Hypothsis testing

Hypothesis Testing

Shakeel Nouman
M.Phil Statistics

Hypothesis Testing By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

Slide 1

7
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Slide 2

Hypothesis Testing

Using Statistics
The Concept of Hypothesis Testing
Computing the p-value
The Hypothesis Tests
Testing population means, proportions and
variances
Pre-Test Decisions


7-1: Using Statistics
•

•

•

Slide 3

A hypothesis is a statement or assertion about
the state of nature (about the true value of an
unknown population parameter):
The accused is innocent
  = 100
Every hypothesis implies its contradiction or
alternative:
The accused is guilty
 100
A hypothesis is either true or false, and you may
fail to reject it or you may reject it on the basis of
information:
Trial testimony and evidence
Sample data

Decision-Making
•
•

Slide 4

One hypothesis is maintained to be true until a
decision is made to reject it as false:
Guilt is proven “beyond a reasonable doubt”
The alternative is highly improbable
A decision to fail to reject or reject a hypothesis
may be:
 Correct
» A true hypothesis may not be rejected

» An innocent defendant may be acquitted

» A false hypothesis may be rejected

» A guilty defendant may be convicted
Incorrect

» A true hypothesis may be rejected

» An innocent defendant may be convicted

» A false hypothesis may not be rejected
» A guilty defendant may be acquitted


Statistical Hypothesis Testing
•

•

Slide 5

A null hypothesis, denoted by H0, is an assertion
about one or more population parameters. This is the
assertion we hold to be true until we have sufficient
statistical evidence to conclude otherwise.

H0:  = 100

The alternative hypothesis, denoted by H1, is the
assertion of all situations not covered by the null
hypothesis.

H1:  100

• H0 and H1 are:
 Mutually exclusive
– Only one can be true.
 Exhaustive
– Together they cover all possibilities, so one or the other must be
true.

Hypothesis about other
Parameters
•

Slide 6

Hypotheses about other parameters such as
population proportions and and population
variances are also possible. For example

H0: p  40%
H1: p < 40%
H0: s2 50
H1: s2  50


The Null Hypothesis, H0
•

Slide 7

The null hypothesis:
Often represents the status quo situation or an existing belief.
Is maintained, or held to be true, until a test leads to its
rejection in favor of the alternative hypothesis.
Is accepted as true or rejected as false on the basis of a
consideration of a test statistic.


7-2 The Concepts of Hypothesis
Testing
•

•

Slide 8

A test statistic is a sample statistic computed from
sample data. The value of the test statistic is used in
determining whether or not we may reject the null
hypothesis.
The decision rule of a statistical hypothesis test is a rule
that specifies the conditions under which the null
hypothesis may be rejected.

Consider H0: = 100. We may have a decision rule that says: “Reject
H0 if the sample mean is less than 95 or more than 105.”
In a courtroom we may say: “The accused is innocent until proven
guilty beyond a reasonable doubt.”

Decision Making

•
•

Slide 9

There are two possible states of nature:
H0 is true
H0 is false
There are two possible decisions:
Fail to reject H0 as true
Reject H0 as false


Decision Making
•
•

Slide 10

A decision may be correct in two ways:
Fail to reject a true H0
Reject a false H0
A decision may be incorrect in two ways:
Type I Error: Reject a true H0
• The Probability of a Type I error is denoted
by 
.
Type II Error: Fail to reject a false H0
• The Probability of a Type II error is denoted
by 
.

Errors in Hypothesis Testing
•

Slide 11

A decision may be incorrect in two ways:
Type I Error: Reject a true H0
» The Probability of a Type I error is denoted by .
» is called the level of significance of the test

Type II Error: Accept a false H0
» The Probability of a Type II error is denoted by .
» 1 - is called the power of the test.

•

and are conditional probabilities:
 = P(Reject H 0 H 0 is true)




= P(Accept H 0 H 0 is false)


Type I and Type II Errors

Slide 12

A contingency table illustrates the possible outcomes
of a statistical hypothesis test.


The p-Value

Slide 13

The p-value is the probability of obtaining a value of the test statistic as extreme as,
or more extreme than, the actual value obtained, when the null hypothesis is true.
The p-value is the smallest level of significance,  at which the null hypothesis
,
may be rejected using the obtained value of the test statistic.
Policy: When the p-value is less than a , reject H0.

