STAT 211
1
Handout 8 (Chapter 9): Inferences Based on Two Samples
Population characteristics (parameter)
Sample characteristics (statistics)
_
1   2
_
x y
^
^
p1  p2
p1  p 2
 12 /  22
s12 / s 22
1. One-sided (One tailed) test:
A. Lower tailed: H0: population characteristics  claimed constant value
(Left-sided) Ha: population characteristics < claimed constant value
B. Upper tailed: H0: population characteristics  claimed constant value
(Right-sided) Ha: population characteristics > claimed constant value
2. Two-sided (Two tailed) test: H0: population characteristics = claimed constant value
Ha: population characteristics  claimed constant value
A. Independent Samples
I.
Population characteristics: Difference between two population means, 1-2.
0 is the claimed constant.
1 and  2 are the population means for X's and Y's, respectively.
_
_
x and y are the sample means for X's and Y's, respectively.
m and n are the sample sizes for X's and Y's, respectively.
 12 and  22 are the population variances for X's and Y's, respectively.
s12 and s 22 are the sample variances for X's and Y's, respectively.

Test statistics:
_ _
 x y    0

z

2
1
m

when both popn. distributions are normal and  12 ,  22 are known
n


 x y    0

z
when there is large sample size (m>40 and n>40) and  12 ,  22 are
2
2
s1 s 2

m n
unknown
_


2
2
_
STAT 211

2
_ _
 x y    0

t
when both popn. distributions are normal and at least one sample size is
2
s1 s 22

m n
small with unknown  12 ,  22 are assumed to be different (  12   22 ) and degrees of freedom,
2
 s12 s 22 
  
m n
is used to look up the critical values. If v is not computed as integer,
v
2
2
s12 / m
s 22 / n

m 1
n 1
it should be rounded down. Your textbook calls this two-sample t test.
_ _
 x y    0
(m  1) s12  (n  1) s 22

, s 2p 
when both popn. distributions are normal and at
t
mn2
1 1
sp

m n
least one sample size is small where unknown  12 ,  22 are assumed to be the same (  12 =  22 )
and degrees of freedom, v  m  n  2 is used to look up the critical values. Your textbook
calls this pooled t test.


 

Decision can be made in one of the two ways in hypothesis testing:
a. Let z* or t* be the computed test statistic values.
if test statistics is z
if test statistics is t
Lower tailed test P-value = P(z<z*)
P-value = P(t<t*)
Upper tailed test P-value = P(z>z*)
P-value = P(t>t*)
Two-tailed test
P-value = 2P(z>|z*|)= 2P(z<-|z*|) P-value = 2P(t > |t*| )= 2P(t <- |t*| )
In each case, you can reject H0 if P-value   and fail to reject H0 (accept H0) if P-value > 
b. Rejection region for level  test:
if test statistics is z
Lower tailed test z  -z
Upper tailed test z  z
Two- tailed test |z |  z/2
if test statistics is t
t  -t;v
t  t;v
|t|  t/2;v
100(1-)% confidence Intervals with the same assumptions,
 12  22
_ _
x

y

z


  /2
m
n


s2 s2
_ _
 x  y   z / 2 1  2
m n


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s12 s 22
_ _
when you assume  12   22

 x  y   t / 2 ; v
m n


1 1
_ _
 when you assume  12 =  22
 x  y   t / 2 ;v s p
m n


Example 1: Two types of plastic are suitable for use by an electronics component manufacturer.
The breaking strength of this plastic is very important. It is known that  1 =  2 =1 psi. Random
_
sample of size n1 =10 and n 2 =12 drawn from a normal distribution, we obtain x 1 =162.5 and
_
x 2 =155. The company will not adopt plastic 1 unless its mean breaking strength exceeds that of
plastic 2 by at least 10 psi. Based on the sample information, would they use plastic 1? Use the
significance level 0.05 in reaching a decision.
H 0 : 1   2  10 (adopt plastic 1) versus H a : 1   2  10 (do not adopt plastic 1)
_ _
 x y    0
(162.5  155)  10


 5.84
test statistics, z  
2
2
1  2
12 12


10 12
m
n
Decision:
(i) reject H0 if z  -z=-1.645. z =-5.84 < -1.645 then reject H0.
(ii) P-value = P(Z<-5.84)=P(Z>5.84)=0. Since the P-value  =0.05, reject H0.
Conclusion: Do not adopt plastic 1.
Example 2 (Exercise 9.2): We will use the given data in this exercise.
 i : true average tread lives for two competing brand of size P205/65R15 radial tires, i=1,2
Test H 0 : 1   2  0 versus H a : 1   2  0
_
m=45
x  42500
s1  2200
_
n=45
y  40400
s2  1900
Notice that sample sizes are large while population variances are unknown and this is a twotailed test.
_ _
 x y    0
(42500  40400)  0


