Chapter 8: Interval Estimation
•
Point Estimation:
is a point estimator of is a point estimator of p
•
Interval Estimation:
: ± Margin of Error
p: ± Margin of Error
known
± ± √
Confidence interval for :
± Margin of Error
unknown
± √
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8.1 Population mean: known
Example:
Given:
Consider marks that are normally distributed with 10.
Let = sample average = 58 with 16.
Question:
Calculate a 95% confidence interval for :
Answer:
1 – 0.95 = 0.05 = = level of significance
0.05
0.025
2
2
± √
58 ± 1.96 10
√16
58
58 ± 4.9
4.9,58 # 4.9
53.1,62.9
95% confident that is between 53.1 and 62.9.
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FOCUSING ON THE MARGIN OF ERROR:
± Margin of Error
± √
1.96 10
√16
= 4.9
95% of the time the sampling error will be 4.9 or less.
|
| |
|
OR
There is a 0.95 probability that the sample mean, , will provide a sampling error of 4.9 or less.
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Example:
Given:
Consider marks that are normally distributed with 10.
Let = sample average = 58 with 16.
Question:
Calculate a 90% confidence interval for :
Answer:
1 – 0.9 = 0.1 = = level of significance
0.1
0.05
2
2
± √
58 ± 1.645 10
√16
58
58 ± 4.1125
4.1125,58 # 4.1125
53.8875,62.1125
90% confident that is between 53.8875 and 62.1125.
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FOCUSING ON THE MARGIN OF ERROR:
± Margin of Error
± √
1.645 10
√16
= 4.1125
90% of the time the sampling error will be 4.1125 or less.
|
| |
|
OR
There is a 0.90 probability that the sample mean, , will provide a sampling error of 4.1125 or less.
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Example:
Given:
Consider marks that are normally distributed with 10.
Let = sample average = 58 with 16.
Question:
Calculate a 99% confidence interval for :
Answer:
1 – 0.99 = 0.01 = = level of significance
0.01
0.005
2
2
± √
58 ± 2.576 10
√16
58
58 ± 6.44
6.44,58 # 6.44
51.56,64.44
99% confident that is between 51.56 and 64.44.
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FOCUSING ON THE MARGIN OF ERROR:
± Margin of Error
± √
2.576 10
√16
= 6.44
99% of the time the sampling error will be 6.44 or less.
|
| |
|
OR
There is a 0.99 probability that the sample mean, , will provide a sampling error of 6.44 or less.
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Confidence Level
Confidence
coefficient
'
'
(
)'
Margin of Error
90%
0.90
0.10
0.05
1.645
1.645
*
95%
0.95
0.05
0.025
1.960
1.960
*
99%
0.99
0.01
0.005
2.576
2.576
*
(
Note: • = level of significance
• level of significance + confidence coefficient = 1
Graph:
8.2 Population mean: unknown
± √
______________________________________________________________________________
t – distribution:
•
•
•
Symmetric around 0.
Degrees of freedom (df) = n – 1.
Table in textbook.
