Lecture 5: Confidence Intervals
Taeyong Park
Carnegie Mellon University in Qatar
October 1, 2019
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Roadmap: Where are we now?
Probability theory; sampling distributions
⇒ Confidence intervals and hypothesis testing
⇒ Applying to mean difference tests (t tests and Anova) and
proportion difference tests (z tests and χ2 tests).
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Sampling distribution of the sample mean
Sampling distribution of the sample mean
For random sampling with a large sample size n , the sampling
distribution of the sample mean y is approximately normal:
σ2
y ∼ N µ,
n
µ
¶
The mean of the distribution is equal to population mean µ.
The standard deviation of the distribution is equal to
σ is a population standard deviation.
pσ ,
n
where
σ
p is also called a standard error.
n
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What is next?
We are hunting a population mean µ.
E.g, what is the average salary of EAI 2,500 employees?
We sample from the population and calculate a sample statistic
y.
We use a sample statistic like y to estimate a population
parameter like µ.
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Point estimation
Point estimation for µ.
Sample mean: y =
Ï
n
1X
yi
n i =1
Terminology: a sample mean is a point estimator; a realized
sample mean value y is a point estimate.
Example: You are the project manager for a company who is
deciding whether or not to place a new product on the market. You
commission a poll of consumers in the region to find out whether
they would need the new product. Your pollster gives you a point
estimate:
54.5 percent of consumers need the product.
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Point estimation
Point estimation for µ.
Sample mean: y =
Ï
n
1X
yi
n i =1
Terminology: a sample mean is a point estimator; a realized
sample mean value y is a point estimate.
Example: You are the project manager for a company who is
deciding whether or not to place a new product on the market. You
commission a poll of consumers in the region to find out whether
they would need the new product. Your pollster gives you a point
estimate:
54.5 percent of consumers need the product.
How much can we believe it to be close to µ?
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Interval estimation
A point estimate is OK, but there is always uncertainty around
point estimates: A point estimate (sample statistic) is hardly the
same as the population parameter.
Alternatively, interval estimation.
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Interval estimation
In the example above, your pollster could give you interval estimates:
We can be 25% confident that the true percent point of
consumers who need the product is between 52 and 57.
We can be 95% confident that the true percent point of
consumers who need the product is between 45.75 and and
62.25.
We can be 100% confident that the true percent point of
consumers who need the product is between 0 and 100.
Trade-off between the width of your interval estimates and the
confidence you can have: The more confident you are, the wider your
interval is, and hence the less precise your estimate becomes.
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An interval estimator for µ
Confidence interval
A confidence interval for a population parameter is a range of
numbers within which a parameter is believed to fall given repeated
sampling.
Confidence interval for µ: [ y ± Margin of error].
How to calculate the margin of error in [ y ± Margin of error]?
Ï
Ï
Use the sampling distribution of y .
Information about the probability that y will be within a given
distance of µ.
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Sampling distribution and confidence interval
Recall the following question from the last lecture:
Ï
Ï
Suppose µ = 51800 and σ = 4000 for the EAI salary data. What
is the probability that y will be within of $500 of µ when the
sample size is 30?
µ
¶
The sampling distribution of the sample mean: y ∼ N µ,
500
500
=
p .
σy
4000/ 30 µ
F
z=
F
Then, we calculate P
Park (CMUQ)
−500
500
p ≤z≤
p
4000/ 30
4000/ 30
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σ2
.
n
¶
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Sampling distribution and confidence interval
Recall the following question from the last lecture:
Ï
Ï
Suppose µ = 51800 and σ = 4000 for the EAI salary data. What
is the probability that y will be within of $500 of µ when the
sample size is 30?
µ
¶
The sampling distribution of the sample mean: y ∼ N µ,
500
500
=
p .
σy
4000/ 30 µ
F
z=
F
Then, we calculate P
−500
500
p ≤z≤
p
4000/ 30
4000/ 30
σ2
.
n
¶
Generally, we can ask the question: P (µ − k ≤ y ≤ µ + k)?
