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Hypothesis testing
Hypothesis testing
A study showed that 30% of men are coffee
drinkers compared to 20% for women!
Possible Interpretation !?????
1. True difference that also exists in the population
from which we drew our sample
2. Difference might be due to chance
3. Difference is due to bias
Choosing a significance test
Bias can be ruled out if we feel confident that our study design and
research implementation techniques were sound
The distinction between differences due to chance or real variation
need significance testing
A significance test
Estimates the likelihood that an observed study result is due to
chance
It is used to find out whether a study result which is observed
in a sample can be considered as a result which exists in the
population from which the sample was drawn
If you are measuring an association between two variables, you
determine how likely it is that your results could have occurred by
chance (i.e: occurred due to sampling variation)
A significance test
If it is unlikely (< 5%) that your result occurred by
chance, you reject the chance explanation and accept
that there is a real difference
The difference is statistically significant
If
it is likely (> 5%) that your result occurred by chance, you
cannot conclude that a real difference exists.
Therefore the difference is not statistically significant
Null hypothesis
The assumption that in the total population no real
difference exists between groups (no real association
exists between variables)
Example: Males do not smoke more than females
Statistical difference
A statistically significant difference does not
necessarily mean that the relationship is an
important one
Developing Null and Alternative
Hypotheses

Hypothesis testing can be used to determine whether a
statement about the value of a population parameter
should or should not be rejected.

The null hypothesis, denoted by H0 , is a tentative
assumption about a population parameter.

The alternative hypothesis, denoted by Ha, is the
opposite of what is stated in the null hypothesis.
A Summary of Forms for Null and Alternative
Hypotheses about a Population Mean

The equality part of the hypotheses always appears in
the null hypothesis.

In general, a hypothesis test about the value of a
population mean  must take one of the following
three forms (where 0 is the hypothesized value of the
population mean).
H0:  > 0
Ha:  < 0
H0:  < 0
Ha:  > 0
H0:  = 0
Ha:  0
Example: Metro EMS

Null and Alternative Hypotheses
A major west coast city provides one of the most comprehensive
emergency medical services in the world.
Operating in a multiple hospital system with approximately 20
mobile medical units, the service goal is to respond to medical
emergencies with a mean time of 12 minutes or less.
The director of medical services wants to formulate a
hypothesis test that could use a sample of emergency
response times to determine whether or not the service
goal of 12 minutes or less is being achieved.
Example: Metro EMS

Null and Alternative Hypotheses
Hypotheses
Conclusion and Action
H0:  
The emergency service is meeting
the response goal; no follow-up
action is necessary.
Ha: 
The emergency service is not
meeting the response goal;
appropriate follow-up action is
necessary.
Where:  = mean response time for the population
of medical emergency requests.
Type I and Type II Errors

Since hypothesis tests are based on sample data, we must allow for
the possibility of errors.

A Type I error is rejecting H0 when it is true.
A Type II error is accepting H0 when it is false.


The person conducting the hypothesis test specifies the maximum
allowable probability of making a Type I error, denoted by  and
called the level of significance.

Generally, we cannot control for the probability of making a Type
II error, denoted by .

Statistician avoids the risk of making a Type II error by using “do
not reject H0” and not “accept H0”.
Example: Metro EMS

Type I and Type II Errors
Conclusion
Population Condition
True status
H0 True
H0 False
(  )
( )
Accept H0
(Conclude  
Correct
Conclusion
Reject H0
(Conclude 
Type I
rror
Type II
Error
Correct
Conclusion
P-Value
It is the likelihood or probability of observing a result by
chance
A p<0.05 means that you would observe a difference in your
data only 5 times or less in every 100 samples examined
Choosing a significance test
Before applying any significance test, state the null
hypothesis in relation to the data to which the test is being
applied. This will enable you to interpret the results of the
test
The Use of p-Values

The p-value is the probability of obtaining a sample result
that is at least as unlikely as what is observed.

