Statistics
Statistical Inference for Two Samples
Contents, figures, and exercises come from the textbook: Applied Statistics and Probability
for Engineers, 5th Edition, by Douglas C. Montgomery, John Wiley & Sons, Inc., 2011.
Inference on the Difference in
Means of Two Normal Distributions,
Variances Known

Assumptions




1. X11 , X 12 , …, X 1n is a random sample from
population 1.
2. X 21 , X 22 , …, X 2n is a random sample from
population 2.
3. The two populations represented by X1 and X 2
are independent
4. Both populations are normal.
1
2
E ( X 1  X 2 )  E ( X 1 )  E ( X 2 )  1  2
V ( X1  X 2 )  V ( X1 )  V ( X 2 ) 

The quantity
Z
X 1  X 2  ( 1   2 )
 12
n1


has a N (0,1) distribution
 22
n2
 12
n1

 22
n2


Hypothesis tests on the difference in means,
variance known
Hypotheses, two-sided alternative
X  X 2  0
 Test statistic:
Z0  1

 12  22

n1 n2



Hypotheses, two-sided alternative
H 0 : 1   2   0
H1 : 1  2   0
P-value: P  2[1  (| z0 |)]
Reject H 0 if z0  z / 2 or z0   z / 2

Hypotheses, upper-tailed alternative
 H 0 : 1   2   0
 H1 : 1  2   0
 P-value: P  1  ( z0 )
 Reject H 0 if z0  z

Hypotheses, lower-tailed alternative
 H 0 : 1   2   0
 H1 : 1  2   0


P-value: P  ( z0 )
Reject H 0 if z0   z

Type II error and choice of sample size
 Finding the probability of type II error 
 Hypotheses, two-sided alternative
 H 0 : 1   2   0
 H1 : 1  2   0
 Suppose the true value of the difference under H1
is 1  2  
 Test statistic:

z0 
X1  X 2  0

2
1
n1


2
2
n2

X1  X 2  

2
1
n1


2
2
n2

  0
 12
n1

 22
n2

Type II error and choice of sample size
 Finding the probability of type II error 
 Hypotheses, two-sided alternative
 Under H1




   0

z0  N 
,1
2
2
 1   2 
 n

n
1
2












  0 
  0 
   z / 2 
    z / 2 

2
2
2
2
1  2 
1  2 








n
n
n
n
1
2
1
2





Type II error and choice of sample size
 Sample size formulas
 If    0



  0
   z / 2 
2
2



1
2


n1 n2






  0 
  z / 2 

2
2
1  2 




n
n
1
2










  0 
    z / 2 

2
2
1  2 





n1 n2 



Type II error and choice of sample size
 Sample size formulas
 If    0
 Let z  be the 100 upper percentile of the
standard normal distribution. Then   ( z  )
 z   z / 2 
  0
 12
n1

 22
n2

Note










 (   0 ) 
 (   0 ) 
   z / 2 
    z / 2 

2
2
2
2
1  2 
1  2 








n
n
n
n
1
2
1
2














  0 
  0 
 1    z / 2 
  1   z / 2 

2
2
2
2
1  2 
1  2 








n
n
n
n
1
2 
1
2 












  0 
  0 
  z / 2 
    z / 2 

2
2
2
2
1  2 
1  2 







n1 n2 
n1 n2 



Sample size for a two-sided test on the difference in
mean with n1  n2 , variance known
n

( z / 2  z  ) 2 ( 12   22 )
(   0 ) 2
Sample size for a one-sided test on the difference in
the mean with n1  n2 , variance known
n
( z  z  ) 2 ( 12   22 )
(   0 ) 2

Operating characteristic (OC) curves
 Curves plotting  against a parameter d for
various sample size n  n1  n2
|   0 |
d
 12  22

n1 n2




See Appendix VII
For a given n and d , find  .
For a given  and d , find n
If n1  n2 , use
 12   22
n 2
 1 / n1   22 / n2

Confidence interval on the difference in means,
variances known
Z
X 1  X 2  ( 1   2 )
 12
n1