A confidence interval is [ y − k ≤ µ ≤ y + k ], where k is Margin of
error. ⇒ Confidence interval for µ is [ y ± Margin of error].
Margin of error is a value like 500 in the above example: z × σ y .
Ï
σ y is from the sampling distribution, and z is from the
confidence level you determine, mostly 95% or 99%.
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Confidence interval for µ with σ known
µ is unknown: Our target.
What about σ? Generally, σ is unknown. But in some
applications, when large amounts of relevant historical data are
available, σ is estimated prior to sampling.
We begin with σ known, and will consider σ unknown later.
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Confidence interval for µ with σ known
µ is unknown: Our target.
What about σ? Generally, σ is unknown. But in some
applications, when large amounts of relevant historical data are
available, σ is estimated prior to sampling.
We begin with σ known, and will consider σ unknown later.
Example:
Ï
Ï
Ï
Lloyd’s Department Store wants to learn about the amount
spent per shopping trip.
Lloyd’s has been using the weekly survey for several years ⇒
Distribution and σ = $20.
It uses a new random sample of 100 customers to compute a
95% confidence interval for the amount spent per shopping trip.
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Confidence interval for µ with σ known
95% of a normal distribution falls within 1.96 standard deviations
from the mean. This holds true for every normal distribution.
µ
¶
σ2
Given x ∼ N µ,
, x falls within 1.96σx with probability 0.95.
n
Ï Whether x or y is a just matter of terminology.
Once a sample selected, if x falls between µ − 1.96σx and
µ + 1.96σx , then the interval x ± 1.96σx will contain µ.
Here, NOTE that x falls between µ − 1.96σx and µ + 1.96σx
with probability 0.95.
That is, with 95 times out of 100 repeated samples, x occurs
such that the interval x ± 1.96σx contains the population mean
µ.
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Confidence interval for µ with σ known
³
2
´
20
Getting back to the Lloyd’s example, x ∼ N µ, 100
and σx = 2.
95% of all x values from repeated samples will be within ±3.92 of µ.
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Interpretation of confidence interval
95% C.I.: Among 100 confidence intervals, 95 intervals contain the
population mean.
We are 95% confident that the population mean lies in the 95% C.I.
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A common mistake when Interpreting confidence
intervals
The probability that a one single confidence interval contains µ?
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A common mistake when Interpreting confidence
intervals
The probability that a one single confidence interval contains µ?
INCORRECT interpretation: The probability that a confidence
interval contains the population mean is 95%.
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Confidence interval for µ with σ known: A how-to
guide
The margin of error in [ y ± Margin of error] is given by z α/2 × σ ȳ ,
where z α/2 is a z-value, and σ ȳ is the standard deviation of the
sampling distribution of sample mean (a.k.a. standard error).
σ
Therefore, [ y ± Margin of error] = y± z α/2 × p .
n
z α/2 is determined by the confidence level (mostly 95% or 99%).
α = 1-the confidence level (e.g., 95% confidence level
⇔ α = 0.05, α/2 = 0.025.)
α/2: how much area we need under the curve to the right.
z α/2 for the 95% confidence level: Find a z-value corresponding
to the probability (1 − 0.95)/2 = 0.025 on the right tail.
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Exercise
z α/2 for the 99% confidence level?
0.005 = 2.58
Given a sample of n = 100 with y = 9.6, find a 95% confidence
interval for the population mean. Suppose σ = 4.
9.6+-0.005 *4/10
Find a 99% confidence interval for the population mean.