Test statistics: z  
=4.8462
2
2
s1 s 2
2200 2 1900 2


45
45
m n
Decision:
(i) Reject H0 if z  -z/2=-1.96 or z  z/2 =1.96. z =4.8462 > 1.96 then reject H0.
(ii) P-value = 2P(Z>4.8462)=2(0)=0. Since the P-value  =0.05, reject H0.
Conclusion: true average tread lives for two competing brand tires are different.
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If you prefer to answer the question computing the confidence interval, 95% confidence interval
s2 s2
2200 2 1900 2
_ _
would become  x y   z / 2 1  2  (42500  40400)  1.96

m n
45
45


=(1250.67, 2949.33). You would see that zero does not fall into interval and you would reject
H0.
What would be different if this was an upper tailed test instead of two tailed test? (hypothesis,
test statistics, decision, conclusion)
What would be different if this was an upper tailed test with the hypothesized value 1000 instead
of two tailed test with the hypothesized value 0? (hypothesis, test statistics, decision, conclusion)
Example 3 (Exercise 9.8): The data is on the tensile strength test of two different grades or wire
rod.
Grade
Sample size
Sample mean
Population
Sample standard
mean
deviation
_
AISI 1064
m=129
s1 =1.3
1
x =107.6
_
AISI 1078
n=129
s2 =2.0
2
y =123.6
(a) Does the data provide compelling evidence for concluding that the true average strength for
1078 grade exceeds that for the 1064 grade by more than 10kg/mm2?
Sample sizes are large while population variances are unknown. . Notice that this is an uppertailed test.
H 0 :  2  1  10 versus H a :  2  1  10
_ _
 y x   0
(123.6  107.6)  10


Test statistics: z  
=28.57
2
2
s1 s 2
1.3 2 2 2


129 129
m n
Decision:
(i) Reject H0 if z  z =1.645 if =0.05. z =28.57 > 1.645 and reject H0.
(ii) P-value = P(Z>28.57)=0. Since the P-value  =0.05, reject H0.
Conclusion: the data provide compelling evidence that the true average strength for the 1078
grade exceeds that for the 1064 grade by more than 10.
Or you can answer the same question using H 0 : 1   2  10 versus H a : 1   2  10
Notice that it became a lower tailed test instead of upper tailed test.
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_ _
 x y    0
(107.6  123.6)  10


Test statistics: z  
= -28.57
2
2
s1 s 2
1.3 2 2 2


129 129
m n
Decision:
(i) Reject H0 if z - z =-1.645 if =0.05. z =-28.57 <- 1.645 and reject H0.
(ii) P-value = P(Z<-28.57)=0. Since the P-value  =0.05, reject H0.
(b) Estimate the difference in a way that provides information about the precision and reliability
s12 s 22
_ _
95% confidence interval for 1   2 is  x  y   z / 2

m n


= 107.6  123.6  1.96
1.3 2 2 2

=(-16.84 , -15.16)
129 129
s2 s2
_ _
95% confidence interval for  2  1 is  y  x   z / 2 1  2 = (15.16 , 16.84)
m n


Example 4 (Exercise 9.18):
 i : true average densities for two different type of brick, i=1,2
Assume normality of two density distributions
Test H 0 : 1   2  0 versus H a : 1   2  0
_
m=6
x  22.73
s1  0.164
_
n=5
y  21.95
s2  0.240
Sample sizes are small and population variances are unknown. . Notice that this is a two-tailed
test.
If the population variances are assumed to be different,
_ _
 x y    0
(22.73  21.95)  0


Test statistics: t  
=6.166
2
2
s1 s 2
0.164 2 0.24 2


6
5
m n
Decision:
(i) Reject H0 if t  -t/2;v= -2.447 or t  t/2;v =2.447 where =0.05 and v=6.886. t =6.166 >
2.447 and reject H0.
(ii) P-value = 2P(t>6.166)=2(0.0005)=0.001. Since the P-value  =0.05, reject H0.
Conclusion: the true average densities for two different types of brick are different.
If the population variances are assumed to be the same,
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_ _
 x y    0
(22.73  21.95)  0