The p-value can be used to make the decision in a
hypothesis test by noting that:
• if the p-value is less than the level of significance , the
value of the test statistic is in the rejection region.
• if the p-value is greater than or equal to , the value of
the test statistic is not in the rejection region.

Reject H0 if the p-value < .
Statistical significance and causation
An association that is statistically significance does not
necessarily imply the existence of a causal relation.
It calls for further investigation to find out whether a
causal relationship does exist
The Steps of Hypothesis Testing






Determine the appropriate hypotheses.
Select the test statistic for deciding whether or not to
reject the null hypothesis.
Specify the level of significance  for the test.
Use to develop the rule for rejecting H0.
Collect the sample data and compute the value of the
test statistic.
a) Compare the test statistic to the critical value(s) in
the rejection rule, or
b) Compute the p-value based on the test statistic and
compare it to to determine whether or not to reject
H0 .
Two-Tailed Tests about a Population Mean:
Large-Sample Case (n > 30)

Hypotheses
H0:  = 
Ha:   

Test Statistic  Known  Unknown
z=

x  0
/ n
Rejection Rule
t = x 
s/ n
0
Reject H0 if |z | > |z1-  /2 |
Reject H0 if |t | > |t1-  /2 |
Example: Glow Toothpaste

Two-Tailed Tests about a Population Mean: Large n
The production line for Glow toothpaste is designed to fill
tubes of toothpaste with a mean weight of 6 ounces.
Periodically, a sample of 30 tubes will be selected in order to
check the filling process. Quality assurance procedures call
for the continuation of the filling process if the sample results
are consistent with the assumption that the mean filling
weight for the population of toothpaste tubes is 6 ounces;
otherwise the filling process will be stopped and adjusted.
Example: Glow Toothpaste

Two-Tailed Tests about a Population Mean: Large n
A hypothesis test about the population mean can
be used to help determine when the filling process
should continue operating and when it should be
stopped and corrected.

Hypotheses
H0: =
Ha: 

Rejection Rule

assuming a .05 level of significance,

Two-Tailed Test about a Population Mean: Large n
Assume that a sample of 30 toothpaste tubes provides a sample
mean of 6.1 ounces and standard deviation of 0.2 ounces.
Let n = 30,
H0: =
Ha:  ≠
= 6.1,
x
s = .2 ounces
x  0 6.1  6
t=
=
= 2.74
s / n .2 / 30
Since 2.74 > Critical value, we reject H0.
The p-value .0062 is less than  = .05, so H0 is rejected
Conclusion: We are 95% confident that the mean filling weight of the toothpaste tubes is not 6
ounces. The filling process should be stopped and the filling mechanism adjusted .
Confidence Interval Approach to a
Two-Tailed Test about a Population Mean

Select a simple random sample from the
population and use the value of the sample
mean x to develop the confidence interval for
the population mean .

If the confidence interval contains the
hypothesized value 0, do not reject H0.
Otherwise, reject H0.
Example: Glow Toothpaste

Confidence Interval Approach to a Two-Tailed
Hypothesis Test
The 95% confidence interval for  is
s
x  t1 / 2
n
For example, if the confidence interval is
(6.01 to 6.2)
Since the hypothesized value for the population mean, 0
= 6, is not in this interval, the hypothesis-testing
conclusion is that the null hypothesis,
H0:  = 6, can be rejected.
Tests about a Population Mean:
Small-Sample Case (n < 30)

Test Statistic
t=

 Unknown
x  0
s/ n
This test statistic has a t distribution with n - 1 degrees of
freedom.
Rejection Rule
One-Tailed
H0: 
H0: 
H0: =
Two-Tailed
Reject H0 if | t| > t
Reject H0 if |t| > t
p -Values and the t Distribution

The format of the t distribution table provided in most statistics
textbooks does not have sufficient detail to determine the exact
p-value for a hypothesis test.

However, we can still use the t distribution table to identify a
range for the p-value.

