1) The document describes methods for constructing confidence intervals for the difference between two population means. It provides formulas and examples for when the population variances are known, unknown but assumed equal, and unknown and not assumed equal. 2) It also describes how to construct a confidence interval for the mean difference between two dependent or matched pair samples. An example is given of estimating the difference in effectiveness of two drugs using a 99% confidence interval from data on cholesterol reductions in paired patients. 3) Key steps shown in examples include calculating sample means and variances, determining the test statistic and degrees of freedom, and reporting the confidence interval range.

1. Lecture 4

2. Confidence interval for difference between two
population mean (σ known)
Confidence interval for (µx –µy):
2
Population X Population Y
Population mean= µx Population mean= µy
Population variance = σ2
x Population variance = σ2
y
Sample size = nx Sample size = ny
Sample mean = Sample mean =
x y
2
2
/2 /2
( )
( ) ( )
y
x
x y
x y
x y z x y z
n n
 



     

Where, 1-α = confidence coefficient; α = level of significance
Z α/2= Z value providing an area of α/2 in the negative tail of the
standard normal prob. distribution
95% confidence level= confidence coefficient, 1-α =0.95i.e. α = 0.05.
/2 /2 1.96 1.96
z z z
  
    

3. Example: Independent random samples of accounting professors
and information systems (IS) professors were asked to provide the
number of hours they spend in preparation for each class. The
sample of 321 IS professors had a mean time of 3.01 preparation
hours and the sample of 94 accounting professors had a mean time
of 2.88 hours. From similar past studies the population standard
deviation for the IS professors and accounting professors are
assumed to be 1.09 and 1.01 respectively. Denoting µx the
population mean for IS professors and µy the population mean for
accounting professors, find 95% confidence interval for (µx –µy)
Solution:95% confidence interval= confidence coefficient, 1-α =0.95
i.e. α = 0.05. Hence
3
/2 /2 1.96 1.96
z z z
  
    
Confidence interval for (µx –µy):
2
2
/2 /2
( )
( ) ( )
y
x
x y
x y
x y z x y z
n n
 



     


4. 4
Here, =3.01 hours, =2.88 hours, nx = 321, ny = 94,
σx =1.09, σy =1.01
2
2
/2 ( )
2 2
)
( ) (3.01 2.88) 1.96
(1.09) (1.01)
0.13 1.96 0.106 ( 0.366
321 94
y
x
x y
x y
x y
x y z
n n
 


     
        

x y
Confidence interval for (µx –µy):

5. Confidence interval for difference between two
population mean (σ unknown but assumed to be
equal)
Confidence interval for (µx –µy):
5
Population X Population Y
Population mean= µx Population mean= µy
Popl. variance = σ2
x = ? Popl. variance = σ2
y = ?
Sample size = nx Sample size = ny
Sample mean = Sample mean =
x y
( 2) . ( 2) .
2 2
/2 /2
( )
( ) ( )
n n D F n n D F
x y x y
p p
x y
x y
s s
x y t x y t
n n
   
 

     

2
2 2
2
where, pooled sample variance, , is given by
( 1) ( 1)
2
p
x x y y
p
x y
s
n s n s
s
n n
  

 

6. Example: The residents of Orange city complain that traffic
speeding fines given in their city are higher than the traffic speeding
fines that are given in nearby Deland city. The assistant to the
county manager agreed to study the problem and to indicate if the
complaints were reasonable. Independent random samples of the
amounts paid by residents for speeding tickets in each of two cities
over the last three months were obtained. These amounts were,
Find a 95% confidence interval for the difference in the mean costs
of speeding tickets in these two cities.
6
Orange city 100 125 135 128 140 142 128 137 156 142
Deland 95 87 100 75 110 105 85 95

7. 7
Solution:95% confidence interval= confidence coefficient, 1-α
=0.95 i.e. α = 0.05.
Here, =133.3 =94, nx = 10, ny = 8,
=218.01, =129.43
x y
2
x
s 2
y
s
2
2 2
2
Pooled sample variance, , is given by
( 1) ( 1)
2
(10 1)(218.01) (8 1)(129.43)
179.256
10 8 2
p
x x y y
p
x y
s
n s n s
s
n n
  