Test statistics: t  
=
=6.396 where
1 1
1 1
sp

0.2014

6 5
m n
(m  1) s12  (n  1) s 22 5(0.164) 2  4(0.24) 2
2
=0.0405
sp 

mn2
652
Decision:
(i) Reject H0 if t  -t/2;v= -2.262 or t  t/2;v =2.262 where =0.05 and v=6+5-2=9. t =6.166 >
2.262 and reject H0.
(ii) P-value = 2P(t>6.166)<2(0.0005)=0.001. Since the P-value  =0.05, reject H0.
Conclusion: the true average densities for two different types of brick are different.
II.
Population characteristics: Difference between two population proportions, p1-p2.
p0 is the claimed constant.
^
^
p 1 and p 2 are the sample proportions for X's and Y's, respectively.
p1 and p 2 are the population proportions for X's and Y's, respectively.
m and n are the large sample sizes for X's and Y's, respectively.
^
X Y
m ^
n ^

p1 
p2
The estimator for p is p 
mn mn
mn
^
^

 p1  p 2   p0

Test statistics: z  
^
^

 1 1 
p1  p   

 m n 
Decision can be made in one of the two ways:
(a) Let z* be the computed test statistic values.
Lower tailed test P-value = P(z<z*)
Upper tailed test P-value = P(z>z*)
Two-tailed test
P-value = 2P(z > |z*| )=2P(z <- |z*| )
In each case, you can reject H0 if P-value   and fail to reject H0 (accept H0) if P-value > 
(b) Rejection region for level  test:
Lower tailed test z  -z
Upper tailed test z  z
Two- tailed test z  -z/2 or z  z/2
100(1-)% large sample confidence Interval:
 ^ ^ 
 p1  p 2   z / 2


^
^
^
^




p1 1  p1  p 2 1  p 2 

 

m
n
STAT 211
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Example 5: Two different types of injection-molding machines are used to form plastic parts. A
part is considered defective if it has excessive shrinkage or is discolored. Two random samples,
each of size 300 are selected and 15 defective parts are found from machine 1 while 8 defective
parts are found in the sample from machine 2. Is it reasonable to conclude that both machines
produce the same fraction of defective parts, using the significance 0.05?
If this analysis done by hand
Sample sizes are large enough to satisfy the assumptions and it is a two-tailed test.
^
^
^
X Y
15  8
15
8
p

=0.0383 where p 1 
=0.05 and p 2 
=0.0267
m  n 300  300
300
300
^
^

 p1  p 2   p0
(0.05  0.0267)  0


Test statistics: z  
=1.49
^
^
0.0383(0.9617)(1 / 300  1 / 300)

 1 1 
p1  p   

 m n 
(i) Reject H0 if z  -z/2=-1.96 or z  z/2 =1.96. -1.96< z=1.49 < 1.96 and fail to reject H0. .
(ii) P-value = 2P(Z>1.49)=2(0.0681)=0.1362. Since the P-value > =0.05, fail to reject H0.
Conclusion: Yes it is reasonable to assume that both machines produce the same fraction of
defective parts
The MINITAB output analyzing such data is
Test and CI for Two Proportions
Sample
X
N Sample p
1
15
300 0.050000
2
8
300 0.026667
Estimate for p(1) - p(2): 0.0233333
95% CI for p(1) - p(2): (-0.00733568, 0.0540023)
Test for p(1) - p(2) = 0 (vs not = 0): Z = 1.49 P-Value = 0.136
Example 6 (Exercise 9.48(a)):
Sample sizes are large enough to satisfy the assumptions and it is a two-tailed test.
^
^
^
X Y
63  75
63
75
p

=0.2875 where p 1 
=0.21 and p 2 
=0.4167
m  n 300  180
300
180
^
^

 p1  p 2   p0
(0.21  0.4167)  0


Test statistics: z  
=-4.844
^
^
0.2875(0.7125)(1 / 300  1 / 180)

 1 1 
p1  p   

 m n 
Decision:
(i) Reject H0 if z  -z/2=-1.96 or z  z/2 =1.96. |z| =4.844 > 1.96 and reject H0.
(ii) P-value = 2P(Z>4.844)=2(0)=0. Since the P-value  =0.05, reject H0.
Conclusion: it is different for two groups of residents.
The MINITAB output analyzing such data is
Test and CI for Two Proportions
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Sample
X
N Sample p
1
63
300 0.210000
2
75
180 0.416667
Estimate for p(1) - p(2): -0.206667
95% CI for p(1) - p(2): (-0.292174, -0.121159)
Test for p(1) - p(2) = 0 (vs not = 0): Z = -4.74
P-Value = 0.000
Population characteristics: Ratio of the two population variances,  12 /  22 or
standard deviations,  1 /  2 .
X and Y's are random sample from a normal distribution.
 12 and  22 are the population variances for X's and Y's, respectively.
s12 and s 22 are the sample variances for X's and Y's, respectively.
m and n are the sample sizes for X's and Y's, respectively.
s2
Test statistics: F  12
s2
Decision can be made in one of the two ways:
III.
(a) Let F* be the computed test statistic values.
Lower tailed test P-value = P(F<F*)
Upper tailed test P-value = P(F>F*)
Two-tailed test
P-value = 2P(F > F*)
In each case, you can reject H0 if P-value   and fail to reject H0 (accept H0) if P-value > 
(b) Rejection region for level  test:
Lower tailed test F  F1-;m-1,n-1
Upper tailed test F  F;m-1,n-1
Two- tailed test F  F1-/2;m-1,n-1 or F  F/2;m-1,n-1
Notice that F1-/2;m-1,n-1 = 1 / F/2;n-1,m-1
100(1-)% confidence Interval for  12 /  22 :
s12 / s 22
F / 2;m 1,n 1

