Chapter 8
Interval Estimation
Learning Objectives
1. Know how to construct and interpret an interval estimate of a population mean and a population
proportion.
2. Understand and be able to compute the margin of error.
3. Learn about the t distribution and its use in constructing an interval estimate for a population mean
when
is unknown.
4. Be able to determine the size of a simple random sample necessary to estimate a population mean and
a population proportion with a specified margin of error.
5. Know the definition of the following terms:
confidence interval margin of error
confidence coefficient degrees of freedom
confidence level
Solutions:
1. a.
/ 5/ 40 .79
xn

= = =
2. a. 32 1.645
( / )6 50
32 1.4 or 30.6 to 33.4
b. 32 1.96
( / )6 50
( / )6 50
3. a. 80 1.96
( / )15 60
80 3.8 or 76.2 to 83.8
b. 80 1.96
( / )15 120
4. Sample mean
Margin of Error = 160 – 156 = 4
1.96( / ) 4n
=
/2 .005 2.576zz
Margin of Error =
2.576 / 2.576(6 / 64) 1.93n
==
.025 ( / )x z n
Using Excel and the webfile TravelTax, the sample mean is
40.31x=
and the sample size is n =
400. The sample standard deviation
8.5
=
is known. The confidence interval is
.025 1.96z=
$600
=
50n=
Margin of error = 1.96(600/
50
) = 166.31
A larger sample size would be needed to reduce the margin of error. Section 8.3 can be used to show
that the sample size would need to be increased to n = 62.
b. Margin of error:
.025 ( / ) 1.96(5.5 / 10) 3.41zn
==
33.5 1.96
(9 / 40)
33.5 2.8 or 30.7 to 36.3 hours
3486 1.645
(650 / 120)
3486 98 or $3388 to $3584
b. 3486 1.96
(650 / 120)
3486 116 or $3370 to $3602
(650 / 120)
c. .05
d. .01
e. 1 – 2(.025) = .95
c. 2.457
d. Use .05 column, -1.708 and 1.708
80 10
i
x
19.5 ± 1.08 or 18.42 to 20.58
b.
.025 ( / )x t s n
df = 39 t.025 = 2.023
9.7063 ± 2.023
(7.9805 / 40)
9.7063 ± 2.5527 or 7.1536 to 12.2590
d.
.025 ( / )x t s n
df = 39 t.025 = 2.023
3.8854 ± 2.023
(1.6194 / 40)
3.8854 ± .5180 or 3.3674 to 4.4034
The 95% confidence interval for the population mean yield is 3.3674 to 4.4034 percent.
Maximum
10
Sum
317
Count
50
Confidence Level(95.0%)
0.6147
b.
.025 ( / )x t s n
c. At 95% confidence, the population mean is between $253 and $293. This is definitely above the
$229 level of 2 years ago. Hotel room rates are increasing.
The point estimate of the increase is $273 – $229 = $44 or 19%.
20. a.
/ 51,020/ 20 2551
i
x x n= = =
b.
22
( ) ( 2551) 301.3077
1 20 1
ii
x x x
sn
− −
= = =
−−
95% confidence interval:
x
t.025
( / )s n
df = 19
2551 2.093
(301.3077 / 20)
$2551 141.01 or $2409.99 to $2692.01
1 20 1
−−
t.025 = 2.093 df = 19
95% confidence interval:
x
t.025
( / )s n
71 2.093
(22.3513 / 20)
Median
10750
Mode
#N/A
Standard Deviation
3962.11
Sample Variance
15698350
Kurtosis
1.6567
Skewness
1.0147
Range
17200
Minimum
4200
Maximum
21400
Sum
272625
Count
25
Confidence Level(95.0%)
1635.48
and $12,540.48.
b. Mean number of customers per theater = 10,905/7.16 = 1523
c. Total number of customers = 3118(1523) = 4,748,714
23.
22 22
.025
22
(1.96) (40) 61.47 Use 62
10
z
nn
E
= = = =
If the normality assumption for the population appears questionable, this should be adjusted upward
to at least 30.
2
(.07)
EIA may choose to increase the sample size to a round number of 50
c.
22
2
(1.96) (.25) 96.04
(.05)
n==
Use 97
For reporting purposes, the US EIA might decide to round up to a sample size of 100.
45,000 30,000 3750
−
2
(200)
c.
22
2
(1.96) (3750) 5402.25 Use 5403
(100)
nn= = =
d. Sampling 5403 college graduates to obtain the $100 margin of error would be viewed as too
expensive and too much effort by most researchers.
22 22
/2
(1.645) (25) 187.92 Use 188
z
d. The sample size gets larger as the confidence level is increased. We would not recommend 99%
confidence. The sample size must be increased by 79 respondents (267 – 188) to go from 90% to
95%. This may be reasonable. However, increasing the sample size by 194 respondents (461 – 267)
to go from 95% to 99% would probably be viewed as too expensive and time consuming for the 4%
gain in confidence level.