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t Distribution
Area or
Probability
0
Degrees
of
freedom
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
31
32
33
34
t
Area in Upper Tail
0.20
1.376
1.061
0.978
0.941
0.920
0.906
0.896
0.889
0.883
0.879
0.876
0.873
0.870
0.868
0.866
0.865
0.863
0.862
0.861
0.860
0.859
0.858
0.858
0.857
0.856
0.856
0.855
0.855
0.854
0.854
0.853
0.853
0.853
0.852
0.10
3.078
1.886
1.638
1.533
1.476
1.440
1.415
1.397
1.383
1.372
1.363
1.356
1.350
1.345
1.341
1.337
1.333
1.330
1.328
1.325
1.323
1.321
1.319
1.318
1.316
1.315
1.314
1.313
1.311
1.310
1.309
1.309
1.308
1.307
0.05
6.314
2.920
2.353
2.132
2.015
1.943
1.895
1.860
1.833
1.812
1.796
1.782
1.771
1.761
1.753
1.746
1.740
1.734
1.729
1.725
1.721
1.717
1.714
1.711
1.708
1.706
1.703
1.701
1.699
1.697
1.696
1.694
1.692
1.691
0.025
12.706
4.303
3.182
2.776
2.571
2.447
2.365
2.306
2.262
2.228
2.201
2.179
2.160
2.145
2.131
2.120
2.110
2.101
2.093
2.086
2.080
2.074
2.069
2.064
2.060
2.056
2.052
2.048
2.045
2.042
2.040
2.037
2.035
2.032
0.01
31.821
6.965
4.541
3.747
3.365
3.143
2.998
2.896
2.821
2.764
2.718
2.681
2.650
2.624
2.602
2.583
2.567
2.552
2.539
2.528
2.518
2.508
2.500
2.492
2.485
2.479
2.473
2.467
2.462
2.457
2.453
2.449
2.445
2.441
0.005
63.657
9.925
5.841
4.604
4.032
3.707
3.499
3.355
3.250
3.169
3.106
3.055
3.012
2.977
2.947
2.921
2.898
2.878
2.861
2.845
2.831
2.819
2.807
2.797
2.787
2.779
2.771
2.763
2.756
2.750
2.744
2.738
2.733
2.728
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t distribution (Continued)
Degrees
of
freedom
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
Area in Upper Tail
0.20
0.852
0.852
0.851
0.851
0.851
0.851
0.850
0.850
0.850
0.850
0.850
0.850
0.849
0.849
0.849
0.849
0.849
0.849
0.848
0.848
0.848
0.848
0.848
0.848
0.848
0.848
0.848
0.847
0.847
0.847
0.847
0.847
0.847
0.847
0.847
0.847
0.847
0.847
0.847
0.847
0.846
0.846
0.846
0.846
0.846
0.10
1.306
1.306
1.305
1.304
1.304
1.303
1.303
1.302
1.302
1.301
1.301
1.300
1.300
1.299
1.299
1.299
1.298
1.298
1.298
1.297
1.297
1.297
1.297
1.296
1.296
1.296
1.296
1.295
1.295
1.295
1.295
1.295
1.294
1.294
1.294
1.294
1.294
1.293
1.293
1.293
1.293
1.293
1.293
1.292
1.292
0.05
1.690
1.688
1.687
1.686
1.685
1.684
1.683
1.682
1.681
1.680
1.679
1.679
1.678
1.677
1.677
1.676
1.675
1.675
1.674
1.674
1.673
1.673
1.672
1.672
1.671
1.671
1.670
1.670
1.669
1.669
1.669
1.668
1.668
1.668
1.667
1.667
1.667
1.666
1.666
1.666
1.665
1.665
1.665
1.665
1.664
0.025
2.030
2.028
2.026
2.024
2.023
2.021
2.020
2.018
2.017
2.015
2.014
2.013
2.012
2.011
2.010
2.009
2.008
2.007
2.006
2.005
2.004
2.003
2.002
2.002
2.001
2.000
2.000
1.999
1.998
1.998
1.997
1.997
1.996
1.995
1.995
1.994
1.994
1.993
1.993
1.993
1.992
1.992
1.991
1.991
1.990
0.01
2.438
2.434
2.431
2.429
2.426
2.423
2.421
2.418
2.416
2.414
2.412
2.410
2.408
2.407
2.405
2.403
2.402
2.400
2.399
2.397
2.396
2.395
2.394
2.392
2.391
2.390
2.389
2.388
2.387
2.386
2.385
2.384
2.383
2.382
2.382
2.381
2.380
2.379
2.379
2.378
2.377
2.376