NOTE: More detailed discussions about the p-value will be
given later in the chapter when examples on hypothesis
tests are presented.


The Power of a Test

Slide 14

The power of a statistical hypothesis test is the
probability of rejecting the null hypothesis when the
null hypothesis is false.
Power = (1 - 
)


The Power Function

Slide 15

The probability of a type II error, and the power of a test, depends on the actual value
of the unknown population parameter. The relationship between the population mean
and the power of the test is called the power function.
Value of m

b

Power = (1 - b)

Power of a One-Tailed Test:=60, 
=0.05
1.0

0.8739
0.7405
0.5577
0.3613
0.1963
0.0877
0.0318
0.0092
0.0021

0.1261
0.2695
0.4423
0.6387
0.8037
0.9123
0.9682
0.9908
0.9972

0.9
0.8

Power

61
62
63
64
65
66
67
68
69

0.7
0.6
0.5
0.4
0.3



0.2
0.1
0.0
60

61

62

63

64

65

66

67

68

69


70



Factors Affecting the Power
Function








Slide 16

The power depends on the distance between the
value of the parameter under the null hypothesis
and the true value of the parameter in question:
the greater this distance, the greater the power.
The power depends on the population standard
deviation: the smaller the population standard
deviation, the greater the power.
The power depends on the sample size used: the
larger the sample, the greater the power.
The power depends on the level of significance of
the test: the smaller the level of significance,,
the smaller the power.

Example
A company that delivers packages within a large metropolitan
area claims that it takes an average of 28 minutes for a package to
be delivered from your door to the destination. Suppose that you
want to carry out a hypothesis test of this claim.
Set the null and alternative hypotheses:
H0: = 28
H1: 28

x  z

. 025

s
5
 315  196
.
.
n
100
 315  .98  30.52, 32.48
.

Collect sample data:
n = 100
x = 31.5
s=5

We can be 95% sure that the average time for
all packages is between 30.52 and 32.48
minutes.

Construct a 95% confidence interval for
the average delivery times of all packages:

Since the asserted value, 28 minutes, is not
in this 95% confidence interval, we may
reasonably reject the null hypothesis.


7-3 1-Tailed and 2-Tailed Tests

Slide 18

The tails of a statistical test are determined by the need for an action. If action
is to be taken if a parameter is greater than some value a, then the alternative
hypothesis is that the parameter is greater than a, and the test is a right-tailed
test.
H0: 50
H1: 50

If action is to be taken if a parameter is less than some value a, then the
alternative hypothesis is that the parameter is less than a, and the test is a lefttailed test.
H0: 50
H1: 50

If action is to be taken if a parameter is either greater than or less than some
value a, then the alternative hypothesis is that the parameter is not equal to a,
and the test is a two-tailed test.
H0: 50
H1: 50


7-4 The Hypothesis Tests

Slide 19

We will see the three different types of hypothesis tests, namely

Tests of hypotheses about population means
Tests of hypotheses about population proportions
Tests of hypotheses about population proportions.


Testing Population Means

Slide 20

• Cases in which the test statistic is Z

s is known and the population is normal.
s is known and the sample size is at least 30. (The population
need not be normal)

The formula for calculating Z is :
x
z
s 


 n

Testing Population Means

Slide 21

• Cases in which the test statistic is t

s is unknown but the sample standard deviation is known and
the population is normal.

The formula for calculating t is :
x
t
 s 


 n

Rejection Region
• The

Slide 22

rejection region of a statistical hypothesis
test is the range of numbers that will lead us to
reject the null hypothesis in case the test
statistic falls within this range. The rejection
region, also called the critical region, is defined
by the critical points. The rejection region is
defined so that, before the sampling takes place,
our test statistic will have a probability  of
falling within the rejection region if the null
hypothesis is true.


Nonrejection Region

Slide 23

• The

nonrejection region is the range of values
(also determined by the critical points) that will
lead us not to reject the null hypothesis if the
test statistic should fall within this region. The
nonrejection region is designed so that, before
the sampling takes place, our test statistic will
have a probability 1- of falling within the
nonrejection region if the null hypothesis is true

In a

two-tailed test, the rejection region consists of
the values in both tails of the sampling distribution.