Lower bound and upper bound
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A z-table
Standard Normal Distribution Table (Right-Tail Probabilities)
z
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
3.1
3.2
3.3
3.4
Park (CMUQ)
.00
.5000
.4602
.4207
.3821
.3446
.3085
.2743
.2420
.2119
.1841
.1587
.1357
.1151
.0968
.0808
.0668
.0548
.0446
.0359
.0287
.0228
.0179
.0139
.0107
.0082
.0062
.0047
.0035
.0026
.0019
.0013
.0010
.0007
.0005
.0003
.01
.4960
.4562
.4168
.3783
.3409
.3050
.2709
.2389
.2090
.1814
.1562
.1335
.1131
.0951
.0793
.0655
.0537
.0436
.0351
.0281
.0222
.0174
.0136
.0104
.0080
.0060
.0045
.0034
.0025
.0018
.0013
.0009
.0007
.0005
.0003
.02
.4920
.4522
.4129
.3745
.3372
.3015
.2676
.2358
.2061
.1788
.1539
.1314
.1112
.0934
.0778
.0643
.0526
.0427
.0344
.0274
.0217
.0170
.0132
.0102
.0078
.0059
.0044
.0033
.0024
.0018
.0013
.0009
.0006
.0005
.0003
.03
.4880
.4483
.4090
.3707
.3336
.2981
.2643
.2327
.2033
.1762
.1515
.1292
.1093
.0918
.0764
.0630
.0516
.0418
.0336
.0268
.0212
.0166
.0129
.0099
.0075
.0057
.0043
.0032
.0023
.0017
.0012
.0009
.0006
.0004
.0003
.04
.4840
.4443
.4052
.3669
.3300
.2946
.2611
.2296
.2005
.1736
.1492
.1271
.1075
.0901
.0749
.0618
.0505
.0409
.0329
.0262
.0207
.0162
.0125
.0096
.0073
.0055
.0041
.0031
.0023
.0016
.0012
.0008
.0006
.0004
.0003
.05
.4801
.4404
.4013
.3632
.3264
.2912
.2578
.2266
.1977
.1711
.1469
.1251
.1056
.0885
.0735
.0606
.0495
.0401
.0322
.0256
.0202
.0158
.0122
.0094
.0071
.0054
.0040
.0030
.0022
.0016
.0011
.0008
.0006
.0004
.0003
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.06
.4761
.4364
.3974
.3594
.3228
.2877
.2546
.2236
.1949
.1685
.1446
.1230
.1038
.0869
.0721
.0594
.0485
.0392
.0314
.0250
.0197
.0154
.0119
.0091
.0069
.0052
.0039
.0029
.0021
.0015
.0011
.0008
.0006
.0004
.0003
.07
.4721
.4325
.3936
.3557
.3192
.2843
.2514
.2206
.1922
.1660
.1423
.1210
.1020
.0853
.0708
.0582
.0475
.0384
.0307
.0244
.0192
.0150
.0116
.0089
.0068
.0051
.0038
.0028
.0021
.0015
.0011
.0008
.0005
.0004
.0003
.08
.4681
.4286
.3897
.3520
.3156
.2810
.2483
.2177
.1894
.1635
.1401
.1190
.1003
.0838
.0694
.0571
.0465
.0375
.0301
.0239
.0188
.0146
.0113
.0087
.0066
.0049
.0037
.0027
.0020
.0014
.0010
.0007
.0005
.0004
.0003
.09
.4641
.4247
.3859
.3483
.3121
.2776
.2451
.2148
.1867
.1611
.1379
.1170
.0985
.0823
.0681
.0559
.0455
.0367
.0294
.0233
.0183
.0143
.0110
.0084
.0064
.0048
.0036
.0026
.0019
.0014
.0010
.0007
.0005
.0003
.0002
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Determining the sample size
Now suppose we want the population mean to be estimated
within a certain margin of error. That is, we want to choose a
sample size large enough to provide a desired margin of error.
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Determining the sample size
Now suppose we want the population mean to be estimated
within a certain margin of error. That is, we want to choose a
sample size large enough to provide a desired margin of error.
Using the example above, y = 9.6, σ = 4, 95% confidence level.
Margin of error = z 0.025 × pσn = 1.96 × p4n .
We want to choose n such that we estimate µ within a margin
of error of 0.5.
Ï
Ï
4
When n = 100 above, the margin of error = 1.96 × p100
= 0.784.
What is your guess? n greater than 100 or smaller than 100?