 22
n2
has a standard normal distribution




X 1  X 2  ( 1   2 )


P  z / 2 
 z / 2   1  
2
2
1  2





n1 n2



 12  22
 12  22 

P  X 1  X 2  z / 2

 1   2  X 1  X 2  z / 2

  1
n1 n2
n1 n2 




Confidence interval on the difference in
means, variances known
X 1  X 2  z / 2
 12
n1

 22
n2
 1   2  X 1  X 2  z / 2
 12
n1

 22
n2
Contents, figures, and exercises come from the textbook: Applied Statistics and Probability
for Engineers, 5th Edition, by Douglas C. Montgomery, John Wiley & Sons, Inc., 2011.


Choice of sample size
From  z / 2 
X 1  X 2  ( 1   2 )

2
1
n1


2
2
 z / 2
n2
| X 1  X 2  ( 1   2 ) | z / 2





 12
n

 22
n
2
we have
 z / 2 
2
2
n
 ( 1   2 )
 E 
if n  n1  n2 .
The sample size n required so that the error in
estimating 1  2 by x1  x2 will be less than E at
100(1   )% confidence.
Contents, figures, and exercises come from the textbook: Applied Statistics and Probability
for Engineers, 5th Edition, by Douglas C. Montgomery, John Wiley & Sons, Inc., 2011.

One-sided confidence bounds on the
differences in means, variance known
◦ A 100(1   )% upper-confidence bound for 
is
 12  22
1   2  X 1  X 2  z
n1

n2
◦ A 100(1   )% lower-confidence bound for 
is
X 1  X 2  z
 12
n1

 22
n2
 1   2
Contents, figures, and exercises come from the textbook: Applied Statistics and Probability
for Engineers, 5th Edition, by Douglas C. Montgomery, John Wiley & Sons, Inc., 2011.


Example 10-1 Paint Drying Time
   8 ,   0.05 , n1  n2  10 , x1  121 , x2  112
 What conclusions can the product developer
draw about the effectiveness of the new
ingredient?
Example 10-2 Paint Drying Time, Sample Size from
OC Curves
 If the true difference in mean drying times is as
much as 10 minutes, find the sample sizes
required to detect this difference with
probability at least 0.90.
 0  0
,   10 , d 
|   0 |
 
2
1
2
2

10
8 8
2
2
 0.88

Example 10-3 Paint Drying Time Sample Size
 If the true difference in mean drying times is as
much as 10 minutes, find the sample sizes
required to detect this difference with
probability at least 0.90.
1    0.90

Exercise 10-9
 The concentration of active ingredient in a liquid
laundry detergent is thought to be affected by
the type of catalyst used in the process. The
standard deviation of active concentration is
known to be 3 grams per liter, regardless of the
catalyst type. Ten observations on concentration
are taken with each catalyst, and the date follow:
 Catalyst 1: 57.9, …
 Catalyst 2: 66.4, …
 (a) Find a 95% confidence interval on the
difference in mean active concentrations for the
two catalysts. Find the P-value.

Exercise 10-9
 (b) Is there any evidence to indicate that the
mean active concentrations depend on the
choice of catalyst? Base your answer on the
results on part (a).
 (c) Suppose that the true mean difference in
active concentration is 5 grams per liter. What is
the power of the test to detect this difference if
   0.05 ?
 (d) If this difference of 5 grams per liter is really
important, do you consider the sample sizes
used by the experimenter to be adequate? Does
the assumption of normality seem reasonable
for both samples?