Picturing Hypothesis Testing
Population
mean under H0

= 28

Slide 24

95% confidence
interval around
observed sample mean

30.52

x = 31.5

32.48

It seems reasonable to reject the null hypothesis, H0: = 28, since the
hypothesized value lies outside the 95% confidence interval. If we’re 95% sure
that the population mean is between 30.52 and 32.58 minutes, it’s very unlikely
that the population mean is actually be 28 minutes.
Note that the population mean may be 28 (the null hypothesis might be true), but
then the observed sample mean, 31.5, would be a very unlikely occurrence. There’s
still the small chance (= .05) that we might reject the true null hypothesis.
represents the level of significance of the test.

Nonrejection Region

Slide 25

If the observed sample mean falls within the nonrejection region, then you fail to
reject the null hypothesis as true. Construct a 95% nonrejection region around
the hypothesized population mean, and compare it with the 95% confidence
interval around the observed sample mean:
 0  z.025

s
5
 28  1.96
n
100
 28.98   27,02 ,28.98

95%
nonrejection region
around
the
population Mean

27.02

 =28
0

28.98

95% Confidence
Interval
around
the
Sample Mean

30.52

x

x  z .025

s
5
 315  1.96
.
n
100
 315.98   30.52 ,32.48
.

32.48

The nonrejection region and the confidence interval are the same width, but
centered on different points. In this instance, the nonrejection region does not
include the observed sample mean, and the confidence interval does not include
the hypothesized population mean.

Picturing the Nonrejection and
Rejection Regions
If the null hypothesis were
true, then the sampling
distribution of the mean
would look something
like this:

Slide 26

T he Hypothesized Sampling Distribution of the Mean
0.8
0.7

.95

0.6
0.5
0.4
0.3
0.2

.025

.025

0.1

We will find 95% of the
sampling distribution between
the critical points 27.02 and 28.98,
and 2.5% below 27.02 and 2.5% above 28.98 (a two-tailed test).
The 95% interval around the hypothesized mean defines the
nonrejection region, with the remaining 5% in two rejection
regions.
0.0

27.02

0=28

28.98


The Decision Rule

Slide 27

The Hypothesized Sampling Distribution of the Mean
0.8
0.7

.95

0.6
0.5
0.4
0.3
0.2

.025

.025

0.1
0.0
27.02

0=28

28.98
x5

•

LwrRt Nrt UrRt
R
R
R

Construct a (1- nonrejection region around the
)
hypothesized population mean.

Do not reject H0 if the sample mean falls within the nonrejection
region (between the critical points).
Reject H0 if the sample mean falls outside the nonrejection region.


Example 7-5

Slide 28

An automatic bottling machine fills cola into two liter (2000 cc) bottles. A consumer advocate wants to test the null
hypothesis that the average amount filled by the machine into a bottle is at least 2000 cc. A random sample of 40
bottles coming out of the machine was selected and the exact content of the selected bottles are recorded. The
sample mean was 1999.6 cc. The population standard deviation is known from past experience to be 1.30 cc.
Test the null hypothesis at the 5% significance level.

H0: 2000
H1: 2000
n = 40
For = 0.05, the critical value
of z is -1.645
x  0
s
Do not reject H0 if: [z  n
-1.645]

The test statistic
z is:

Reject H0 if:  
z
]

n = 40
x = 1999.6

s = 1.3

z

x

s
n

0 = 1999.6 - 2000

1.3
40

=  1.95  Reject H
0


Example 7-5: p-value approach

Slide 29

An automatic bottling machine fills cola into two liter (2000 cc) bottles. A consumer advocate wants to test the null
hypothesis that the average amount filled by the machine into a bottle is at least 2000 cc. A random sample of 40
bottles coming out of the machine was selected and the exact content of the selected bottles are recorded. The
sample mean was 1999.6 cc. The population standard deviation is known from past experience to be 1.30 cc.
Test the null hypothesis at the 5% significance level.

H0: 2000
H1: 2000
n = 40
of z is -1.645
x
The test statistic is: 0
z
s
n
Do not reject H if: [p-value 0.05]
0

Reject H0 if: 
p-value 
0.0
]

z

x

s

0 = 1999.6 - 2000

n

1.3
40

=  1.95
p - value  P(Z  - 1.95)
 0.5000 - 0.4744
 0.0256  Reject H since 0.0256  0.05
0


Example 7-5: Using the
Template

Slide 30

Use when s
is known

Use when s
is unknown


Example 7-6: Using the Template
with Sample Data

Use when s
is known

Use when s
is unknown


Testing Population Proportions

Slide 32

• Cases in which the binomial distribution can be used

The binomial distribution can be used whenever we are able to
calculate the necessary binomial probabilities. This means that
for calculations using tables, the sample size n and the population
proportion p should have been tabulated.
Note: For calculations using spreadsheet templates, sample
sizes up to 500 are feasible.