 12
s12 / s 22

 22 F1 / 2;m 1,n 1
Example 7 (Exercise 9.57):
(a) On the F-table, column for 5 and row for 8 will give the area on the right 0.05 with F0.05:5,8 =
3.69
(d) F0.95:8,5 = 1/ F0.05:5,8 = 1/3.69 =0.271
(e) The percentile is the area on the left of the value and it means the area on the right of the
value is 0.01. On the F-table, look at column for 10 and row for 12 with the area on the right
0.01.
P( F  F0.01:10,12 )= 0.99 then F0.01:10,12 = 4.30
(h) P(0.177  F  4.74) = P(F0.99:10,5  F  F0.05:10,5) =1-(0.01+0.05)=0.94 where F0.99:10,5 =
1/F0.01:5,10 = 1/5.64=0.177
STAT 211
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Example 8: A study was performed to determine whether men and women differ in their
repeatability in assembling components on printed circuit boards. Two samples of 26 men and
21 women were selected and each subject assembled the units. The two sample standard
deviations of assembly time were smen=0.98 min and swomen=1.02 min. Is there evidence to
support the claim that men and women differ in repeatability for this assembly task? Use the
significance level 0.02 and state any necessary assumptions about underlying distribution of the
data.
2
2
H 0 :  men
  women
2
2
H a :  men
  women
=0.02, m=26, n=21
s2
0.98 2
Test statistics: F  2men 
 0.9231
s women 1.02 2
Decision: Reject H0 if F  F1-/2;m-1,n-1 = F0.99;25,20 =1/2.70=0.37 or F  F/2;m-1,n-1= F0.01;25,20
=2.84. Since 0.37 < F=0.9231 <2.84, fail to reject H0.
Conclusion: men and women do not differ in repeatability for this assembly task
Example 9 (Exercise 9.62):
H 0 :  12   22
H a :  12   22
=0.10, m=48, n=45
s12 21.45 2
 1.2219
Test statistics: F  2 
s 2 19.45 2
Decision: Reject H0 if F  F1-/2;m-1,n-1 = F0.95;47,44 =1/1.6336=0.6121 or F  F/2;m-1,n-1=
F0.05;47,44 =1.6412 (The F-table on your book is not very detailed, you can use both degrees of
freedom as 40 to look at the table. I gave you the real values using the computer programs.)
Since 1.2219 is not larger than 1.6412 or smaller than 0.6121, fail to reject H0.
Conclusion: the variances are the same.
B. Dependent Samples- Paired Data
Population characteristics: Difference between two population means, D =1-2.
0 is the claimed constant.
Assumption: the difference distribution should be normal.
_
Test statistics: t 
d  0
_
where D=X-Y and d and s D are the corresponding sample average
sD / n
and the standard deviation of D. Both X and Y must have n observations.
Decision can be made in one of the two ways:
(a) Let t* be the computed test statistic values.
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10
Lower tailed test P-value = P(t<t*)
Upper tailed test P-value = P(t>t*)
Two-tailed test
P-value = 2P(t > |t*| )=2P(t <- |t*| )
In each case, you can reject H0 if P-value   and fail to reject H0 (accept H0) if P-value > 
(b) Rejection region for level  test:
Lower tailed test t  -t;n-1
Upper tailed test t  t;n-1
Two- tailed test t  -t/2;n-1 or t  t/2;n-1
100(1-)% confidence Intervals with the same assumptions,
_
s
d  t / 2;n 1 D
n
Example 10: The manager of a fleet of automobiles is testing two brands of radial tires. He
assigns one tire of each brand at random to the two rear wheels of eight cars and runs the cars
until the tires wear out. The descriptive statistics for the data are shown below (in kilometers).
Find the 99% confidence interval on the difference in mean life. Which brand would you prefer
based on this calculation? Is there an alternative method to answer this question instead of
computing the confidence interval?
Variable N Mean Median StDev SE Mean Minimum Maximum Q1
Q3
Brand1 8 38479 37067 5590
1976
32100
48360 34185 43525
Brand2
8 37611 36655 5244
1854
31950
47800 33491 41214
Difference 8 868
475 1290
456.1
-805
3020
N/A
N/A
Brand1
Brand2
Difference
36925 45300 36240 32100 37210 48360 38200 33500
34318 42280 35500 31950 38015 47800 37810 33215
2607 3020 740 150 -805 560 390 285
_
99% confidence interval for 1   2 is d  t / 2;n 1
sD
 868  3.499
1290
=(-727.84 , 2463.84)
n
8
Confidence interval only tells us that there is no difference between those brands.
The MINITAB output analyzing such data is
B1
B2
Difference
N
8
8
8
Mean
38479
37611
868
StDev
5590
5244
1290
SE Mean
1976
1854
456
Test for equal true means in independent samples assuming equal true variances
Difference = mu B1 - mu B2
Estimate for difference: 868
95% CI for difference: (-4944, 6681)
T-Test of difference = 0 (vs not =): T-Value = 0.32 P-Value = 0.753 DF = 14
Both use Pooled StDev = 5420
STAT 211
11
Test for equal true means in independent samples assuming unequal true variances
Difference = mu B1 - mu B2
Estimate for difference: 868
95% CI for difference: (-4986, 6723)
T-Test of difference = 0 (vs not =): T-Value = 0.32 P-Value = 0.754 DF = 13
Test for equal true means in dependent samples
95% CI for mean difference: (-211, 1948)
99% CI for mean difference: (-728, 2465)
T-Test of mean difference = 0 (vs not = 0): T-Value = 1.90
P-Value = 0.099
Test for equal variances: B1 versus B2
F-Test (normal distribution)
Test Statistic: 1.136
P-Value
: 0.870
Example 11 : An experiment to compare the yield (kg/ha) of Sundance winter wheat and
Manitou spring wheat is considered. Data from nine different plots is given in the following
table. Is there sufficient evidence to conclude that true average yield for the Sundance winter
wheat is more than 500 kg/ha than the Manitou spring wheat? Check the plausibility of any
assumptions needed to carry out an appropriate test of hypothesis.
1
2
3
4
5
6
7
8
9
S
3201 3095 3297 3644 3604 2860 3470 2042 3689
M
2386 2011 2616 3094 3069 2074 2308 1525 2779
D=S-M 815 1084 681 550 535 786 1162 517 910
H 0 :  S   M  500 or H 0 :  S   M  500
H a :  S   M  500 or H a :  S   M  500
The difference distribution should be normal.
_
Use the differences and compute the sample mean, d =782.2222
s D =236.736.
_
Test statistics: t 
d  0
sD / n