2.376
2.375
2.374
0.005
2.724
2.719
2.715
2.712
2.708
2.704
2.701
2.698
2.695
2.692
2.690
2.687
2.685
2.682
2.680
2.678
2.676
2.674
2.672
2.670
2.668
2.667
2.665
2.663
2.662
2.660
2.659
2.657
2.656
2.655
2.654
2.652
2.651
2.650
2.649
2.648
2.647
2.646
2.645
2.644
2.643
2.642
2.641
2.640
2.639
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t distribution (Continued)
Degrees
of
freedom
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
∞
Area in Upper Tail
0.20
0.846
0.846
0.846
0.846
0.846
0.846
0.846
0.846
0.846
0.846
0.846
0.846
0.846
0.846
0.845
0.845
0.845
0.845
0.845
0.845
0.845
0.842
0.10
1.292
1.292
1.292
1.292
1.292
1.292
1.291
1.291
1.291
1.291
1.291
1.291
1.291
1.291
1.291
1.291
1.290
1.290
1.290
1.290
1.290
1.282
0.05
1.664
1.664
1.664
1.663
1.663
1.663
1.663
1.663
1.662
1.662
1.662
1.662
1.662
1.661
1.661
1.661
1.661
1.661
1.661
1.660
1.660
1.645
0.025
1.990
1.990
1.989
1.989
1.989
1.988
1.988
1.988
1.987
1.987
1.987
1.986
1.986
1.986
1.986
1.985
1.985
1.985
1.984
1.984
1.984
1.960
0.01
2.374
2.373
2.373
2.372
2.372
2.371
2.370
2.370
2.369
2.369
2.368
2.368
2.368
2.367
2.367
2.366
2.366
2.365
2.365
2.365
2.364
2.326
0.005
2.639
2.638
2.637
2.636
2.636
2.635
2.634
2.634
2.633
2.632
2.632
2.631
2.630
2.630
2.629
2.629
2.628
2.627
2.627
2.626
2.626
2.576
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Example:
Given: 15, 53.87 and 6.82. Take note: is unknown
Question:
Calculate a 95% confidence interval for :
1 – 0.95 = 0.05 = = level of significance
Answer:
0.05
0.025
2
2
df = n – 1 = 15 – 1 = 14
± √
53.87 ± 2.145 6.82
√15
53.87
53.87 ± 3.78
3.78,53.87 # 3.78
50.09,57.65
95% confident that is between 50.09 and 57.65.
Obtaining + using Excel:
,
= TINV(, df)
= TINV(0.05, 14)
= 2.144787
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Obtaining the level of significance using Excel:
= TDIST(+ , df, tails)
,
= TDIST(2.145, 14, 2)
= 0.04998 ≈ 0.05
FOCUSING ON THE MARGIN OF ERROR:
± Margin of Error
± √
2.145 6.82
√15
= 3.78
95% of the time the sampling error will be 3.78 or less.
|
| |
|
OR
There is a 0.95 probability that the sample mean, , will provide a sampling error of 3.78 or less.
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Example:
Given: 15, 53.87 and 6.82. Take note: is unknown
Question:
Calculate a 90% confidence interval for :
1 – 0.9 = 0.1 = = level of significance
Answer:
0.1
0.05
2
2
df = n – 1 = 15 – 1 = 14
± √
53.87 ± 1.761 6.82
√15
53.87
53.87 ± 3.1
3.1,53.87 # 3.1
50.77,56.97
90% confident that is between 50.77 and 56.97.
Obtaining + using Excel:
,
= TINV(, df)
= TINV(0.1, 14)
= 1.76131
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Obtaining the level of significance using Excel:
= TDIST(+ , df, tails)
,
= TDIST(1.761, 14, 2)
= 0.100054 ≈ 0.1
FOCUSING ON THE MARGIN OF ERROR:
± Margin of Error
± √
1.761 6.82
√15
= 3.1
90% of the time the sampling error will be 3.1 or less.
|
| |
|
OR
There is a 0.90 probability that the sample mean, , will provide a sampling error of 3.1 or less.