Testing Population Proportions

Slide 33

• Cases in which the normal approximation is to be used

If the sample size n is too large (n > 500) to calculate binomial
probabilities then the normal approximation can be used.and the
population proportion p should have been tabulated.


Example 7-7: p-value approach

Slide 34

A coin is to tested for fairness. It is tossed 25 times and only 8 Heads are observed. Test
if the coin is fair at an a of 5% (significance level).

Let p denote the probability of a Head
H0: p = 0.5
H1: p 0.5
Because this is a 2-tailed test, the p-value = 2*P(X 8)
From the binomial tables, with n = 25, p = 0.5, this value
2*0.054 = 0.108.s
Since 0.108 > = 0.05, then
do not reject H0


with the Binomial Distribution


with the Normal Distribution


Testing Population Variances

Slide 37

• For testing hypotheses about population variances, the test
statistic (chi-square) is:
n  1s
 

2

2

s

2
0

where s is the claimed value of the population variance in the
null hypothesis. The degrees of freedom for this chi-square
random variable is (n – 1).
2

0

Note: Since the chi-square table only provides the critical values, it cannot
be used to calculate exact p-values. As in the case of the t-tables, only a
range of possible values can be inferred.


Example 7-8

Slide 38

A manufacturer of golf balls claims that they control the weights of the golf balls
accurately so that the variance of the weights is not more than 1 mg2. A random sample
of 31 golf balls yields a sample variance of 1.62 mg2. Is that sufficient evidence to reject
the claim at an a of 5%?

Let s2 denote the population variance. Then
H 0 : s2 < 1
H1: s2 > 1
In the template (see next slide), enter 31 for the sample size
and 1.62 for the sample variance. Enter the hypothesized value
of 1 in cell D11. The p-value of 0.0173 appears in cell E13. Since
This value is less than the a of 5%, we reject the null hypothesis.


Example 7-8

Slide 39


Additional Examples (a)

Slide 40

As part of a survey to determine the extent of required in-cabin storage capacity, a
researcher needs to test the null hypothesis that the average weight of carry-on baggage
per person is 0 = 12 pounds, versus the alternative hypothesis that the average weight
is not 12 pounds. The analyst wants to test the null hypothesis at = 0.05.

H0: = 12
H1: 12

The Standard Normal Distribution
0.8
0.7

.95

0.6

For = 0.05, critical values of z are 1.96
x  0
z
The test statistic is:
s
n
Do not reject H0 if: [-1.96 z 
1.96]
Reject H0 if: [z <-1.96] or  
z 1.96]

0.5
0.4
0.3
0.2

.025

.025

0.1
0.0
-1.96

Lower Rejection
Region



z

1.96

Nonrejection
Region

Upper Rejection
Region


Additional Examples (a):
Solution
n = 144

Slide 41

0.8

x = 14.6

0.7

.95

0.6
0.5

s = 7.8

0.4
0.3

z

x   0 14.6-12
=
s
7.8
n
144

2.6
=
4
0.65

0.2

.025

.025

0.1
0.0
-1.96



z

1.96


Lower Rejection
Region

Nonrejection
Region

Upper Rejection
Region

Since the test statistic falls in the upper rejection region, H0 is rejected, and we may
conclude that the average amount of carry-on baggage is more than 12 pounds.

Additional Examples (b)

Slide 42

An insurance company believes that, over the last few years, the average liability
insurance per board seat in companies defined as “small companies” has been $2000.
Using = 0.01, test this hypothesis using Growth Resources, Inc. survey data.
n = 100

H0: = 2000
H1: 2000

x = 2700
s = 947


x  0
z
s
n

Do not reject H0 if: [-2.576 z 2.576]
Reject H0 if: [z <-2.576] or  
z 2.576]

z

x  0

2700 - 2000
=

s

947

n

100

700
=
94.7

 7 .39  Reject H
0


Additional Examples (b) :
Continued
0.8
0.7

.99

0.6
0.5
0.4
0.3
0.2

.005

.005

0.1
0.0
-2.576



z

2.576

Slide 43

Since the test statistic falls in
the upper rejection region, H0
is rejected, and we may
conclude that the average
insurance liability per board
seat in “small companies” is
more than $2000.