An advantage of computer software packages is that the
computer output will provide the p-value for the t distribution.
One-Tailed Tests about a Population Mean:
Large-Sample Case (n > 30)

Hypotheses
H0:   
Ha: 

H0:   
Ha: 
Test Statistic
 Known
z=

or
x  0
/ n
 Unknown
t = x 
s/ n
0
Rejection Rule
Reject H0 if |z| > |z1-|
Accept H0 if |z| < |z|

One-Tailed Test about a Population Mean:
Large n
Let n = 40,
x= 13.25 minutes,
s = 3.2 minutes
x   13.25  12
=
= 2.47
s / sncan3.be
2 / used
40 to
(The sample standard deviation
t=
estimate the population standard deviation .)
Since 2.47 > critical value, we reject H0.
Conclusion: We are 95% confident that Metro EMS is not meeting the
response goal of 12 minutes; appropriate action should be taken to improve
service.
Example: Highway Patrol

One-Tailed Test about a Population Mean:
Small n
A State Highway Patrol periodically samples vehicle speeds
at various locations on a particular roadway. The sample of
vehicle speeds is used to test the hypothesis
H0:  < 65.
The locations where H0 is rejected are deemed the best locations
for radar traps.
At Location F, a sample of 16 vehicles shows a mean speed
of 68.2 mph with a standard deviation of 3.8 mph. Use an  =
.05 to test the hypothesis.
Example: Highway Patrol

One-Tailed Test about a Population Mean:
Small n
Let n = 16,
= 68.2 mph, s = 3.8 mph
x = 16-1 = 15, t = 1.753
 = .05, d.f.

x  0 68.2  65
t=
=
= 3.37
s / n 3.8 / 16
Since 3.37 > 1.753, we reject H0.
Conclusion: We are 95% confident that the mean speed of
vehicles at Location F is greater than 65 mph. Location F is a
good candidate for a radar trap.
Summary of Test Statistics to be Used in a
Hypothesis Test about a Population Mean
Yes
s known ?
Yes
n > 30 ?
No
Yes
Use s to
estimate s
s known ?
Yes
x 
z=
/ n
No
x 
t=
s/ n
x 
z=
/ n
No
Popul.
approx.
normal
?
No
Use s to
estimate s
x 
t=
s/ n
Increase n
to > 30
A Summary of Forms for Null and Alternative
Hypotheses about a Population Proportion


The equality part of the hypotheses always appears in
the null hypothesis.
In general, a hypothesis test about the value of a
population proportion p must take one of the following
three forms (where p0 is the hypothesized value of the
population proportion).

H0: p > p0
H0: p < p0
H0: p = p0

Ha: p < p0
Ha: p > p0
Ha: p
p0
Tests about a Population Proportion:
Large-Sample Case (np > 5 and n(1 - p) > 5)

Test Statistic
z=
p =
where:
p  p0
p
p0 (1  p0 )
n
Rejection Rule
One-Tailed
H0: p  p
H0: p  p
H0: p=p
Two-Tailed
Reject H0 if z > z
Reject H0 if z < -z
Reject H0 if |z| > z
Example: NSC

Two-Tailed Test about a Population Proportion: Large
n
For a Christmas and New Year’s week, the
National Safety Council estimated that 500 people
would be killed and 25,000 injured on the nation’s
roads. The NSC claimed that 50% of the accidents
would be caused by drunk driving.
A sample of 120 accidents showed that 67 were
caused by drunk driving. Use these data to test the
NSC’s claim with  = 0.05.
Example: NSC

Two-Tailed Test about a Population Proportion:
Large n

Hypothesis
H0: p = .5
Ha: p  .5

Test Statistic
p0 (1  p0 )
.5(1  .5)
p =
=
= .045644
n
120
z=
p  p0
p
=
(67 /120)  .5
= 1.278
.045644
Example: NSC

Two-Tailed Test about a Population Proportion:
Large n
Rejection Rule
Reject H0 if z < -1.96 or z > 1.96
 Conclusion
Do not reject H0.
For z = 1.278, the p-value is .201. If we reject
H0, we exceed the maximum allowed risk of
committing a Type I error (p-value > .050).