Exercise 10-9
Catalyst 1, normal probability plot


Exercise 10-9
Catalyst 2, normal probability plot
Inference on the Difference in
Means of Two Normal Distributions,
Variances Unknown
E ( X 1  X 2 )  E ( X 1 )  E ( X 2 )  1  2
V ( X1  X 2 )  V ( X1 )  V ( X 2 ) 
 12
n1

 22
n2
Hypothesis tests on the difference in means,
variances unknown
2
2
2
 Case 1:  1   2  
2
2
S

 The pooled estimator of
, denoted by p

2
2
(
n

1
)
S

(
n

1
)
S
1
2
2
S p2  1
n1  n2  2
We know that
X 1  X 2  ( 1   2 )
Z0 
1 1


n1 n2
 has a N (0,1) distribution. Then

T

X 1  X 2  ( 1   2 )
1 1
Sp

n1 n2
has a t distribution with n1  n2  2 degrees of
freedom



Hypothesis tests on the difference in means,
variances unknown
Case 1:  12   22   2
Hypotheses, two-sided alternative
X1  X 2  0
 Test statistic:
T0 
1 1

Sp

n1 n2
H 0 : 1   2   0
H1 : 1  2   0

P-value: P  2 P(Tn  n 2 | t0 |)
Reject H 0 if t0  t / 2,n  n  2 or t0  t / 2,n  n
1

2
1
2
1
2 2

Hypotheses, upper-tailed alternative
 H 0 : 1   2   0
 H1 : 1  2   0
 P-value: P  P (Tn  n  2  t0 )
 Reject H 0 if t0  t , n  n  2
Hypotheses, lower-tailed alternative
 H 0 : 1   2   0
 H1 : 1  2   0
1
2
1



2
P-value: P  P(Tn  n  2  t0 )
Reject H 0 if t0  t ,n  n  2
1
2
1
2
Hypothesis tests on the difference in means,
variances unknown
2
2
 Case 2:  1   2
 If H 0 : 1   2   0 is true, the statistic
X  X 2  0
T0*  1
S12 S 22

n1 n2


is distributed approximately as t with degrees of
freedom given
2
s
s 
  
n1 n2 

v 2
( s1 / n1 ) 2 ( s22 / n2 ) 2

n1  1
n2  1
2
1
2
2

Type II error and choice of sample size
 Finding the probability of type II error 
 Case 1:  12   22   2
 Hypotheses, two-sided alternative
 H 0 : 1   2   0
 H1 : 1  2     0

Test statistic:

t0 
  0
X1  X 2  

1 1
1 1




n1 n2
n1 n2
(n1  n2  2) S p2
1
2
(n1  n2  2)
 Under H1 , t 0 is of the noncentral t distribution
with n1  n2  2 degrees of freedom and
 . 0
noncentrality parameter
 1 / n1  1 / n2

Type II error and choice of sample size
 Finding the probability of type II error 
 Case 1:  12   22   2
 Hypotheses, two-sided alternative
 H 0 : 1   2   0
 H1 : 1  2     0
  P{t / 2,n  n 2  T0  t / 2,n  n 2 | H1}
1
2
1
2
 P{t / 2,n1  n2 2  T0 '  t / 2,n1  n2 2 }

where T0 ' denotes the noncentral t random
variable with n1  n2  2 degrees of freedom and
noncentrality
  0
 1 / n1  1 / n2

Type II error and choice of sample size
 Finding the probability of type II error 
 Case 1:  12   22   2
 Hypotheses, two-sided alternative
 H 0 : 1   2   0
 H1 : 1  2     0





Operating characteristic (OC) curves
Curves plotting  against a parameter d for
various sample size n*
|    0 | n*  2n  1
n1  n2  n
d
2
,
,
See Appendix VII
Note that d depends on the unknown
parameter 
Confidence interval on the difference in means,
variances unknown
2
2
2
 Case 1:  1   2  
2
2
S

 The pooled estimator of
, denoted by p

2
2
(
n

1
)
S

(
n

1
)
S
1
2
2
S p2  1
n1  n2  2


We know that
X 1  X 2  ( 1   2 )
T
1 1
Sp

n1 n2
has a t distribution with n1  n2  2 degrees of
freedom
P(t / 2,n1  n2  2  T  t / 2,n1  n2  2 )  1  