Lower Rejection
Region

Nonrejection
Region

Upper Rejection
Region


Additional Examples (c)

Slide 44

The average time it takes a computer to perform a certain task is believed to be 3.24
seconds. It was decided to test the statistical hypothesis that the average performance
time of the task using the new algorithm is the same, against the alternative that the
average performance time is no longer the same, at the 0.05 level of significance.

H0: = 3.24
H1: 3.24
x 0
The test statisticis:
z

n = 200
x = 3.48
s = 2.8

z

s

s
n

=

3.48 - 3.24
2.8

n

Do not reject H0 if: [-1.96 z 
1.96]
Reject H0 if: [z < -1.96] or  
z 1.96]

x  0

=

0.24
 1.21
0.20

200
 Do not reject H
0


Additional Examples (c) :
Continued

the nonrejection region, H0 is
not rejected, and we may
performance time has not
changed from 3.24 seconds.

0.8
0.7

.95

0.6
0.5
0.4
0.3
0.2

.025

.025

0.1

0.0
-1.96



1.96

Slide 45

z



Lower Rejection
Region

Nonrejection
Region

Upper Rejection
Region


Additional Examples (d)

Slide 46

According to the Japanese National Land Agency, average land prices in central Tokyo
soared 49% in the first six months of 1995. An international real estate investment
company wants to test this claim against the alternative that the average price did not rise
by 49%, at a 0.01 level of significance.

H0: = 49
H1: 49
n = 18
For = 0.01 and (18-1) = 17 df ,
critical values of t are 2.898
The test

x  0
statistic sis:
n
t

Do not reject H0 if: [-2.898 t 2.898]
Reject H0 if: [t < -2.898] or  2.898]
t

n = 18
x = 38
s = 14

t 

x 
s
n

=

0 =

38 - 49
14
18

- 11
 3.33  Reject H
0
3.3


Additional Examples (d) :
Continued
The t Distribution
0.8
0.7

.99

0.6
0.5
0.4
0.3
0.2

.005

.005

0.1
0.0
-2.898



2.898

t



Lower Rejection
Region

Nonrejection
Region

Upper Rejection
Region

Slide 47

the rejection region, H0 is
rejected, and we may conclude
that the average price has not
risen by 49%. Since the test
statistic is in the lower
rejection region, we may
price has risen by less than
49%.


Additional Examples (e)

Slide 48

Canon, Inc,. has introduced a copying machine that features two-color copying capability
in a compact system copier. The average speed of the standard compact system copier is
27 copies per minute. Suppose the company wants to test whether the new copier has the
same average speed as its standard compact copier. Conduct a test at an = 0.05 level
of significance.

H0: = 27
H1: 27
n = 24
For = 0.05 and (24-1) = 23 df ,
critical values of t are 2.069

t 

n = 24
x = 24.6
s = 7.4

t 

s

x  0
s
n

Do not reject H0 if: [-2.069 t 2.069]
Reject H0 if: [t < -2.069] or  2.069]
t

x  0

n

=

=

24.6 - 27
7.4
24

-2.4
 1.59
1.51

 Do not reject H
0


Additional Examples (e) :
Continued
The t Distribution

not rejected, and we may not
speed is different from 27
copies per minute.

0.8
0.7

.95

0.6
0.5
0.4
0.3
0.2

.025

.025

0.1
0.0

-2.069



2.069

Slide 49

t



Lower Rejection
Region

Nonrejection
Region

Upper Rejection
Region


Statistical Significance

Slide 50

Wtuytssstttrutruut
ytsstst,utsttrt,t
ytsstststtsrvtuytssvr
ttrtvytssTssuswtr
rut,trtyTyIrrr,ts
sswsr,sus00r005Tus,wwrt
uytss,wvvurs,
swwtrssrtyttwv
rrr
Avswttrtu
ytssusstsutstrtrt
tstTts,trtrusss
ttrtstyrt,sttstss,r
tytszTrtrs,tur,t
vusssttrstyrt,
sttstss,rtytsz

Additional Examples (f)
AvsttystrGSsCywtttstt
ytssyBrtssurtsxrtstt70%r
vstrstBrtsrtwrArTysttr
rs20utsrvstrsLu
tt0wrwyUStzsAtt 005vs,s
trvtrtttBrtssurtsxrts
n = 210
H0:070
130
p=