Hypothesis Testing and Decision
Making

In many decision-making situations the decision maker
may want, and in some cases may be forced, to take
action with both the conclusion do not reject H0 and
the conclusion reject H0.

In such situations, it is recommended that the
hypothesis-testing procedure be extended to include
consideration of making a Type II error.
Determining Differences
Between Groups
Introduction


t-test is used for numerical data when
comparing the means of two groups (mean
of income, age, weight...etc for males and
females)
2 is used for categorical data when
comparing proportions of events occurring
in two groups (prevalence rate in rural and
urban)
Student’s t-test or
t-test
 Is
the observed difference between
means of two groups statistically
?
significant
Comparisons Involving Means




Estimation of the Difference Between the Means of
Two Populations: Independent Samples
Hypothesis Tests about the Difference between the
Means of Two Populations: Independent Samples
Inferences about the Difference between the Means
of Two Populations: Matched Samples
Inferences about the Difference between the
Proportions of Two Populations:
Estimation of the Difference Between the Means
of Two Populations: Independent Samples




Point Estimator of the Difference between the
Means of Two Populations
Sampling Distribution
x1  x2
Interval Estimate of Large-Sample Case
Interval Estimate of Small-Sample Case
Point Estimator of the Difference Between
the Means of Two Populations





Let 1 equal the mean of population 1 and 2 equal
the mean of population 2.
The difference between the two population means is
1 - 2.
To estimate 1 - 2, we will select a simple random
sample of size n1 from population 1 and a simple
random sample of size n2 from population 2.
Let equal the mean of sample 1 and
equal the
x2
meanx1 of sample 2.
The point estimator of the difference between the
means of the populations 1 and 2 is
x1  x2
Sampling Distribution of x1  x2

Properties of the Sampling Distribution of

x1  x2
Standard Deviation
 x1  x2 =
12
n1

 22
n2
where: 1 = standard deviation of population 1
2 = standard deviation of population 2
n1 = sample size from population 1
n2 = sample size from population 2
Interval Estimate of 1 - 2:
Large-Sample Case (n1 > 30 and n2 > 30)

Interval Estimate with 1 and 2 Known
where:

x1  x2  z / 2  x1  x2
1 -  is the confidence coefficient
Interval Estimate with 1 and 2 Unknown
x1  x2  z / 2 sx1  x2
where:
sx1  x2
s12 s22
=

n1 n2
Example: Par, Inc.
 Interval
Estimate of 1 - 2: Large-Sample
Case
Par, Inc. is a manufacturer of golf
equipment and has developed a new golf ball
that has been designed to provide “extra
distance.” In a test of driving distance using a
mechanical driving device, a sample of Par golf
balls was compared with a sample of golf balls
made by Rap, Ltd., a competitor.
The sample statistics appear on the next
slide.
Example: Par, Inc.

Interval Estimate of 1 - 2: Large-Sample
Case

Sample Statistics
Sample #1
Sample Size
Mean
Standard Dev.
Sample #2
Par, Inc.
Rap, Ltd.
n1 = 120 balls
n2 = 80 balls
x1 = 235 yards
x2= 218 yards
s1 = 15 yards s2 = 20 yards
Example: Par, Inc.

Point Estimate of the Difference Between
Two Population Means
1 = mean distance for the population of
Par, Inc. golf balls
2 = mean distance for the population of
Rap, Ltd. golf balls
Point estimate of 1 - 2 = x1  x2 = 235 - 218
= 17 yards.
Point Estimator of the Difference
Between the Means of Two Populations
Population 1
Par, Inc. Golf Balls
Population 2
Rap, Ltd. Golf Balls
1 = mean driving
2 = mean driving
distance of Par
golf balls
distance of Rap
golf balls
m1 – 2 = difference between
the mean distances
Simple random sample
of n1 Par golf balls
Simple random sample
of n2 Rap golf balls
x1 = sample mean distance
for sample of Par golf ball
x2 = sample mean distance
for sample of Rap golf ball
x1 - x2 = Point Estimate of m1 – 2
Example: Par, Inc.

