Transcript No Slide Title

Dr. Ka-fu Wong
ECON1003
Analysis of Economic Data
Ka-fu Wong © 2003
Chap 11- 1
Overview
To test the effect of an herbal treatment on improvement
of memory you randomly select two samples, one to
receive the treatment and one to receive a placebo.
Results of a memory test taken one month later are given.
Sample
1
x1  77
s1  15
n1  95
x 2  73
s 2  12
Sample
2
n 2  105
Experimental Group
Control Group
Treatment
Placebo
The resulting test statistic is 77 - 73 = 4. Is
this difference significant or is it due to chance
(sampling error)?
Ka-fu Wong © 2003
Chap 11- 2
Chapter Eleven
Two Sample Tests of Hypothesis
GOALS
1.
Understand the difference between dependent and
independent samples.
2.
Conduct a test of hypothesis about the difference
between two independent population means when
both samples have 30 or more observations.
3.
Conduct a test of hypothesis about the difference
between two independent population means when at
least one sample has less than 30 observations.
Conduct a test of hypothesis about the mean
difference between paired or dependent observations.
Conduct a test of hypothesis regarding the difference
in two population proportions.
4.
5.
l
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Chap 11- 3
Two Sample Tests
TEST FOR EQUAL VARIANCES
Ho
Population 1
TEST FOR EQUAL MEANS
Ho
Population 2
Population 2
H1
Population 1
Population 2
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Population 1
H1
Population 1 Population 2
Chap 11- 4
The formula of general test statistic
 Suppose we are interested in testing the population
parameter () is equal to k.
 H 0:  = k
 H 1:   k
 First, we need to get a sample estimate (q) of the
population parameter ().
 Second, we know in most cases, the test statistics will be
in the following form:
 t=(q-k)/q
 The form of q depends on what q is.
 Sample size and the null at hand determine the
distribution of the statistic.
 If  is population mean, and the sample size is larger
than 30, t is approximately normal.
Ka-fu Wong © 2003
Chap 11- 5
Comparing two populations
 We wish to know whether the distribution of
the differences in sample means has a mean
of 0.
 If both samples contain at least 30
observations we use the z distribution as the
test statistic.
Ka-fu Wong © 2003
Chap 11- 6
Hypothesis Tests for Two Population
Means
Preferred
Two-Tailed
Test
Format 1
Upper OneTailed Test
Lower OneTailed Test
H 0 : 1   2  0.0
H 0 : 1   2  0.0
H 0 : 1   2  0.0
H A : 1   2  0.0
H A : 1   2  0.0
H A : 1   2  0.0
Format 2
H 0 : 1   2
H 0 : 1   2
H 0 : 1   2
H A : 1   2
H A : 1   2
H A : 1   2
Ka-fu Wong © 2003
Chap 11- 7
Two Independent Populations: Examples
1.
An economist wishes to determine whether there is a
difference in mean family income for households in
two socioeconomic groups.
 Do HKU students come from families with higher
income than CUHK students?
2.
An admissions officer of a small liberal arts college
wants to compare the mean SAT scores of applicants
educated in rural high schools & in urban high
schools.
 Do students from rural high schools have lower
A-level exam score than from urban high schools?
Ka-fu Wong © 2003
Chap 11- 8
Two Dependent Populations: Examples
1.
An analyst for Educational Testing Service wants to
compare the mean GMAT scores of students before & after
taking a GMAT review course.

2.
Get HKU graduates to take A-Level English and Chinese
exam again. Do they get a higher A-Level English and
Chinese exam score than at the time they enter HKU?
Nike wants to see if there is a difference in durability of 2
sole materials. One type is placed on one shoe, the other
type on the other shoe of the same pair.
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Chap 11- 9
Thinking Challenge
Are they independent or dependent?
1. Miles per gallon ratings of cars before & after
mounting radial tires dependent
2. The life expectancies of light bulbs made in
two different factories independent
3. Difference in hardness between 2 metals: one
contains an alloy, one doesn’t independent
4. Tread life of two different motorcycle tires:
one on the front, the other on the back dependent
Ka-fu Wong © 2003
Chap 11- 10
Comparing two populations
 No assumptions about the shape of the
populations are required.
 The samples are from independent populations.
 Values in one sample have no influence on the
values in the other sample(s).
 Variance formula for independent random
variables A and B: V(A-B) = V(A) + V(B)
 The formula for computing the value of z is:
z
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X1  X 2
s12 s22