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8.4 Population proportion
p = population proportion
= sample proportion
Interval estimate of a population proportion:
± Margin of error
± .
± /
1
To use this expression to develop an interval estimate of a population proportion, p, the value of p
would have to be known. But, the value of p is what we are trying to estimate, so we simply
substitute the sample proportion for p. Therefore,
± /
1
Example: Women Golfers
Given:
A national survey of 902 women golfers was taken to learn how women golfers view themselves
as being treated at golf courses.
The survey found that 397 if the women golfers felt that they were treated fairly.
Question:
Calculate a 95% confidence interval for p.
Answer:
397
0.4401
902
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± /
0.4401 ± 1.96/
0.4401
1
0.44011 0.4401
902
0.4401 ± 0.0324
0.0324,0.4401 # 0.0324
0.4077,0.4725
95% confident that p is between 0.4077 and 0.4725.
FOCUSING ON THE MARGIN OF ERROR:
± Margin of error
± /
1.96/
1
0.44011 0.4401
902
= 0.0324
95% of the time the sampling error will be 0.0324 or less.
|
| |
|
OR
There is a 0.95 probability that the sample proportion, , will provide a sampling error of 0.0324 or
less.
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Question:
Calculate a 90% confidence interval for p.
Answer:
± /
1
0.44011 0.4401
0.4401 ± 1.645/
902
0.4401
0.4401 ± 0.027189
0.027189,0.4401 # 0.027189
0.4129,0.467
90% confident that p is between 0.4129 and 0.467.
FOCUSING ON THE MARGIN OF ERROR:
± Margin of error
± /
1
0.44011 0.4401
1.645/
902
= 0.027189
90% of the time the sampling error will be 0.027189 or less.
|
| |
|
OR
There is a 0.9 probability that the sample proportion, , will provide a sampling error of 0.027189 or
less.
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Relationships between the z and t – distributions
Characteristics of the t-distribution:
Symmetric around 0.
Has one degree of freedom, namely n – 1.
As the degrees of freedom increase, the t-distribution tends to the standard normal
distribution.
For “large sample cases” ( ≥ 30 the t-distribution approaches the standard normal
distribution.
In Figure 1 it is illustrated that as the degrees of freedom increase, i.e. as n – 1 increases, i.e. as the
sample size n increases, the t-distribution tends to the standard normal distribution.
Figure 1
•
•
•
The red curve represents the t-distribution with 4 degrees of freedom (denoted t(4));
the blue curve represents the t-distribution with 10 degrees of freedom (denoted t(10));
the black curve represents both the t-distribution with df tending to infinity (denoted t(∞))
and the standard normal distribution (denoted z-distribution).
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The importance of this? Take note of the following:
95% confidence interval
When working with a 95% confidence interval using the standard normal distribution we have:
The + value is obtained using the standard normal table.
,
When working with a 95% confidence interval using the t-distribution where the df tends to
infinity we have:
The + value is obtained using the t-table with area in the upper tail = 0.025 and df = ∞.
,
Note: The + and + values are the same, since the t-distribution tends to the standard normal
,
,
distribution as the df increase.
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90% confidence interval
When working with a 90% confidence interval using the standard normal distribution we have:
The + value is obtained using the standard normal table.
,
When working with a 90% confidence interval using the t-distribution where the df tends to
infinity we have:
The + value is obtained using the t-table with area in the upper tail =0.05 and df = ∞.
,
Note: The + and + values are the same, since the t-distribution tends to the standard normal
,
,
distribution as the df increase.
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99% confidence interval
When working with a 99% confidence interval using the standard normal distribution we have:
The + value is obtained using the standard normal table.
,
When working with a 99% confidence interval using the t-distribution where the df tends to
infinity we have:
The + value is obtained using the t-table with area in the upper tail =0.005 and df = ∞.
