2. Confidence intervals for the difference between
two population proportions
Confidence interval for (Px –Py):
( x y)
/2
ˆ
ˆ
ˆ ˆ
( )
y y
x x
x y
p q
p q
p p z
n n
x y
n n
Example: During a presidential election year many forecasts are
made to determine how voters perceive a particular candidate. In a
made to determine how voters perceive a particular candidate. In a
random sample of 120 registered voters in sector A, 107 indicated
that they supported the candidate in question. In an independent
random sample of 141 registered voters in sector B, only 73
indicated support for the same candidate. If the respective
population proportions are denoted P and P find a 95% confidence
population proportions are denoted PA and PB find a 95% confidence
interval for the population difference, (PA - PB).
2
3. Solution: Here, nA = 120; PA =107/120=0.892; qA =0.108
, A ; A ; qA
nB = 141; PA =73/141=0.518; qB =0.482
B ; A ; qB
95% confidence level= confidence coefficient, 1-α =0.95i.e. α = 0.05.
/2 /2 1.96 1.96
z z z
Th i d 95% fid l l
The required 95% confidence level,
(0.892)(0.108) (0.518)(0.482)
(0 892 0 518) z
/2
(0.892 0.518)
120 141
0.374 1.96 0.0507
z
x
0.374 0.0994
0 274 0 473
P P
3
0.274 0.473
A B
P P
4. Sampling Distributions of Sample Variances
p g p
• Sometimes the primary parameter of interest is not the
pop lation mean b t rather the pop lation ariance 2
population mean µ but rather the population variance σ2.
• We choose a random sample of size n from a normal
distribution The sample variance s2 can be used in its
distribution. The sample variance s2 can be used in its
standardized form:
2
( )
n
x x
2
2 1
2 2
( )
( 1) i
i
x x
n s
2
This is called (chi-square) distribution
with (n-1) degrees of freedom.
4
5. Example: George Samson is responsible for quality assurance at
Integrated Electronics. He has asked you to establish a quality
i i f h f f l d i A Th
monitoring process for the manufacture of control device A. The
variability of the electrical resistance, measured in ohms, is critical
for this device Manufacturing Standards specify a standard
for this device. Manufacturing Standards specify a standard
deviation of 3.6 and the population distribution of resistance
measures is normal. The monitoring process requires that a random
g p q
sample of n=6 observations be obtained from the population of
devices and the sample variance be computed. Determine an upper
li i f h l i h h h b bili f di
limit for the sample variance such that the probability of exceeding
this limit, given a population standard deviation of 3.6, is less than
0 05
0.05.
5
6. Solution: Here, n = 6; σ2 = 3.62 =12.96
, ;
2
(
2
) 0.05
P s k
2
2
( 1) . . 2
( 1)
, [ ] 0.05
n D F
n s
or P
2
5 . .
5
, [ ] 0.05
12 96
D F
xk
or P
2
0 05) 5
12.96
5
, 11.07
D F
xk
or
0.05),5 . .
, 11.07
12.96
, 28.69
D F
or
or k
,
If the sample variance, S2, from a random sample of size n=6
exceeds 28.69, there is strong evidence to suspect that the
l ti i d 12 96 d th f t i
6
population variance exceeds 12.96 and the manufacturing process
should be halted and appropriate adjustments performed.
7. Confidence intervals for variance
2 2
2 2
2
2 2
( 1) . ., ( 1) . ., (1-
( 1) ( 1)
n D F n D F
n s n s
( ) , ( ) , (
Example: The manager of Northern Steel, Inc. wants to assess the
temperature variation in the firm’s new electric furnace. A random
l f 25 t t 1 k i d i bt i d d th
sample of 25 temperatures over a 1-week period is obtained, and the
sample variance is found to be s2 =100. Find a 95% confidence
interval for the population variance temperature
interval for the population variance temperature.
Solution: Here, n = 25; s2 = 100 =12.96. 95% confidence
interval= confidence coefficient, 1-α =0.95 i.e. α = 0.05.
e v co de ce coe c e , α .9 .e. α . .
2 2
2
2 2
( 1) . ., ( 1) . ., (1-
( 1) ( 1)
n D F n D F
n s n s
( 1) . ., ( 1) . ., (1
2 2
2 2
24 100 24 100 2400 2400
39 364 12 401
n D F n D F
x x
7
24 . ., 24 . ., 0.975
2
39.364 12.401
60.97 193.53
D F D F
8. Sample size for confidence interval estimation of
population mean
2 2
/2
( )
; here ME Margin of error or sampling error
z
/2
2
( )
; where ME= Margin of error or sampling error
n
ME
Example: The lengths of metal rods produced by an industrial
p g p y
process are normally distributed with a standard deviation of 1.8
millimeters. Based on a random sample of nine observations from
this population, the 99% confidence interval, 194.65<µ<197.75 was
found for the population mean length.
S th t d ti b li th t th i t l i t
Suppose that a production manager believes that the interval is too
wide for practical use and instead requires a 99% confidence
interval extending no further than 0 50 more on each side of the
interval extending no further than 0.50 more on each side of the
mean. How large a sample is needed to achieve such an interval?
8
9. Solution: Here, ME= 0.50; σ2 = 1.82 =3.24.
2 575 2 575
z z z
2 2
/2
( )
h ME M i f li
z
/2 /2 2.575 2.575
z z z
/2
2
2
( )
; where ME= Margin of error or sampling error
(2 575) 3 24
n
ME
x
2
(2.575) 3.24
86
0.50
x
Th f t ti f th ’ i t l f t l t
Therefore, to satisfy the manager’s requirement, a sample of at least
86 observations is needed.
9
10. Sample size for confidence interval estimation of
population proportion
2
/2
0.25( )
; here ME Margin of error or sampling error
z/2
2
( )
; where ME= Margin of error or sampling error
n
ME
Example: A 95% confidence interval for the proportion of graduate
p p p g
admissions personnel, who viewed scores on standardized exams as
very important in the consideration of a candidate, is
0.533<P<0.693.
Suppose, instead, that it must be ensured that a 95% confidence
i t l f th l ti ti t d f th th 0 06
interval for the population proportion extends no further than 0.06
on each side of the sample proportion. How large a sample must be
taken?
taken?
10
11. Solution: Here, ME= 0.06;
1 96 1 96
z z z
2
/2
0.25( )
h ME M i f li
z
/2 /2 1.96 1.96
z z z
/2
2
2
( )
; where ME= Margin of error or sampling error
0 25(1 96)
n
ME
2
0.25(1.96)
266.78
0.06
Th f t hi thi fid i t l l f
Therefore, to achieve this narrower confidence interval, a sample of
at least 267 observations is needed.
11