n1 n2
Chap 11- 11
EXAMPLE 1
Two cities, Bradford and Kane are separated only
by the Conewango River. There is competition
between the two cities. The local paper recently
reported that the mean household income in
Bradford is $38,000 with a standard deviation of
$6,000 for a sample of 40 households. The same
article reported the mean income in Kane is
$35,000 with a standard deviation of $7,000 for a
sample of 35 households. At the .01 significance
level can we conclude the mean income in
Bradford is more?
Ka-fu Wong © 2003
Chap 11- 12
EXAMPLE 1
continued
 Step 1: State the null and alternate hypotheses.
H0: µB ≤ µK ; H1: µB > µK
 Step 2: State the level of significance. The .01
significance level is stated in the problem.
 Step 3: Find the appropriate test statistic.
Because both samples are more than 30, we can
use z as the test statistic.
Ka-fu Wong © 2003
Chap 11- 13
Example 1
continued
 Step 4: State the decision rule. The null
hypothesis is rejected if z is greater than 2.33.
Probability density of z
statistic : N(0,1)
H 0: µB ≤ µK ;
H 1: µB > µK
Rejection
Region  = 0.01
0
z  2.33
Acceptance Region  = 0.01
Ka-fu Wong © 2003
Chap 11- 14
Example 1
continued
 Step 5: Compute the value of z and make a
decision.
z
$38,000  $35,000
($6,000)2 ($7,000)2

40
35
H 0: µB ≤ µK ;
H 1: µB > µK
Acceptance Region 0 = 0.01 1.98
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 1.98
Rejection
Region  = 0.01
z  2.33
Chap 11- 15
Example 1
continued
 The decision is to not reject the null
hypothesis. We cannot conclude that
the mean household income in
Bradford is larger.
Ka-fu Wong © 2003
Chap 11- 16
Example 1
continued
 The p-value is:
P(z > 1.98) = .5000 - .4761 = .0239
P-value = 0.0239
H 0 : µB ≤ µK ;
H 1: µB > µK
Rejection
Region  = 0.01
z  2.33
0
1.98
Ka-fu Wong © 2003
Chap 11- 17
Small Sample Tests of Means
 The t distribution is used as the test statistic
if one or more of the samples have less than
30 observations.
 The required assumptions are:
1. Both populations must follow the normal
distribution.
2. The populations must have equal
standard deviations.
3. The samples are from independent
populations.
Ka-fu Wong © 2003
Chap 11- 18
Small sample test of means
continued
 Finding the value of the test statistic requires
two steps.
Step 1: Pool the sample standard deviations.
2
2
(
n

1
)
s

(
n

1
)
s
1
2
2
sp2  1
n1  n2  2
Step 2: Determine the value of t from the
following formula.
t
X1  X 2
1 1
s   
 n1 n2 
2
p
Ka-fu Wong © 2003
Chap 11- 19
EXAMPLE 2
 A recent EPA study compared the highway fuel
economy of domestic and imported passenger
cars. A sample of 15 domestic cars revealed a
mean of 33.7 mpg with a standard deviation of
2.4 mpg. A sample of 12 imported cars
revealed a mean of 35.7 mpg with a standard
deviation of 3.9.
 At the .05 significance level can the EPA
conclude that the mpg is higher on the
imported cars?
Ka-fu Wong © 2003
Chap 11- 20
Example 2
continued
 Step 1: State the null and alternate hypotheses.
H0: µD ≥ µI ; H1: µD < µI
 Step 2: State the level of significance. The .05
significance level is stated in the problem.
 Step 3: Find the appropriate test statistic. Both
samples are less than 30, so we use the t
distribution.
Ka-fu Wong © 2003
Chap 11- 21
EXAMPLE 2
continued
Step 4: The decision rule is to reject H0 if t<-1.708.
There are 25 degrees of freedom.
H0 : D  I
H A : D  I
Probability density of t
statistic : t (df=25)
  0.05
Rejection
Region  = 0.05
t  1.708
Ka-fu Wong © 2003
0
Chap 11- 22
EXAMPLE 2
Step 5:
continued
We compute the pooled variance:
2
2
(
n

1
)(
s
)

(
n

1
)(
s
2
1
1
2
2)
sp 
n1  n2  2
(15  1)(2.4)2  (12  1)(3.9)2

 9.918
15  12  2
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Chap 11- 23
Example 2
continued
We compute the value of t as follows.
t 