95% Confidence Interval Estimate of the Difference Between
Two Population Means: Large-Sample Case, 1 and 2
Unknown
Substituting the sample standard deviations for the
population standard deviation:
x1  x2  z / 2
12
 22
(15) 2 ( 20) 2

= 17  1. 96

n1 n2
120
80
= 17 + 5.14 or 11.86 yards to 22.14 yards.
We are 95% confident that the difference between the mean
driving distances of Par, Inc. balls and Rap, Ltd. balls lies in
the interval of 11.86 to 22.14 yards.
Interval Estimate of 1 - 2:
Small-Sample Case (n1 < 30 and/or n2 < 30)

Interval Estimate with  2 Known
x1  x2  z / 2 x1  x2
where:
 x1  x2
1 1
=  (  )
n1 n2
2
Interval Estimate of 1 - 2:
Small-Sample Case (n1 < 30 and/or n2 < 30)

Interval Estimate with  2 Unknown
x1  x2  t / 2 sx1  x2
where:
sx1  x2
1 1
= sp (  )
n1 n2
2
2
2
(
n

1
)
s

(
n

1
)
s
1
2
2
sp 2 = 1
n1  n2  2
Example: Specific Motors
Specific Motors of Detroit has developed a new
automobile known as the M car. 12 M cars and 8 J
cars (from Japan) were road tested to compare
miles-per-gallon (mpg) performance. The sample
statistics are:
Sample Size
Mean
Standard Deviation
Sample #1
M Cars
n1 = 12 cars
x1 = 29.8 mpg
s1 = 2.56 mpg
Sample #2
J Cars
n2 = 8 cars
x2= 27.3 mpg
s2 = 1.81 mpg
Example: Specific Motors

Point Estimate of the Difference Between
Two Population Means
1 = mean miles-per-gallon for the population of M cars
2 = mean miles-per-gallon for the population of J cars
Point estimate of 1 - 2 =
mpg.
x1  x2= 29.8 - 27.3 = 2.5
Example: Specific Motors

95% Confidence Interval Estimate of the Difference
Between Two Population Means: Small-Sample Case
We will make the following assumptions:
The miles per gallon rating must be normally
distributed for both the M car and the J car.
 The variance in the miles per gallon rating must
be the same for both the M car and the J car.
Using the t distribution with n1 + n2 - 2 = 18 degrees
of freedom, the appropriate t value is t.025 = 2.101.
We will use a weighted average of the two sample
variances as the pooled estimator of  2.

Example: Specific Motors

95% Confidence Interval Estimate of the Difference
Between Two Population Means: Small-Sample Case
2
2
2
2
(
n

1
)
s

(
n

1
)
s
11
(
2
.
56
)

7
(
1
.
81
)
1
2
2
s2 = 1
=
= 5. 28
n1  n2  2
12  8  2
x1  x2  t.025
1 1
1 1
s (  ) = 2.5  2.101 5. 28(  )
n1 n2
12 8
2
= 2.5 + 2.2 or .3 to 4.7 miles per gallon.
We are 95% confident that the difference between the
mean mpg ratings of the two car types is from .3 to 4.7 mpg
(with the M car having the higher mpg).
Hypothesis Tests About the Difference
Between the Means of Two Populations: Independent
Samples


Hypotheses
H0: 1 - 2 < 0
H0: 1 - 2 > 0
H0: 1 - 2 = 0
Ha: 1 - 2 > 0
Ha: 1 - 2 < 0
Ha: 1 - 2  0
Test Statistic
Large-Sample
z=
( x1  x2 )  ( 1   2 )
12
n1   22
n2
Small-Sample
t=
( x1  x2 )  ( 1   2 )
s2 (1 n1  1 n2 )
Example: Par, Inc.