Confidence interval on the difference in means,
variances unknown and equal
◦ Case 1:  12   22   2
X 1  X 2  t / 2,n1  n2  2 S p
1 1

n1 n2
 1   2  X 1  X 2  t / 2,n1  n2  2 S p
1 1

n1 n2
Contents, figures, and exercises come from the textbook: Applied Statistics and Probability
for Engineers, 5th Edition, by Douglas C. Montgomery, John Wiley & Sons, Inc., 2011.
Confidence interval on the difference in means,
variances unknown and unequal
2
2
 Case 2:  1   2
 If H 0 : 1   2   0 is true, the statistic

T0*  ( X 1  X 2   0 ) / S12 / n1  S 22 / n2


is distributed approximately as t with degrees of
freedom given by
2
2 2
 s1 s2 
  
n1 n2 

v 2
( s1 / n1 ) 2 ( s22 / n2 ) 2

n1  1
n2  1
Confidence interval
X 1  X 2  t / 2,
S12 S 22
S12 S 22

 1   2  X 1  X 2  t / 2,

n1 n2
n1 n2


Example 10-5 Yield from a Catalyst
   0.05 , n1  n2  8 , x1  92.255 , x2  92.733 ,
 s1  2.39 , s2  2.98
 Is there any difference between the mean yields?
Example 10-6 Arsenic in Drinking Water
   0.05, n1  n2  10, x1  12.5 , x2  27.5 ,
 s1  7.63 , s2  15.3
 Is there any difference in mean arsenic
concentrations? (It is unlikely that the population
variances are the same)

Example 10-7 Yield from a Catalyst Sample Size
 If catalyst 2 produces a mean yield that differs
from the mean yield of catalyst 1 by 4.0%, we
would like to reject the null hypothesis with
probability at least 0.85. What sample size is
required?
s p  2.70
d |  | / 2 | 4.0 | /[( 2)( 2.70)]  0.74

Example 10-8 Cement Hydration
   0.05 , n1  10 , n2  15
, x1  90.0
 x2  87.0 , s1  5.0 , s2  4.0
 Find the confidence interval for 1  2 .

Exercise 10-29
 The overall distance traveled by a golf ball is
tested by hitting the ball with Iron Byron, a
mechanical golfer with a swing that is said to
emulate the legendary champion, Byron Nelson.
Ten randomly selected balls of two different
brands are tested and the overall distance
measured. The data follow:
 Brand 1: 251, …
 Brand 2: 236, …
 (a) Is there evidence that overall distance is
approximately normally distributed? Is an
assumption of equal variances justified?

Exercise 10-29
 (b) Test the hypothesis that both brands of ball
have equal mean overall distance. Use   0.05 .
What is the P-value?
 (c) Construct a 95% two-sided CI on the mean
difference in overall distance between the two
brands of golf balls.
 (d) What is the power of the statistical test in
part (b) to detect a true difference in mean
overall distance of 4.5 m?
 (e) What sample size would be required to
detect a true difference in mean overall distance
of 2.75 m with power of approximately 0.75?

Exercise 10-29
 Normal probability plot
A Nonparametric Test for the
Difference in Two Means

Wilcoxon rank-sum test
 Appendix Table X ( w )
 Two samples, n1  n2
 Arrange all n1  n2 observations in ascending order
of magnitude and assign ranks to them. If two or
more observations are tied (identical), use the mean
of the ranks that would have been assigned if the
observations differed.
 W1 : the sum of the ranks in the smaller sample
 W2  (n1  n2 )(n1  n2  1) / 2  W1


Wilcoxon rank-sum test
 Appendix Table X ( w )
 Hypotheses, two-sided alternative
H 0 : 1  2
H1 : 1  2
 Reject H 0 if min( w1 , w2 )  w
 Hypotheses, upper-tailed alternative
 H 0 : 1   2
 H1 : 1  2
Reject H 0 if w2  w
 Hypotheses, lower-tailed alternative

 H 0 : 1   2
H1 : 1  2
 Reject H 0 if w1  w


Normal approximation for Wilcoxon rank-sum test
statistic ( n1  8 and n2  8 )
n1 (n1  n2  1)
1

2
n n (n  n  1)
 W21  1 2 1 2
12
W1  W1
Z0 

W 
W
1
Reject H 0 if | z0 | z / 2 for H1 : 1  2

or if z0  z for H1 : 1  2

or if z0   z for H :   
1
1
2


Example 10-9 Axial Stress
 n  n  10 ,   0.05
1
2
 We wish to test the hypothesis that the means of
the two stress distributions are identical.

