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Chapter 8

Tests of Hypotheses

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Relationship: Confidence Interval

and Test of Hypothesis

Two general methods for making inferenceabout population parameters

confidence interval (estimating their values)

hypothesis testing (making decisions about

them)

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Elements of a Statistical Test

1) Null Hypothesisto be rejected or not rejected

2) Alternative Hypothesis- not equal, greater

than, less than. Used mainly to define rejection

region as one-sided or two-sided

3) Test Statisticcalculated from sample data

4) Rejection Regionvalues of test statistic that

will imply rejection5) Conclusiondecision made to whether accept

or reject the null hypothesis

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Conclusions and Consequences

DecisionTruth

Ho is true Ho is false

Retain Ho (fail toreject) Correct decision Type II error ()

Reject Ho Type I error () Correct decision

You are trying to detect a departure from Ho.

You cant prove Ho is true.

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Hypotheses

H0: equal

Ha: less than, not

equal, or greater than

One-tailed or two-tailed

is a decision basedon assumed

alternative

Example:

H0: m = m0

Ha: m m0or

Ha: m m0

orHa: m m0

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Reject Region

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Note

Failure to Reject is NOT Equivalent to Accept(different than what book says)

Just as a court decision of NOT GUILTY does notmean INNOCENT

The powerof a statistical test, (1-), is the

probability of rejecting the null hypothesis Howhen in fact, Hois false

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Example 8.5 (p.344)1. Null Hypothesis H0 = 72

2. Alternative Hypothesis. Concerned that traffic volume

might be greater so Ha: > 72

3. Test Statistic

calculated from sample data

4. Rejection Region Z0> 1.28 (from Normal table)

Decision: reject or not reject H0?

12.1

50

3.13

721.7472

0 =

=

=

ns

yZ

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Note

Choose H0 and Ha before obtaining data

to avoid bias

Statistical Significance does NOT equal to

PRACTICALSignificance

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Testing Population Mean

Large Sample (n 30)

1. Define null hypothesis

2. Define alternativethis specifies one-tailed or two-tailed & location of rejection

region

3. Calculate Z values for rejection region4. Calculate average of data set

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Testing Population Mean -

Large Sample continued

5. Calculate standard deviat ion of data

6. Convert observed average into Z-score

7. Compare to Z rejection region. Make

decision

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Example 8.8 page 348

1. Define null hypothesis H0: m= 8.5

2. Define alternativethis specifies one-tailed or

two-tailed & location of rejection region

Ha: m8.5

3. Look up Z values for rejection region.For = .01, find z = 2.58

4. Calculate average o f data. 2579.y=

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Example 8.8 continued

5. Calculate standard deviat ion of data.

s = 1.203

6. Convert observed data into Z-score

7. z

8. Compare to Z rejection region

Decision is to reject or not reject H0

?

03.4

41

203.1

5.8256.90 =

=

ns

y m

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Decision

Figure 8.6

Rejection region for

Example 8.8

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The Observed Significance Level

for a Test

The observed sign i f icance level orp-value,

for a specific statistical test is the probability

(assuming Hois true) of observing a value of thetest statistic that is at least as contradictory to

the null hypothesis, and supportive of the

alternative hypothesis, as the one computed

from the sample data

Not to be confused withpfrom Binomial

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Large Sample Tests

Let z0= computed value of test statistic.

Identify probability that corresponds to zc

Determine: p-value =

P (z > z0) if upper-tailed

P (z < z0) if lower-tailed

2 P (z > |z0|) if two-tailed

Ifp-value exceeds , there is insufficientevidence to reject Ho

Ifp-value is less than the maximum value ,

reject Ho16

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Example 8.10

Determine p-value = P (z > z0) for upper-

tailed test.

P (z > 1.12) = .5 - .3686 = 0.1314

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Review - Example 8.9 page 351

1. Define Hypothesis H0: m= 1

2. Define alternativethis specifies one-tailed ortwo-tailed & location of rejection region.

Ha:m> 1

3.Look up t values for rejection region and

degrees of freedom. For = .05 and samplesize of 20, find t.05, 19= 1.729

4. Calculate average o f data. ybar = 2.14318

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Example 8.9 continued

5. Calculate standard deviat ion of data.

s = 1.736

6. Convert observed average into t-score

= (2.143 - 1)/(1.7/20) = 2.957. Compare to t rejection region. Make decision.