 0.619
H: 070
210
20
Fr 005rtvusz r96
p-p

0.619 - 0.70
0
Ttststtsts:
z=
=
p  p0

p q
(0.70)(0.30)
z
0 0
p0 q 0
210
n
n
DtrtH0 :96 z  96
-0.081
RtH0 :z96rz 96
=
 2.5614  Reject H
0.0316
0

Additional Examples (g)

Slide 52

The EPA sets limits on the concentrations of pollutants emitted by various industries. Suppose that the
upper allowable limit on the emission of vinyl chloride is set at an average of 55 ppm within a range of two
miles around the plant emitting this chemical. To check compliance with this rule, the EPA collects a
random sample of 100 readings at different times and dates within the two-mile range around the plant. The
findings are that the sample average concentration is 60 ppm and the sample standard deviation is 20 ppm.
Is there evidence to conclude that the plant in question is violating the law?

H0: 55
H1: 
55
n = 100
of z is 2.326

n = 100
x = 60
s = 20

z

s

The test statistic  x   0
z is:

s
n
[z 2.326]

Do not reject H0 if:
Reject H0 if:  
z 2.326]

x  0

n

=

5
 2.5
2

=

60 - 55
20
100
 Reject H
0


Additional Examples (g) :
Continued
Critical Point for a Right-Tailed Test

the rejection region, H0 is
rejected, and we may conclude
that the average concentration
of vinyl chloride is more than
55 ppm.

0 .4

0.99

f(z)

0 .3

0 .2



0 .1

0 .0
-5

0

z

Slide 53

5

2.326
2.5

Nonrejection
Region

Rejection
Region


Additional Examples (h)

Slide 54

A certain kind of packaged food bears the following statement on the package: “Average net weight 12 oz.”
Suppose that a consumer group has been receiving complaints from users of the product who believe that they are
getting smaller quantities than the manufacturer states on the package. The consumer group wants, therefore, to
test the hypothesis that the average net weight of the product in question is 12 oz. versus the alternative that the
packages are, on average, underfilled. A random sample of 144 packages of the food product is collected, and it is
found that the average net weight in the sample is 11.8 oz. and the sample standard deviation is 6 oz. Given these
findings, is there evidence the manufacturer is underfilling the packages?

H0: 12
H1: 12
n = 144
of z is -1.645 x   0
z


s
n

Do not reject H0 if: [z 
-1.645]
Reject H0 if:  
z
]

n = 144
x = 11.8
s = 6

z

x
s
n

=

0 = 11.8 -12

6
144

-.2
 0.4  Do not reject H
0
.5


Additional Examples (h) :
Continued
Critical Point for a Left-Tailed Test

conclude that the manufacturer
is underfilling packages on
average.

0 .4

0.95

f(z)

0 .3

0 .2


0 .1

0 .0
-5

0

Slide 55

5

z

-1.645
-0.4

Rejection
Region

Nonrejection
Region


Additional Examples (i)

Slide 56

A floodlight is said to last an average of 65 hours. A competitor believes that the average life of the
floodlight is less than that stated by the manufacturer and sets out to prove that the manufacturer’s
claim is false. A random sample of 21 floodlight elements is chosen and shows that the sample
average is 62.5 hours and the sample standard deviation is 3. Using 
=0.01, determine whether
there is evidence to conclude that the manufacturer’s claim is false.

H0: 65
H1: 65
n = 21
For = 0.01 an (21-1) = 20 df, the
critical value -2.528

Do not reject H0 if: [t 
-2.528]
Reject H0 if:  
z
]

Additional Examples (i) :
Continued
Critical Point for a Left-Tailed Test
0 .4

095

f(t)

0 .3

0 .2

005
0 .1

0 .0
-5

0

-3.82

Rt
R

5

t

-2.528

Nrt
R

Slide 57

Sttststtst
strt
r,H0 srt,
wyuttt
uturrss
s,tttvr
tsss
t65urs


Additional Examples (j)
“After looking at 1349 hotels nationwide, we’ve found 13 that meet our standards.” This statement by the Small Luxury Hotels
Association implies that the proportion of all hotels in the United States that meet the association’s standards is 13/1349=0.0096. The
management of a hotel that was denied acceptance to the association wanted to prove that the standards are not as stringent as claimed
and that, in fact, the proportion of all hotels in the United States that would qualify is higher than 0.0096. The management hired an
independent research agency, which visited a random sample of 600 hotels nationwide and found that 7 of them satisfied the exact
standards set by the association. Is there evidence to conclude that the population proportion of all hotels in the country satisfying the
standards set by the Small Luxury hotels Association is greater than 0.0096?