Exercise 10-33
 The manufacturer of a hot tub is interested in
testing two different heating elements for his
product. The element that produces the maximum
heat gain after 15 minutes would be preferable. He
obtains 10 samples of each heating unit and tests
each one. The heat gain after 15 minutes (in ℃ (K))
follows.
 Unit 1: 25, …
 Unit 2: 31, …
 (a) Is there any raeson to suspect that one unit is
superior to the other? Use   0.05 and the
Wilcoxon rank-sum test.
 (b) Use the normal approximation for the Wilcoxon
rank-sum test. Assume that   0.05 . What is the
approximate P-value for the test statistic?
Paired t-Test


Define the differences between each pair of
observations as D j  X 1 j  X 2 j , j  1,2,..., n . The D j‘s
are assumed to be normally distributed with mean
D  E( X1  X 2 )  E( X1 )  E( X 2 )  1  2
and variance  D2
 Test statistic: T  D   0
0
SD / n
H 0 : D  0
H1 :  D   0


P-value: P  2P(Tn1 | t0 |)
Reject H 0 if t0  t / 2,n 1 or t0  t / 2,n 1


Hypotheses, upper-tailed alternative
 H 0 : D  0
 H1 :  D   0
 P-value: P  P(T  t )
n 1
0
 Reject H 0 if t 0  t , n 1
Hypotheses, lower-tailed alternative
 H 0 : D  0
 H1 :  D   0


P-value: P  P(Tn1  t0 )
Reject H 0 if t0  t ,n 1

Paired versus unpaired comparisons
 The two-sample t -statistic is
X1  X 2  0
1 1
Sp

n n
 which would be compared to t 2 n  2
T0 

The paired t -statistic is
D  0
T0 
SD / n

which is compared to
t n 1
D  X1  X 2
V ( D )  V ( X 1  X 2   0 )  V ( X 1 )  V ( X 2 )  2 cov( X 1 , X 2 )
2 2 (1   ) 2 2
1
2 1


 Sp(  )
n
n
n n

Guidelines
 If the experimental units are relatively homogeneous
(small  ) and the correlation within pairs is small,
the gain in precision attributable to pairing will be
offset by the loss of degrees of freedom, so an
independent-sample experiment should be used.
 If the experimental units are relatively
hererogeneous (large  ) and there is large
positive correlation within pairs, the paired
experiment should be used. Typically, this case
occurs when the experimental units are the same
for both treatments.

Confidence interval for  D
D  D
T
SD / n
 has a t distribution with n  1 degrees of freedom


D  D
P  t / 2,n 1 
 t / 2,n 1   1  
SD / n



Confidence interval on the difference in means
d  t / 2,n1sD / n   D  d  t / 2,n1sD / n

Example 10-10 Shear Strength of Steel Girders
◦ n  9 ,   0.05, d  0.2769 , sd  0.1350
◦ We wish to determine whether there is any difference (on the
average) between the two methods.

Example 10-11 Parallel Park Cars
◦ n  14 ,  0.05 , d  1.21 , sd  12.68
◦ The 90% confidence interval for D  1  2 is
d  t0.05,13sD / n  D d  t0.05,13sD / n
Contents, figures, and exercises come from the textbook: Applied Statistics and Probability
for Engineers, 5th Edition, by Douglas C. Montgomery, John Wiley & Sons, Inc., 2011.