Decision is to reject or not reject H0?

ns

yT 01

m=

19

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Observed Significance Level

Test statistic t = 2.95

Find corresponding p-value

What portion of the curve is beyond t19= 2.95?From Table 7, its between 2.861 and 3.579 for

.005 and .001 respectively.

May approximate to the higher value of .005 Sincep-value (.005) is less than value (.05),

reject Ho

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Small Sample Test

Let t0= computed value of test statistic.

Identify probability that corresponds to t0

t tables are incomplete.Determine p-value =

P (t > t0) if upper-tailed

P (t < t0) if lower-tailed

2 P (t > |t0|) if two-tailed

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Testing the Difference Between Two

Population Means: Independent Samples Hypothesis

H0: 1- 2= 0

Ha: 1> 2 or 1- 2> 0 Could have 1- 2= D0 non-zero value Key challenge as in estimation

Determine standard deviation of difference We assume Independent Samples

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Large Sample Test

One-Tailed Test

H0: 1- 2= D0

Ha: 1- 2> D0 (or Ha: 1- 2< D0)

Two-Tailed Test

H0: 1- 2= D0

Ha: 1- 2D0

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Hypothesis About 1- 2 continued

Test Statistic: Z

Rejection region:One-Tailed Test z > z

(or z < -z)

Two-Tailed Test z > z/2

Assumptions

Large samples

Independent samples

=

2

2

2

1

2

1

021

021

21

n

s

n

sDyy

DyyZ

yy

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Example 8.13

New vs. Old Process

H0: 1- 2= D0

Ha: 1- 2< D0 Find -z.05 = -1.645

Calculate Z:

Decision?Accept Ho

=

2

2

2

1

2

1

021

ns

ns

DyyZ

25

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Small Sample Test:

t Distribution One-Tailed Test

H0: 1- 2= D0

Ha: 1- 2> D0 (or Ha: 1- 2< D0)

Two-Tailed Test

H0: 1- 2= D0

Ha: 1- 2D0

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Small Sample Test: t Distributioncontinued

Test Statistic:

Rejection regionOne-Tailed Test

t > t, n1 + n2 - 2

(or t < -t) Two-Tailed Test

t > t/2, n1 + n2 - 2

=

21

2

021

11

nns

DyyT

p

Assumptions:

1. normal populations

2. independent samples

3. equal variance

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Pooled Standard Deviation sp

2

2

2

1assumewewhere =

= 2

11

21

2

22

2

112

nn

snsn

sp

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No Common Variance

Page 361: no common variance but

equal sized samplesuse unpooled

estimate of standard deviation butcomplicated formula for degrees of

freedom

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Test: Difference MATCHED pairs

Relatively straightforward:

Convert all data to the difference d and

treat it as a single variable.Use either large or small sample formula.

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MATCHEDPairs

Convert all of your data to difference

between two values of matched pair.

Carry out all of your analysis on this newcolumn of data.

Calculate its average value

Calculate its standard deviation

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Hypothesis Matched Pair

One-Tailed Test

H0: d = D0

Ha: d > D0 (or Ha: d < D0)

Two-Tailed Test

H0: d = D0

Ha: d D0

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Matched Pair Large Sample

Test Statistic for n>30:

Rejection region

One-Tailed Test z > z (or z < -z)

Two-Tailed Test z > z/2

ns

DdZ

d

0=

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Matched Pair Small Sample

Test Statistic for n t, n-1 (or t < -t)

Two-Tailed Test t > t/2, n-1

ns

DdT

d

0

=

34

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Example 8.15H0: d = D0

Ha: d > D0

One-Tailed Test t > t, 5 = 2.015

Decision?