H0: p 0.0096
H1: p 0.0096
n = 600
For = 0.10 the critical value 1.282

Do not reject H0 if: [z 
1.282]
Reject H0 if:  
z
]

Additional Examples (j) :
Continued
Critical Point for a Right-Tailed Test
0 .4

0.90

f(z)

0 .3

0 .2



0 .1

0 .0
-5

0

z

5

1.282

Slide 59

conclude that proportion of all
hotels in the country that meet
the association’s standards is
greater than 0.0096.

0.519

Nonrejection
Region

Rejection
Region


The p-Value Revisited
Standard Normal Distribution

Standard Normal Distribution

0.4

0.4

p-value=area to
right of the test statistic
=0.0062

0.3

0.2

0.2

0.1

f(z)

p-value=area to
right of the test statistic
=0.3018

0.3

f(z)

Slide 60

0.1

0.0

0.0
-5

0

0.519

Additional Example k

5

-5

z

0

5
2.5

z

Additional Example g

The p-value is the probability of obtaining a value of the test statistic as extreme as,
or more extreme than, the actual value obtained, when the null hypothesis is true.
The p-value is the smallest level of significance,  at which the null hypothesis
,
may be rejected using the obtained value of the test statistic.

The p-Value: Rules of Thumb

Slide 61

When the p-value is smaller than 0.01, the result is called very
significant.
When the p-value is between 0.01 and 0.05, the result is called
significant.

When the p-value is between 0.05 and 0.10, the result is considered
by some as marginally significant (and by most as not significant).
When the p-value is greater than 0.10, the result is considered not
significant.


p-Value: Two-Tailed Tests

Slide 62

p-value=double the area to
left of the test statistic
=2(0.3446)=0.6892
0.4

f(z)

0.3

0.2

0.1

0.0
-5

-0.4

0

0.4

5

z

In a two-tailed test, we find the p-value by doubling the area in
the tail of the distribution beyond the value of the test statistic.

The p-Value and Hypothesis
Testing

Slide 63

The further away in the tail of the distribution the test statistic falls, the smaller
is the p-value and, hence, the more convinced we are that the null hypothesis is
false and should be rejected.
In a right-tailed test, the p-value is the area to the right of the test statistic if the
test statistic is positive.
In a left-tailed test, the p-value is the area to the left of the test statistic if the
test statistic is negative.
In a two-tailed test, the p-value is twice the area to the right of a positive test
statistic or to the left of a negative test statistic.
For a given level of significance,
:
Reject the null hypothesis if and only if
p-value


7-5: Pre-Test Decisions

Slide 64

One can consider the following:
Sample Sizes
b versus a for various sample sizes
The Power Curve
The Operating Characteristic Curve

Note: You can use the different templates that come
with the text to investigate these concepts.



Slide 65

Note: Similar
analysis can
be done when
testing for a
population
proportion.

Computing and
Plotting Required
Sample size.


Template

Slide 66

Plot of b
versus a for
various n.

Note: Similar
analysis can
be done when
testing for a
population
proportion.


Template

Slide 67

The Power
Curve

Note: Similar
analysis can
be done when
testing for a
population
proportion.


Template

Slide 68

The Operating
Characteristic
Curve for
H0:m >= 75;
s = 10; n = 40;
a = 10%
Note:
Similar
analysis can be
done
when
testing
a
population
proportion.

Name
Religion
Domicile
Contact #
E.Mail
M.Phil (Statistics)

Shakeel Nouman
Christian
Punjab (Lahore)
0332-4462527. 0321-9898767
sn_gcu@yahoo.com
sn_gcu@hotmail.com
GC University, .
(Degree awarded by GC University)

M.Sc (Statistics)
Statitical Officer
(BS-17)
(Economics & Marketing
Division)

GC University, .
(Degree awarded by GC University)

Livestock Production Research Institute
Bahadurnagar (Okara), Livestock & Dairy Development
Department, Govt. of Punjab


Hypothsis testing

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