Exercise 10-45
◦ An article in Neurology (1988,Vol. 50, pp. 1246-1252) discussed
that monozygotic twins share numerous physical, psychological,
and pathological traits. The investigators measured an intelligence
score of 10 pairs of twins, and the data are follows: ….
◦ (a) Is the assumption that the difference in score is normally
distributed reasonable? Show results to support your answer.
◦ (b) Find a 95% confidence interval on the difference in mean
score. Is there any evidence that mean score depends on birth
order?
◦ (c) It is important to detect a mean difference in score of one
point, with a probability of at least 0.90. Was the use of 10 pairs a
adequate sample size? If not, how many pairs should have been
used?
Contents, figures, and exercises come from the textbook: Applied Statistics and Probability
for Engineers, 5th Edition, by Douglas C. Montgomery, John Wiley & Sons, Inc., 2011.
Inference on the Variances of Two
Normal Distributions
 F

distribution
Let W and Y be independent chi-square
random variables with u and v degrees of
freedom, respectively. Then the ratio
W /u
F


Y /v
has the probability density function
u  v u u / 2 (u / 2) 1
(
)( ) x
2
v
f ( x) 
, 0 x
(u v ) / 2
u
v  u

( )( ) ( ) x  1
2
2  v

and is said to follow the F distribution with u
degrees of freedom in the numerator and v
degrees of freedom in the denominator

PDF of the F distribution
From Wikipedia, http://www.wikipedia.org.

CDF of the F distribution
From Wikipedia, http://www.wikipedia.org.
 F
distribution
  v /( v  2)
2v 2 (u  v  2)
 
u (v  2) 2 (v  4)
2

P ( F  f  ,u , v ) 
 f ( x)dx  
f . u . v
f1 ,u ,v 
1
f  ,u , v

(n  1) S
F
(n  1)S
1
2

2
1
2
2


/  12 /( n1  1) S12 /  12
 2 2
2
/  2 /( n2  1) S 2 /  2
Test on the ration of variances from two normal
distributions
 Test statistic:
S12
F 2
S2

Hypotheses, two-sided alternative
H 0 :  12   22
H1 :  12   22
 P-value: P  2 min( P ( Fn1 1, n2 1  f 0 ), P ( Fn1 1, n2 1  f 0 ))
 Reject H 0 if f 0  f / 2,n1 1,n2 1 or f 0  f1 / 2, n1 1, n2 1

Hypotheses, upper-tailed alternative
2
2
 H 0 : 1   2
2
2
H
:




1
1
2
P-value: P  P( Fn 1,n 1  f 0 )
 Reject H 0 if f 0  f / 2,n 1,n 1
Hypotheses, lower-tailed alternative
2
2
 H 0 : 1   2
2
2
 H1 :  1   2
 P-value:
P  P( Fn 1,n 1  f 0 )
 Reject H 0 if f 0  f1 ,n 1,n 1

1
2
1

1
2
2
1
2
f 0  f1 / 2,n1 1,n2 1

Type II error and choice of sample size
 Finding the probability of type II error 
 Hypotheses, two-sided alternative

H 0 :  12   22

H1 :  12   22
(n1  1) S12  12
n1  1
 12
 f / 2,n1 1,n2 1 | H1}
  P{ f1 / 2,n1 1,n2 1 
2
2
(n2  1) S 2  2
n2  1
 22
 22
 22
 P{
n2  1

2
1
n1  1
f1 / 2,n1 1,n2 1  Fn1 1,n2 1 
n2  1

2
1
n1  1
f / 2,n1 1,n2 1}

Type II error and choice of sample size
 Finding the probability of type II error 
 Hypotheses, two-sided alternative

H 0 :  12   22
H1 :  12   22
 Operating characteristic (OC) curves
1
 Curves plotting  against a parameter  
2
n

n

n
for various sample size 1 2


See Appendix VII

Confidence interval on the ratio of two variances
S 22 /  22
F 2 2
S1 /  1
 has a F distribution with n2  1 and n1  1 degrees
of freedom


P f1 / 2,n2 1,n1 1  F  f / 2,n2 1,n1 1  1  

Confidence interval on the ratio of variances from
two normal distributions
s12
 12 s12
f
 2  2 f / 2,n2 1,n1 1
2 1 / 2, n2 1, n1 1
s2
 2 s2

Example 10-12 Semiconductor Etch Variability
◦ n  16 , s1  1.96 , s2  2.13
◦ Is there any evidence to indicate that either gas is preferable?
Use a fixed-level test with   0.05 .