Reject H0

See Figure 8.17 p.366

- review Minitab printout

01.30 == nsDdT d

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Testing a Population Proportion

One-Tailed Test

H0: p = p0

Ha: p > p0 (or Ha: p < p0)

Two-Tailed Test

H0: p = p0

Ha: p p0

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Population Proportion

Test Statistic:

Rejection region

One-Tailed Test z > z (or z < -z)

Two-Tailed Test z > z/2

Assumption:

n

qp

ppz

00

0

=

37

4,4 qnpn

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Example 8.16

Hypothesis

H0: p = .95

Ha: p < .95 Test Statistic:

= (.9 - .95)/[(.95*.05)/60] = -1.78 Rejection region z < -z= -1.645 Decision?

Reject Ho

nqp

ppz

00

0

=

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Testing the Difference between two

Population Proportions One-Tailed Test

H0: p1- p2= D0

Ha: p1- p2> D0 (or Ha: p1- p2< D0)

Two-Tailed Test

H0: p1- p2= D0Ha: p1- p2D0

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Difference in Population Proportions

Test Statistic:

where

and rejection region is

|z| > z /2

21

021

pp

Dppz

=

2

22

1

11

21 n

qp

n

qp

pp

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Difference in Population Proportions

Rejection region

One-Tailed Test: z > z (or z < -z)

Two-Tailed Test: z > z/2

To justify normality assumption:

41

4,44

,4

2222

1111

qnpnqnpn

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If D0= 0

When D0= 0p1= p2= p standard deviation formula simplifies

21

11

21 nnqppp

Test Statistic:

where

and rejection region is

|z| > z /2

21

021

pp

Dpp

z

=

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Example 8.17

Hypothesis

H0: p1- p2= 0

Ha: p1- p2< 0 Test Statistic:

where

21

21 0

pp

ppz

=

21

11

21 nnqppp

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Example 8.17 continued

so test statistic z = -3.56

35115002000576652

3841500576

3262000652

2

1

.))/((p

./p

./p

==

==

==

44

0163.01500

1

2000

1649.351.

11

21

21=

=

nn

qppp

56.3

0163.0

058.0

0163.0

0384.326.0

21

21 ====pp

ppz

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Example 8.17 continued

Rejection region

One-Tailed Test z < -z= -1.645

Test statistic z = -3.56

Decision? Reject Hosince Z is less than -Z

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Testing a Population Variance

Chi-square 2 when sample is selectedfrom a normal population

From last chapter:

2= (n-1)s2/2

Chi-Square distribution (Asymmetric) with

n-1 degrees of freedom

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Confidence Intervals and Hypothesis

Testing on Population Variance

(1-)100% Confidence Interval for 2 is

(n-1)s2/ 2/2 2(n-1)s2/ 2(1-/2)

(page 306)

Assumption: Population has approximate

normal distribution

What would you expect test statistic to be forH0: 2= 02 Ha: 2> 02 ? (see p.376)

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Confidence Intervals and Hypothesis

Testing on Population Variance

22

0

2

0

2

2

0

2

0

:

:

=

aH

H

2

1

2

0

2

0

2

2

0

2

0

:

:

=

aH

H

222

1

2

2

0

2

2

0

2

0

22

0

or0

:

:

=

aH

H

Where:

20 = (n-1)s2/ 02

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Example 8.18

H0: 02= .01 Ha: 2 < .01

20 = (n-1)s2/ 02

Smaller value of s2 we observe strongerthe evidence in favor of alternate

hypothesis

So, reject H0 for small values of teststatistic

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Testing the Ratio of Two Population

Variances

Hypothesis

One-Tailed H0: 12/ 22= 1

Ha: 12/ 22> 1 (or Ha: 12/ 22< 1)Two-Tailed H0: 12/ 22= 1

H0: 12/ 221

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Test Statistic

F-Distribution

One-sided F = s12/s2

2 (orF = s22/s1

2)

Two-sided F =Larger s2/Smaller s2

Rejection region

One-sided F > F

Two-sided F > F/2

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Example 8.19 (p.380)

H0: 12

/ 22

= 1H0: 12/ 221

Only upper-tail values of F are in table so always

put larger variance on top

F =larger s2/smaller s2 = s22/s1

2= 2.58

What are degrees of freedom for this example?

n1= 18 n2= 13

Review the rest of the problem on your own