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X1  X 2
 1
1 

s p2 

n

n
2 
 1
33.7  35.7
1 
 1
8.312


 15 12 
 1.640
Chap 11- 24
Example 2
continued
H0 : D  I
H A : D  I
Rejection
Region  = 0.05
  0.05
t  1.708
0
-1.640
H0 is not rejected. There is insufficient sample evidence
to claim a higher mpg on the imported cars.
Ka-fu Wong © 2003
Chap 11- 25
Hypothesis Testing Involving Paired
Observations
 Independent samples are samples that are not
related in any way.
 Dependent samples are samples that are paired
or related in some fashion. For example:
 If you wished to buy a car you would look at
the same car at two (or more) different
dealerships and compare the prices.
 If you wished to measure the effectiveness of
a new diet you would weigh the dieters at
the start and at the finish of the program.
Ka-fu Wong © 2003
Chap 11- 26
Hypothesis Testing Involving Paired
Observations
 Use the following test when the samples are
dependent:
t
d
sd
n
 where d is the mean of the differences
 sd is the standard deviation of the differences
 n is the number of pairs (differences)
Ka-fu Wong © 2003
Chap 11- 27
EXAMPLE 3
 An independent testing agency is comparing
the daily rental cost for renting a compact car
from Hertz and Avis. A random sample of
eight cities revealed the following information.
At the .05 significance level can the testing
agency conclude that there is a difference in
the rental charged?
Ka-fu Wong © 2003
Chap 11- 28
EXAMPLE 3
City
continued
Hertz ($)
Avis ($)
Atlanta
42
40
Chicago
56
52
Cleveland
45
43
Denver
48
48
Honolulu
37
32
Kansas City
45
48
Miami
41
39
Seattle
46
50
Ka-fu Wong © 2003
Chap 11- 29
EXAMPLE 3
continued
 Step 1: State the null and alternate hypotheses.
H0: µd = 0 ; H1: µd ≠ 0
 Step 2: State the level of significance. The .05
significance level is stated in the problem.
 Step 3: Find the appropriate test statistic. We
can use t as the test statistic.
Ka-fu Wong © 2003
Chap 11- 30
EXAMPLE 3
continued
 Step 4: State the decision rule. H0 is rejected if
t < -2.365 or t > 2.365. We use the t
distribution with 7 degrees of freedom.
H 0: µB ≤ µK ;
H 1: µB > µK
Probability density of t
statistic : t (df=7)
Rejection Region I
Probability =0.025
Rejection Region II
probability=0.025
Acceptance Region  = 0.01
t / 2  2.365
Ka-fu Wong © 2003
t / 2  2.365
Chap 11- 31
Example 3
City
continued
Hertz ($)
Avis ($)
d
d2
Atlanta
42
40
2
4
Chicago
56
52
4
16
Cleveland
45
43
2
4
Denver
48
48
0
0
Honolulu
37
32
5
25
Kansas City
45
48
-3
9
Miami
41
39
2
4
Seattle
46
50
-4
16
Ka-fu Wong © 2003
Chap 11- 32
Example 3
d 
sd 
t 
Ka-fu Wong © 2003
continued
d
8.0

 1.00
n
8
2



d
d 2 
n
n 1
d
sd
n

82
78 
8  3.1623

8 1
1.00
 0.894
3.1623 8
Chap 11- 33
Example 3
continued
 Step 5: Because 0.894 is less than the critical
value, do not reject the null hypothesis. There is
no difference in the mean amount charged by
Hertz and Avis.
H 0: µB ≤ µK ;
H 1: µB > µK
0.894
Rejection Region I
Probability =0.025
Rejection Region II
probability=0.025
Acceptance Region  = 0.01
t / 2  2.365
Ka-fu Wong © 2003
t / 2  2.365
Chap 11- 34
Two Sample Tests of Proportions
 We investigate whether two samples came from
populations with an equal proportion of
successes.
 The two samples are pooled using the following
formula.
X1  X 2
pc 
n1  n2
where X1 and X2 refer to the number of successes
in the respective samples of n1 and n2.
Ka-fu Wong © 2003
Chap 11- 35
Two Sample Tests of Proportions
continued
 The value of the test statistic is computed from
the following formula.
z
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p1  p2
pc (1  pc ) pc (1  pc )

n1
n2
Chap 11- 36
Example 4
 Are unmarried workers more likely to be absent
from work than married workers? A sample of
250 married workers showed 22 missed more
than 5 days last year, while a sample of 300
unmarried workers showed 35 missed more than
five days. Use a .05 significance level.
Ka-fu Wong © 2003
Chap 11- 37
Example 4
continued
 The null and the alternate hypothesis are:
H0: U ≤  M
H1 :  U >  M
The null hypothesis is rejected if the computed
value of z is greater than 1.65.
Ka-fu Wong © 2003
Chap 11- 38
Example 4
continued
 The pooled proportion is
35  22
pc 
 .1036
300  250
The value of the test statistic is
z
Ka-fu Wong © 2003
35
22

300 250
 1.10
.1036(1  .1036) .1036(1  .1036)

300
250
Chap 11- 39
Example 4
continued
 The null hypothesis is not rejected. We cannot
conclude that a higher proportion of unmarried
workers miss more days in a year than the
married workers.
 The p-value is:
P(z > 1.10) = .5000 - .3643 = .1457
Ka-fu Wong © 2003
Chap 11- 40
Chapter Eleven
Two Sample Tests of Hypothesis
- END -
Ka-fu Wong © 2003
Chap 11- 41