Example 10-13 Semiconductor Etch Variability Sample
Size
◦ Suppose that one gas resulted in a standard deviation of oxide
thickness that is half the standard deviation of oxide thickness of
the other gas. If we wish to detect such a situation with
probability at least 0.80, is the sample size n1  n2  20
adequate?
1

2
2
Contents, figures, and exercises come from the textbook: Applied Statistics and Probability
for Engineers, 5th Edition, by Douglas C. Montgomery, John Wiley & Sons, Inc., 2011.

Example 10-14 Surface Finish for Titanium Alloy
◦ n1  11 , s1  0.13 , n2  16 , s2  0.12
◦ Find a 90% confidence interval on the ratio of the two standard
deviations, 1 /  2 .
s12
 12 s12
f
 2  2 f 0.05,15,10
2 0.95,15,10
s2
 2 s2
Contents, figures, and exercises come from the textbook: Applied Statistics and Probability
for Engineers, 5th Edition, by Douglas C. Montgomery, John Wiley & Sons, Inc., 2011.

Exercise 10-45
◦ An article in Neurology (1988,Vol. 50, pp. 1246-1252) discussed
that monozygotic twins share numerous physical, psychological,
and pathological traits. The investigators measured an intelligence
score of 10 pairs of twins, and the data are follows: ….
◦ (a) Is the assumption that the difference in score is normally
distributed reasonable? Show results to support your answer.
◦ (b) Find a 95% confidence interval on the difference in mean
score. Is there any evidence that mean score depends on birth
order?
◦ (c) It is important to detect a mean difference in score of one
point, with a probability of at least 0.90. Was the use of 10 pairs a
adequate sample size? If not, how many pairs should have been
used?
Contents, figures, and exercises come from the textbook: Applied Statistics and Probability
for Engineers, 5th Edition, by Douglas C. Montgomery, John Wiley & Sons, Inc., 2011.
Inference on Two Population
Proportions

Hypothesis
H 0 : p1  p2
H1 : p1  p2
 The statistic

Z

Pˆ1  Pˆ2  ( p1  p2 )
p1 (1  p1 ) p2 (1  p2 )

n1
n2
is distributed approximately as standard normal

If
p1  p2  p
Z


Pˆ1  Pˆ2
1 1
p (1  p )(  )
n1 n2
is distributed approximately N (0,1) .
A pooled estimator of the common parameter p is
X1  X 2
ˆ
P
n1  n2

The test statistic for H 0 : p1  p2 is then
Z
Pˆ1  Pˆ2
1 1
Pˆ (1  Pˆ )(  )
n1 n2

Approximate tests on the difference of two
population proportions
 Test statistic:
Z0 

Pˆ1  Pˆ2
1 1
Pˆ (1  Pˆ )(  )
n1 n2
Hypotheses, two-sided alternative
H 0 : p1  p2
H1 : p1  p2
 P-value: P  2(1  (| z0 |))
 Reject H 0 if
or
z0  z / 2
z0   z / 2


Hypotheses, upper-tailed alternative
 H 0 : p1  p2
 H1 : p1  p2
P  1  ( z0 )
 P-value:
 Reject H 0 if
z0  z
Hypotheses, lower-tailed alternative

H 0 : p1  p2
H1 : p1  p2
 P-value:
P  ( z0 )
 Reject H 0 if z0   z


Type II error and choice of sample size
 Finding the probability of type II error 
 Hypotheses, two-sided alternative
H 0 : p1  p2
H1 : p1  p2
  P{ z / 2 
Pˆ1  Pˆ2
 z / 2 | H1}
1 1
Pˆ (1  Pˆ )(  )
n1 n2
 P{ z / 2 pq (1 / n1  1 / n2 )  Pˆ1  Pˆ2  z / 2 pq (1 / n1  1 / n2 ) | H1}
n1 p1  n2 p2
 p
n1  n2
and
q
n1 (1  p1 )  n2 (1  p2 )
n1  n2
Type II error and choice of sample size
 Finding the probability of type II error 
 Hypotheses, two-sided alternative
 H 0 : p1  p2
 H1 : p1  p2
 z / 2 pq (1 / n1  1 / n2 )  ( p1  p2 )
  P{
 Pˆ  Pˆ