Hypothesis Tests About the Difference Between the
Means of Two Populations: Large-Sample Case
Par, Inc. is a manufacturer of golf equipment and
has developed a new golf ball that has been designed
to provide “extra distance.” In a test of driving
distance using a mechanical driving device, a sample
of Par golf balls was compared with a sample of golf
balls made by Rap, Ltd., a competitor. The sample
statistics appear on the next slide.
Example: Par, Inc.

Hypothesis Tests About the Difference Between the
Means of Two Populations: Large-Sample Case
 Sample Statistics
Sample #1
Sample Size
Mean
Standard Dev.
Par, Inc.
n1 = 120 balls
x1 = 235 yards
s1 = 15 yards
Sample #2
Rap, Ltd.
n2 = 80 balls
x2 = 218 yards
s2 = 20 yards
Example: Par, Inc.

Hypothesis Tests About the Difference Between the
Means of Two Populations: Large-Sample Case
Can we conclude, using a .01 level of significance, that
the mean driving distance of Par, Inc. golf balls is greater
than the mean driving distance of Rap, Ltd. golf balls?
1 = mean distance for the population of Par, Inc.
golf balls
2 = mean distance for the population of Rap, Ltd.
golf balls

Hypotheses
H0: 1 - 2 < 0
Ha: 1 - 2 > 0
Example: Par, Inc.

Hypothesis Tests About the Difference Between the Means
of Two Populations: Large-Sample Case

Rejection Rule Reject H0 if z > 2.33
z=

( x1  x2 )  ( 1   2 )
s12 s22

n1 n2
(235  218)  0
17
=
=
= 6.49
2
2
2.62
(15) (20)

120
80
Conclusion
Reject H0. We are at least 99% confident that the
mean driving distance of Par, Inc. golf balls is greater than
the mean driving distance of Rap, Ltd. golf balls.
Example: Specific Motors

Hypothesis Tests About the Difference Between the Means
of Two Populations: Small-Sample Case
Can we conclude, using a .05 level of
significance, that the miles-per-gallon (mpg)
performance of M cars is greater than the miles-pergallon performance of J cars?
1 = mean mpg for the population of M cars
2 = mean mpg for the population of J cars

Hypotheses
H0: 1 - 2 < 0
Ha: 1 - 2 > 0
Example: Specific Motors

Hypothesis Tests About the Difference Between the Means of
Two Populations: Small-Sample Case
 Rejection Rule
Reject H0 if t > 1.734
(a = .05, d.f. = 18)

Test Statistic
where:
t=
( x1  x2 )  ( 1   2 )
sp 2 (1 n1  1 n2 )
2
2
(
n

1
)
s

(
n

1
)
s
1
2
2
sp 2 = 1
n1  n2  2
Mean Heights of Males and
Females in X School
Gender Number of Mean Height
students
in cm
Standard
deviation
Males
60
156
3.1
Females
52
154
2.8
Steps to follow

Calculate the t-value

Use the t-table and find out the tabulated t

Interpret the results
Mean Heights of Males and
Females in Hala School
Gender
Number of
students
Mean Height in
cm
Standard
deviation
Males
60
156
3.1
Females
52
154
2.8
Using a T-Table
1- Decide the significance level (p value) you want to use (in
health studies we usually 0.05)
 Do not forget that:


if p = or < 0.05 then we reject the null hypothesis and
accept the alternative one concluding the presence of
statistically significant difference
if p > 0.05 then we accept the null hypothesis and
conclude that there is no statistically significant
difference.
Using a t-Table
2- Determine the degrees of freedom (sum of
observations for the two groups minus 2)
3- Look for the t-value belonging to the chosen
significance level
4- Compare tabulated t value and the calculated value
5-Apply the rule of accepting or rejecting the null
hypothesis
Using a t-Table
6
- if the calculated t-value is larger than the table value
then p < than .05 and you end up rejecting the null
hypothesis and concluding the presence of difference
7
- if the calculated t-value is smaller than the table
value, then p > .05 and you end up accepting the null
hypothesis and concluding the absence of difference
Example: Express Deliveries