 
  
 
 

8. Confidence interval for (µx –µy):
8
( 2) .
(16) .
/2 ( )
0.025
( )
179.256 179.256
(133.3 94)
10 8
39.3 2.12(6.35) 25.83 ( ) 52.76
n n D F
x y
D F
x y
x y
x y t
t
 
 
 
   
      


9. Confidence interval for difference between two
population mean (σ unknown: not assumed to be
equal)
Confidence interval for (µx –µy):
9
v . .
2
2
/2 /2
( )
( ) ( )
D F v D F
y
x
x y
x y
s
s
x y t x y t
n n
 

     

   
2
2
2
2
2 2
2
the degrees of freedom, v, is given by
/ 1 / 1
y
x
x y
y
x
x y
x y
s
s
n n
v
s
s
n n
n n
 

 
 
 

 
 
  
 
   
   

10. Example: Master’s Accounting firm conducted a random sample of
the accounts payable for the east and the west offices of
amalgamated distributors. From these two independent samples
they wanted to estimate the difference between the population mean
values of the payables. The sample statistics obtained were as
follows:
We do not assume that the unknown population variances are equal.
Estimate the difference between the mean values of the payables for
the two offices using a 95% confidence level.
10
East Office (X) West Office (Y)
Sample mean $290 $250
Sample size 16 11
Sample standard
deviation
15 50

11. Confidence interval for (µx –µy):
11
v . .
2
2
/2 /2
( )
( ) ( )
D F v D F
y
x
x y
x y
s
s
x y t x y t
n n
 

     

       
2 2
2
2 2 2
2 2 2
2 2 2
2
2
the degrees of freedom, v, is given by
15 50
16 11
11
15 50
/ 16 1 / 11 1
/ 1 / 1
16 11
y
x
x y
y
x
x y
x y
s
s
n n
v
s
s
n n
n n
   
 
   
 
   
  
     
 
  
  
     
       
   
95% confidence interval= confidence coefficient, 1-α =0.95 i.e. α = 0.05.
. 11 .
2
2 2 2
/2 0.025
15 50
( ) (290 250) 40 34.19
16 11
v D F D F
y
x
x y
s
s
x y t t
n n

        
5.81 ( ) 74.19
x y
    

12. Confidence interval for difference between two
population mean (matched pair or dependent
samples)
Confidence interval for µd :
12
(n-1) .
2
/2
( )
, where
D F
i
d
d
d
d d
s
d t s
n
n


 

Population X Population Y
Population mean= µx Population mean= µy
x1, x2, x3, ....... xn obs. y1, y2, y3, ....... yn obs.
here, d and =
i i i
x y
d x y
    

13. Example: A medical study was conducted to compare the
difference in effectiveness of two particular drugs in lowering
cholesterol levels. The research team used a paired sample approach
to control variation in reduction that might be due to factors other
than the drug itself. Each member of a pair was matched by age,
weight, lifestyle, and other pertinent factors. Drug X was given to
one person randomly selected in each pair, and drug Y was given to
the other individual in the pair. After a specified amount of time
each person’s cholesterol level was measured again.
Suppose that a random sample of eight pairs of patients with known
cholesterol problems is selected from the large populations of
participants. Table in the following slide gives the number of points
by which each person’s cholesterol level was reduced, as well as the
differences for each pair. Estimate 99% confidence interval for the
mean difference in the effectiveness of the two drugs, X and Y, to
lower cholesterol.
13

14. 14
Pair Drug(X) Drug (Y) Differences
(di =xi – yi)
1 29 26 3
2 32 27 5
3 31 28 3
4 32 27 5
5 32 30 2
6 29 26 3
7 31 33 -2
8 30 36 -6
(n-1) . 7 .
/2 0.005
3.777 3.777
1.625 1.625 3.499
8 8
D F D F
d
s
d t t
n

    
1.625 and s 3.777
d
d  
3.05 ( ) 6.30
x y
     