,
Note: The + and + values are the same, since the t-distribution tends to the standard normal
,
,
distribution as the df increase.
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Typical exam questions
Questions 1 and 2 are based on the following information:
The proportion of business travellers who is dissatisfied with the service of an airline is investigated.
A manager selects a systematic sample of 50 business travellers, from which 10 said that they are
dissatisfied with the service.
Let: p = population proportion of dissatisfied business travellers
Question 1
The lower limit of a 95% confidence interval for the population proportion is:
Answer 1
/ 3
.45.
6
0.2
1.963
7.7.8
97
47
0.089 with 97 0.2.
Question 2
If the confidence coefficient of a confidence interval decreases from 0.95 to 0.90 the:
(A) sample size increases.
(B) interval is narrower.
(C) significance level is smaller.
(D) margin of error is larger.
(E) standard error is larger.
Answer 2
If the confidence coefficient of a confidence interval decreases from 0.95 to 0.90, the / value
decreases from 1.96 to 1.645 and, consequently, the interval is narrower. Answer = B.
Question 3 is based on the following information:
A Business travel magazine rates the service of airlines on a regular basis (the rating scale with a low
score of 0 and a high score of 10 was used). It is known that 1.05. An airline was rated by 30
randomly selected business travellers which provided a sample mean of 7.5.
Question 3
The upper limit of a 99% confidence interval for the population mean is:
Answer 3
:
4.79
# / 6 7.5 # 2.576
7.99
√
√;7
Questions 4 to 10 are based on the following information:
The management team of a soccer stadium wants to estimate the average amount (in Rand) spent on
snacks and cool drinks per spectator. It is known that the amount (in Rand) is normally distributed.
They are also interested in the method of payment used for the purchase namely, credit card or cash.
Let the population mean of the amount (in Rand) spent on snacks and cool drinks per spectator.
the population proportion of spectators who paid with cash.
̅ the average amount (in Rand) spent on snacks and cool drinks.
Consider the following results in Excel:
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Formula worksheet:
Note: Rows 10 to 40 are hidden.
Value worksheet:
Note: Rows 10 to 40 are hidden
Question 4
The point estimate for the population mean of the amount (in Rand) spent is:
Answer 4
∑ >
6
494?.;@
A7
37.93
Question 5
The point estimate of the population proportion is:
Answer 5
In cell D3 of Excel the =COUNTIF(B2:B41, B3) function counts the number of payments made
using a credit card, i.e. 30 out of 40 payments were made using a credit card. Therefore, 40 – 30 = 10
payments were made using cash.
47
A7 0.25
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Question 6
When is used to estimate , the interval estimate for the population mean is based on the:
(A) standard normal distribution
(B) binomial distribution
(C) normal distribution
(D) B-distribution
(E) uniform distribution
Question 7
The margin of error of a 95% confidence interval for is:
Answer 7
To obtain the value of / we use the TINV function of Excel =TINV(, df) = TINV(0.05, 39) =
2.023 (this is given in cell D6 of Excel) with df = n – 1 = 40 – 1 = 39. To obtain the sample standard
deviation, we take the square root of the variance (this is given in cell D4 of Excel).
The margin of error is equal to /
C
√6
2.023 D
?
E 2.2391.
√A7
Question 8
The lower limit of a 99% confidence interval for the population proportion is:
Answer 8
/ 3
.45.
6
0.25
2.5763
7.97.?9
A7
0.0736
Question 9
The margin of error of a 95% confidence interval for the population proportion is:
Answer 9
/ 3
.45.
6
1.963
7.97.?9
A7
0.1342
Question 10
If the confidence coefficient of a confidence interval for is decreased from 95% to 90%, then:
(A) the standard error decreases, which implies a narrower interval.
(B) the standard error increases, which implies a wider interval.
(C) the sample size increases, which implies a narrower interval.
(D) lower limit increases, which implies a wider interval.
(E) the margin of error decreases, which implies a narrower interval.
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