782.2222  500
236.736 / 9
and the standard deviation,
=3.5764
Decision:
(i) Since this is an upper tailed test, reject H0 if t  t;n-1= t0.05;8 =1.86 or if the Pvalue=P(t>3.5764) is less than =0.05. Notice that test statistics, 3.5764 is large than 1.86.
(ii) 0.001 <P-value < 0.005 by looking at the t-table. Reject H0
Conclusion: there is sufficient evidence to conclude that the true average yield in winter wheat is
more than 500 higher than for spring wheat.
STAT 211
12
Normal Probability Plot for difference
ML Estimates - 95% CI
99
ML Estimates
95
Mean
782.222
StDev
223.197
90
Goodness of Fit
Percent
80
AD*
70
60
50
40
30
1.416
20
10
5
1
0
500
1000
1500
Data
Results from the MINITAB program
Paired T for muS - muM
N
9
9
9
S
M
Difference
Mean
3211
2429
782.2
StDev
518
518
236.7
SE Mean
173
173
78.9
95% lower bound for mean difference: 635.5
T-Test of mean difference = 500 (vs > 500): T-Value = 3.58
P-Value = 0.004
More to discuss
Two-sample T for muS vs muM
S
M
N
9
9
Mean
3211
2429
StDev
518
518
SE Mean
173
173
Difference = muS - muM
Estimate for difference: 782
95% CI for difference: (265, 1300)
T-Test of difference = 0 (vs not =): T-Value = 3.20
Both use Pooled StDev = 518
F-Test (normal distribution)
Test Statistic: 1.002
P-Value
: 0.998
Levene's Test (any continuous distribution)
Test Statistic: 0.094
P-Value
: 0.763
P-Value = 0.006
DF = 16