6-4 Large-Sample Confidence Intervals
for the Population Proportion, p
The estimator of the population proportion, p , is the sample proportion, p . If the
sample size is large, p has an approximately normal distribution, with E( p ) = p and
pq
V( p ) =
, where q = (1 - p). When the population proportion is unknown, use the
n
estimated value, p , to estimate the standard deviation of p .
For estimating p , a sample is considered large enough when both n  p an n  q are greater
than 5.
6-4 Large-Sample Confidence Intervals
for the Population Proportion, p
A large - sample (1 -  )100% confidence interval for the population proportion, p :
pˆ  z
α
2
pˆ qˆ
n
where the sample proportion, p̂, is equal to the number of successes in the sample, x,
divided by the number of trials (the sample size), n, and q̂ = 1 - p̂.
Large-Sample Confidence Interval for the
Population Proportion, p (Example 6-4)
A marketing research firm wants to estimate the share that foreign companies
have in the American market for certain products. A random sample of 100
consumers is obtained, and it is found that 34 people in the sample are users
of foreign-made products; the rest are users of domestic products. Give a
95% confidence interval for the share of foreign products in this market.
p  z
2
pq
( 0.34 )( 0.66)

 0.34  1.96
n
100
 0.34  (1.96)( 0.04737 )
 0.34  0.0928
  0.2472 ,0.4328
Thus, the firm may be 95% confident that foreign manufacturers control
anywhere from 24.72% to 43.28% of the market.
Large-Sample Confidence Interval for the Population
Proportion, p (Example 6-4) – Using the Template
Reducing the Width of Confidence
Intervals - The Value of Information
The width of a confidence interval can be reduced only at the
price of:
• a lower level of confidence, or
• a larger sample.
Lower Level of Confidence
Larger Sample Size
Sample Size, n = 200
90% Confidence Interval
p  z 
2
pq

(0.34)(0.66)
 0.34  1645
.
n
100
 0.34  (1645
. )(0.04737)
 0.34  0.07792
 0.2621,0.4197
p  z
2
pq

(0.34)(0.66)
 0.34  196
.
n
200
 0.34  (196
. )(0.03350)
 0.34  0.0657
  0.2743,0.4057
6-5 Confidence Intervals for the Population
Variance: The Chi-Square (2) Distribution
• The sample variance, s2, is an unbiased estimator
•
of the population variance, 2.
Confidence intervals for the population variance
are based on the chi-square (2) distribution.
The chi-square distribution is the probability
distribution of the sum of several independent,
squared standard normal random variables.
The mean of the chi-square distribution is equal to the
degrees of freedom parameter, (E[2] = df). The
variance of a chi-square is equal to twice the number
of degrees of freedom, (V[2] = 2df).
The Chi-Square (2) Distribution

C hi-S q ua re D is trib utio n: d f= 1 0 , d f= 3 0 , d f= 5 0
0 .1 0
df = 10
0 .0 9
0 .0 8
0 .0 7
2

The chi-square random variable cannot
be negative, so it is bound by zero on
the left.
The chi-square distribution is skewed
to the right.
The chi-square distribution approaches
a normal as the degrees of freedom
increase.
f(  )

0 .0 6
df = 30
0 .0 5
0 .0 4
df = 50
0 .0 3
0 .0 2
0 .0 1
0 .0 0
0
50

2
In sampling from a normal population, the random variable:
 
2
( n  1) s 2
2
has a chi - square distribution with (n - 1) degrees of freedom.
100
Values and Probabilities of Chi-Square
Distributions
Area in Right Tail
.995
.990
.975
.950
.900
.100
.050
.025
.010
.005
.900
.950
.975
.990
.995
Area in Left Tail
df
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
.005
0.0000393
0.0100
0.0717
0.207
0.412
0.676
0.989
1.34
1.73
2.16
2.60
3.07
3.57
4.07
4.60
5.14
5.70
6.26
6.84
7.43
8.03
8.64
9.26
9.89
10.52
11.16
11.81
12.46
13.12
13.79
.010
.025
.050
.100
0.000157
0.0201
0.115
0.297
0.554
0.872
1.24
1.65
2.09
2.56
3.05
3.57
4.11
4.66
5.23
5.81
6.41
7.01
7.63
8.26
8.90
9.54
10.20
10.86
11.52
12.20
12.88
13.56
14.26
14.95
0.000982
0.0506
0.216
0.484
0.831
1.24
1.69
2.18
2.70
3.25
3.82
4.40
5.01
5.63
6.26
6.91
7.56
8.23
8.91
9.59
10.28
10.98
11.69
12.40
13.12
13.84
14.57
15.31
16.05
16.79
0.000393
0.103
0.352
0.711
1.15
1.64
2.17
2.73
3.33
3.94
4.57
5.23
5.89
6.57
7.26
7.96
8.67
9.39
10.12
10.85
11.59
12.34
13.09
13.85
14.61
15.38
16.15
16.93
17.71
18.49
0.0158
0.211
0.584
1.06
1.61
2.20
2.83
3.49
4.17
4.87
5.58
6.30
7.04
7.79
8.55
9.31
10.09
10.86
11.65
12.44
13.24
14.04
14.85
15.66
16.47
17.29
18.11
18.94
19.77
20.60
2.71
4.61
6.25
7.78
9.24
10.64
12.02
13.36
14.68
15.99
17.28
18.55
19.81
21.06
22.31
23.54
24.77
25.99
27.20
28.41
29.62
30.81
32.01
33.20
34.38
35.56
36.74
37.92
39.09
40.26
3.84
5.99
7.81
9.49
11.07
12.59
14.07
15.51
16.92
18.31
19.68
21.03
22.36
23.68
25.00
26.30
27.59
28.87
30.14
31.41
32.67
33.92
35.17
36.42
37.65
38.89
40.11
41.34
42.56
43.77
5.02
7.38
9.35
11.14
12.83
14.45
16.01
17.53
19.02
20.48
21.92
23.34
24.74
26.12
27.49
28.85
30.19
31.53
32.85
34.17
35.48
36.78
38.08
39.36
40.65
41.92
43.19
44.46
45.72
46.98
6.63
9.21
11.34
13.28
15.09
16.81
18.48
20.09
21.67
23.21
24.72
26.22
27.69
29.14
30.58
32.00
33.41
34.81
36.19
37.57
38.93
40.29
41.64
42.98
44.31
45.64
46.96
48.28
49.59
50.89
7.88
10.60
12.84
14.86
16.75
18.55
20.28
21.95
23.59
25.19
26.76
28.30
29.82
31.32
32.80
34.27
35.72
37.16
38.58
40.00
41.40
42.80
44.18
45.56
46.93
48.29
49.65
50.99
52.34
53.67
Template with Values and Probabilities
of Chi-Square Distributions
Confidence Interval for the Population
Variance
A (1-)100% confidence interval for the population variance * (where the population is
assumed normal) is:

2
2
 ( n  1) s , ( n  1) s 
  2
2  
1


2
2
2

where is the value of the chi-square distribution with n - 1 degrees of freedom that
2
2


cuts off an area
to its right and
is the value
of the distribution that cuts off an

area of
2
to its left (equivalently, an area of
2
1
2
to its right).
1

2
* Note: Because the chi-square distribution is skewed, the confidence interval for the population
variance is not symmetric
Confidence Interval for the Population
Variance - Example 6-5
In an automated process, a machine fills cans of coffee. If the average amount
filled is different from what it should be, the machine may be adjusted to
correct the mean. If the variance of the filling process is too high, however, the
machine is out of control and needs to be repaired. Therefore, from time to
time regular checks of the variance of the filling process are made. This is done
by randomly sampling filled cans, measuring their amounts, and computing the
sample variance. A random sample of 30 cans gives an estimate s2 = 18,540.
Give a 95% confidence interval for the population variance, 2.

2
2
 ( n  12 ) s , ( n 21) s    ( 30  1)18540 , ( 30  1)18540   11765,33604

457
.
16.0
   
 
1


2
2
Example 6-5 (continued)
Area in Right Tail
.
.
.
12.46
13.12
13.79
.990
.
.
.
13.56
14.26
14.95
.975
.950
.
.
.
15.31
16.05
16.79
.900
.
.
.
16.93
17.71
18.49
.
.
.
18.94
19.77
20.60
.100
.050
.
.
.
37.92
39.09
40.26
.
.
.
41.34
42.56
43.77
C hi-S quare Distribution: df = 29
0.06
0.05
0.95
0.04
2
.
.
.
28
29
30
.995
f( )
df
0.03
0.02
0.025
0.01
0.025
0.00
0
10
20
 20.975  16.05
30
40
2
50
60
 20.025  4572
.
70
.025
.
.
.
44.46
45.72
46.98
.010
.
.
.
48.28
49.59
50.89
.005
.
.
.
50.99
52.34
53.67
Example 6-5 Using the Template
Using Data
6-6 Sample-Size Determination
Before determining the necessary sample size, three questions must be
answered:
• How close do you want your sample estimate to be to the unknown
•
•
parameter? (What is the desired bound, B?)
What do you want the desired confidence level (1-) to be so that the
distance between your estimate and the parameter is less than or equal
to B?
What is your estimate of the variance (or standard deviation) of the
population in question?

n
}
For example: A (1-  ) Confidence Interval for : x  z 
2
Bound, B
Sample Size and Standard Error
The sample size determines the bound of a statistic, since the standard
error of a statistic shrinks as the sample size increases:
Sample size = 2n
Standard error
of statistic
Sample size = n
Standard error
of statistic

Minimum Sample Size: Mean and
Proportion
Minimum required sample size in estimating the population
mean, :
z2 2
n 2 2
B
Bound of estimate:
B = z
2

n
Minimum required sample size in estimating the population
proportion, p
z2 pq
n 2 2
B
Sample-Size Determination:
Example 6-6
A marketing research firm wants to conduct a survey to estimate the average
amount spent on entertainment by each person visiting a popular resort. The
people who plan the survey would like to determine the average amount spent by
all people visiting the resort to within $120, with 95% confidence. From past
operation of the resort, an estimate of the population standard deviation is
s = $400. What is the minimum required sample size?
z 
2
n
2
2
B
2
2
(1.96) ( 400)

120
2
 42.684  43
2
Sample-Size for Proportion:
Example 6-7
The manufacturers of a sports car want to estimate the proportion of people in a
given income bracket who are interested in the model. The company wants to
know the population proportion, p, to within 0.01 with 99% confidence. Current
company records indicate that the proportion p may be around 0.25. What is the
minimum required sample size for this survey?
n
z2 pq
2
B2
2.5762 (0.25)(0.75)

010
. 2
 124.42  125
6-7 The Templates – Optimizing
Population Mean Estimates
6-7 The Templates – Optimizing Population
Proportion Estimates