1

Pˆ1  Pˆ2  ( p1  p2 )
 Pˆ  Pˆ
1
2

2
z / 2 pq (1 / n1  1 / n2 )  ( p1  p2 )
 Pˆ  Pˆ
1
2
 z / 2 pq (1 / n1  1 / n2 )  ( p1  p2 ) 
 

 Pˆ1  Pˆ2


  z / 2 pq (1 / n1  1 / n2 )  ( p1  p2 ) 


 Pˆ1  Pˆ2


| H 1}

Type II error and choice of sample size
 Finding the probability of type II error 
 Hypotheses, upper-tailed alternative
H 0 : p1  p2
H1 : p1  p2
 z pq (1 / n1  1 / n2 )  ( p1  p2 ) 
  

 Pˆ1  Pˆ2


 Hypotheses, upper-tailed alternative

H 0 : p1  p2

H1 : p1  p2
  z pq (1 / n1  1 / n2 )  ( p1  p2 ) 
  1  

 Pˆ1  Pˆ2



Type II error and choice of sample size
 Finding the probability of type II error 
 Hypotheses, two-sided alternative
 H 0 : p1  p2
 H1 : p1  p2
 z 
z / 2 pq (1 / n1  1 / n2 )  ( p1  p2 )
 Pˆ  Pˆ
1
2
z ( p1q1  p2 q2 ) / n  z / 2 ( p1  p2 )( q1  q2 ) /( 2n)  p1  p2

z
n

( p1q1  p2 q2 )  z / 2 ( p1  p2 )( q1  q2 ) / 2
( p1  p2 ) 2

2

Confidence interval on the difference in population
proportions
Z
Pˆ1  Pˆ2  ( p1  p2 )
p1 (1  p1 ) p2 (1  p2 )

n1
n2
P z / 2  Z  z / 2   1  
pˆ 1  pˆ 2  z / 2
p1 (1  p1 ) p2 (1  p2 )

 p1  p2 
n1
n2
pˆ 1  pˆ 2  z / 2
p1 (1  p1 ) p2 (1  p2 )

n1
n2

Example 10-15 St. John’s Wort
◦ n1  n2  100 , p
ˆ1  27 / 100 , pˆ 2  19 / 100
◦ Is there any reason to believe that St. John’s Wort is effective in
treating major depression? Use   0.05 .

Example 10-13 Defective Bearings
ˆ1  0.12 , pˆ 2  8 / 85  0.09
◦ n1  n2  85 , p
◦ Find the approximate 95% confidence interval on the difference
in the proportion of defective bearings produced under the two
processes.
Contents, figures, and exercises come from the textbook: Applied Statistics and Probability
for Engineers, 5th Edition, by Douglas C. Montgomery, John Wiley & Sons, Inc., 2011.

Exercise 10-71
◦ Two different types of polishing solutions are being evaluated for
possible use in a tumble-polish operation for manufacturing
interocular lenses used in the human eye following cataract
surgery. Three hundred lenses were tumble polished using the
first polishing solution, and of this number 253 had no polishinginduced effects. Another 300 lenses were tumble-polishing using
the second polishing solution, and 196 lenses were satisfactory
upon completion.
◦ (a) Is there any reason to believe that the two polishing solutions
differ? Use   0.01 . What is the P-value for this test?
◦ (b) Discuss how this question could be answered with a
confidence interval on p1  p2 .
Contents, figures, and exercises come from the textbook: Applied Statistics and Probability
for Engineers, 5th Edition, by Douglas C. Montgomery, John Wiley & Sons, Inc., 2011.