternative Ha.

˜To test H0 using , we need to know the distribution of the _

_12

22

test statistic under the null.

It can be shown that if H0 is true:

SS

where n rees of freedom for samplen rees of freedom for sample

12

22

1 1

2 2

1 2

1 11 2

~_ν ν

νν

,

: deg .deg .

= − == − =

71

Things To Note:

1) An _ random variable has ___ different degrees of freedom associated with it.

___

( )

( )= =

−−

−−

=

=

∑

∑12

22

11 1

2

1

22 2

2

1

11

11

1

2

nX X

nX X

ii

n

ii

n

_-values contains a random variable with (n1-1)=ν1 d.f. in the numeratorand (n2-1)= ν2 in the denominator.

2) Since the test statistic is a _____ of two numbers which are $ 0, an _ random variable is always $__.

3) _- distribution is __________ skewed.

Looks like a ___-square distribution because the _-distribution is theratio of 2 independent _2 distributions.

4) Use _-tables to find critical values associated with a given ν1 and ν2.

72

Example: Let ν1 =_ and ν2=___. We want the 5% critical _ value: Want F* such that P(F5,10$F*)=0.05.(Notice that the tables are only for _% and _% upper tail regions.)

f (F5,10)

5%

F*From Table VIII (a) in H&M, page 896, we have

F*=____ þ P( F5,10$____)=0.05.

Example: Let ν1 = _ and ν2= _. We want the 1% critical F value: Want F* such that P(F3,8$F*)=0.01. From Table VIII (b) in H&M, page 898, we have

F*= þ P( F3,8$____)=0.01.

So how do we determine the _____ tail critical values,since this is not a symmetrical distribution?

73

Low Side F-Test

A lower-tail critical value can be found by the following:

FFLower critical value

critical value, ,

_____ , ,ν ν

ν ν1 2

2 1

1=

“A _____-tail critical value of F can always be found byreversing the _______ of freedom of the numerator and thedenominator, determining the corresponding value in the_____ tail of the F-distribution, and then taking thereciprocal of this number.”

Suppose we want F* such that P(F3,12 #F*)= 0.05:

f (F3,12)

F3,12 0.___ 8.74Since where F*Upper, 12,3 is the F-value such thatF =

1F

(Lower 3,12)*

(Upper 12, 3)*

P(F 12,3 $ F* upper, 12,3)=0.05. Value þ _.__

74

Hence:

F =1

F=

1_ ._ _

= 0._ _ _

P(F

(Lower 3,12)*

(Upper 12, 3)*

3,12 ≤ =0 0 05._ _ _ ) .

0.___ 8.74

4) It does ___ matter if your ____ statistic is :

SS

SS

SS F

22

12

12

22

22

12

2 1

instead of

Under the null is distributed

.

.,ν ν

Simply put the d.f. in the same order as the _____ of the variances.

75

Returning to the two-sided alternative test: versus .H0 1

222:σ σ= Ha:σ σ1

222≠

Let α= 0.02. P(Reject H0|H0 is true) =0.02.

For a 2-sided alternative put 0.01 (α/2=0.01) in each tail and find thecritical values of F with ν1 and ν2 degrees of freedom.

f (Fv1,v2)

1% 1%

Fv1,v2 F*low F*up

¸Finally calculate the value of the test statistic: , and see if it S

S12

22

falls in the critical region or the acceptance region.

Decision Rule

(I) ______ H0 if: F or FC C< F > Flower*

upper* .

(II) _____t H0 if: F Flower*

upper*≤ ≤FC .

76

Example: Suppose you wish to test the variability in ______ times between Vancouver and _______ and between _______ and Vancouver. A sample of __ flights is taken for the route originating in Vancouver (n1 =__) and a sample of __ flights is taken for the route originating in _______ is taken (n2=__). We find that and S1

2 225= minutes

Assume that the flight times are S22 216= minutes .

normally distributed. Let α=0.10.

Test versus .H0 12

22:σ σ= Ha:σ σ1

222≠

Test Statistic is:

_ ,

: _ _ ._ _ .

,=

= − == − =

SS under the null

where n degrees of freedom for sample 1n degrees of freedom for sample

12

22

1 1

2 2

1 2

11 2

~ Fν ν

νν

P(Reject H0 | H0 is true )=0.10.

So, [ ] [ ]P F F P F Flower upper≤ = ≥ =, ,*

, ,* .15 20 15 20 0 05

From the cumulative F-distribution table (α=0.05) :

F FFupper lower

upper, ,

*, ,_

*

, ,*_ _ _ _

_ _ _ _._ _ _ _ .15 20 15

20 15

1 10= = = =and

Calculating FC:

77

FSSC = = =1

2

22

2516

15625. .

Since Fc is 1.5625, which lies _______ 0.___ and _.__, we cannot _____the null and conclude the two samples are from a population with the____ variability.

We could also use:

F SS under the null

where n degrees of freedom for sample 1n degrees of freedom for sample

=

= − == − =

22

12

1 2

2 1

2 1

1 201 15 2

~ Fν ν

νν

, ,

: ..

So, [ ] [ ]P F F P F Flower upper≤ = ≥ =, ,*

, ,* .20 15 20 15 0 05

From the cumulative F-distribution table (α=0.05) :

78

F and FFupper lower

upper, ,

*, ,

*

, ,*_ _ _ _

_ _ _ _._ _ _ .20 15 20 15

15 20

1 10= = = =

Calculating Fc:

FSSC = = =2

2

12

1625

0 64. .

Since Fc is 0.64, which lies _______0.___ and 2.__, we cannot ____ thenull and conclude the two samples are from a population with the ____variability.