Inference About the Difference Between the Means
of Two Populations: Matched Samples
 di ( 7  6... 5)
d =
=
= 2. 7
n
10
2
76.1
 ( di  d )
sd =
=
= 2. 9
n 1
9
d  d
2. 7  0
t=
=
= 2. 94
sd n 2. 9 10
Conclusion
Reject H0.
There is a significant difference between the mean
delivery times for the two services.
Inferences About the Difference
Between the Proportions of Two Populations
p1  p2



Sampling Distribution of
Interval Estimation of p1 - p2
Hypothesis Tests about p1 - p2

Sampling Distribution of p1  p2
Expected Value
E ( p1  p2 ) = p1  p2

Standard Deviation
 p1  p2 =
p1 (1  p1 ) p2 (1  p2 )

n1
n2
Distribution Form
If the sample sizes are large (n1p1, n1(1 - p1), n2p2,
and n2(1 - p2) are all greater than or equal to 5), the
sampling distribution of p1  pcan
be approximated by a
2
normal probability distribution.
Interval Estimation of p1 - p2

Interval Estimate

Point Estimator of  p1  p2
s p1  p2 =
p1  p2  z / 2  p1  p2
p1 (1  p1 ) p2 (1  p2 )

n1
n2
Example: MRA
MRA (Market Research Associates) is conducting
research to evaluate the effectiveness of a client’s new
advertising campaign. Before the new campaign began, a
telephone survey of 150 households in the test market area
showed 60 households “aware” of the client’s product. The
new campaign has been initiated with TV and newspaper
advertisements running for three weeks. A survey conducted
immediately after the new campaign showed 120 of 250
households “aware” of the client’s product.
Does the data support the position that the advertising
campaign has provided an increased awareness of the client’s
product?
Example: MRA

Point Estimator of the Difference Between the Proportions of
Two Populations
p1  p2 = p1  p2 =
120 60

=. 48. 40 =. 08
250 150
p1 = proportion of the population of households
“aware” of the product after the new campaign
p2 = proportion of the population of households
“aware” of the product before the new campaign
p1= sample proportion of households “aware” of the
product after the new campaign
sample proportion of households “aware” of the
p=
2
product before the new campaign
Example: MRA

Interval Estimate of p1 - p2: Large-Sample Case
For = .05, z.025 = 1.96:
. 48. 40  1. 96
. 48(.52) . 40(. 60)

250
150
.08 + 1.96(.0510)
.08 + .10
or -.02 to +.18

Conclusion
At a 95% confidence level, the interval estimate of the
difference between the proportion of households aware of
the client’s product before and after the new advertising
campaign is -.02 to +.18.
Hypothesis Tests about p1 - p2

Hypotheses
H0: p1 - p2 < 0
Ha: p1 - p2 > 0

Test statistic
z=
( p1  p2 )  ( p1  p2 )
 p1  p2
s p1  p2 = p (1  p )(1 n1  1 n2 )

Point Estimator of  p1  p2 where p1 = p2
where:
n1 p1  n2 p2
p=
n1  n2
Example: MRA

Hypothesis Tests about p1 - p2
Can we conclude, using a .05 level of significance, that
the proportion of households aware of the client’s product
increased after the new advertising campaign?
p1 = proportion of the population of households
“aware” of the product after the new campaign
p2 = proportion of the population of households
“aware” of the product before the new campaign

Hypotheses
H0: p1 - p2 < 0
Ha: p1 - p2 > 0
Example: MRA

Hypothesis Tests about p1 - p2
Rejection Rule
Reject H0 if z > 1.645
 Test Statistic
250(. 48)  150(. 40) 180
p=
=
=. 45
250  150
400

s p1  p2 = . 45(.55)( 1
 1 ) =. 0514
250 150
(. 48. 40)  0 . 08
z=
=
= 1.56
. 0514
. 0514

Conclusion
Do not reject H0.