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Determining the sample size
Now suppose we want the population mean to be estimated
within a certain margin of error. That is, we want to choose a
sample size large enough to provide a desired margin of error.
Using the example above, y = 9.6, σ = 4, 95% confidence level.
Margin of error = z 0.025 × pσn = 1.96 × p4n .
We want to choose n such that we estimate µ within a margin
of error of 0.5.
Ï
Ï
4
When n = 100 above, the margin of error = 1.96 × p100
= 0.784.
What is your guess? n greater than 100 or smaller than 100?
Solve 1.96 × p4n ≤ 0.5.
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Confidence interval for µ with σ unknown
σ is generally unknown in reality.
With σ unknown, we use a sample standard deviation s to
estimate σ, and use a t distribution instead of relying on a
normal sampling distribution.
CI for µ with σ known
Using a normal distribution directly from the sampling distribution of
the sample mean, y ± z α/2 pσn .
CI for µ with σ unknown
Using s =
qP
(y i −y)2
n−1
Park (CMUQ)
and a t distribution, y ± t α/2 psn .
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Confidence interval for µ with σ unknown
We use a t distribution (t score) instead of the standard normal
distribution (z score) to account for the error produced by
estimating pσn using psn .
Especially, when the sample size is small, the error is large.
Then, how does a t distribution account for such an increased
error?
Ï
It makes you less confident about your inference by increasing
the margin of error.
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The t distribution and the standard normal
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Notes on the t distribution
Symmetric, bell-shaped, zero mean.
It has “thicker” tails than the standard normal distribution ⇒ A
larger margin of error and a wider CI.
Unlike the standard normal distribution (σ = 1), the t
distribution’s dispersion depends on degrees of freedom (n − 1),
sometimes listed as df. Text
As df increases, the t distribution becomes closer and closer to
the standard normal distribution. ⇒ When n is large, no big
difference between z α/2 and t α/2 .
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Exercise
Given a sample y = 9.6, S = 4, n = 25, find a 95% confidence
interval for the population mean µ. Use the t distribution and
the t table in the next page.
Given a sample y = 9.6, S = 4, n = 31, find a 99% confidence
interval for the population mean µ. Use the t distribution and
the t table in the next page. 0.005, n = 30
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Exercise
n - 1 = 25 -1
0.025
Given a sample y = 9.6, S = 4, n = 25, find a 95% confidence
interval for the population mean µ. Use the t distribution and
the t table in the next page.
Ï
4
= [7.9488, 11.2512]
9.6 ± 2.064 × p
25
Given a sample y = 9.6, S = 4, n = 31, find a 99% confidence
interval for the population mean µ. Use the t distribution and
the t table in the next page.
Ï
4
= [7.624342, 11.57566]
9.6 ± 2.75 × p
31
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t Table
t Table
cum. prob
t .50
t .75
t .80
t .85
t .90
t .95
t .975
t .99
t .995
t .999
t .9995
one-tail
0.50
1.00
0.25
0.50
0.20
0.40
0.15
0.30
0.10
0.20
0.05
0.10
0.025
0.05
0.01
0.02
0.005
0.01
0.001
0.002
0.0005
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.000
0.816
0.765
0.741
0.727
0.718
0.711
0.706
0.703
0.700
0.697
0.695
0.694
0.692
0.691
0.690
0.689
0.688
0.688
0.687
0.686
0.686
0.685
0.685
0.684
0.684
0.684
0.683
0.683
0.683
0.681
0.679
0.678
0.677
0.675
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
0.851
0.848
0.846
0.845
0.842
1.963
1.386
1.250
1.190
1.156
1.134
1.119
1.108
1.100
1.093
1.088
1.083
1.079
1.076
1.074
1.071
1.069
1.067
1.066
1.064
1.063
1.061
1.060
1.059
1.058
1.058
1.057
1.056
1.055
1.055
1.050
1.045
1.043
1.042
1.037
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
1.303
1.296
1.292
1.290
1.282
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
1.684
1.671
1.664
1.660
1.646
12.71
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
2.021
2.000
1.990
1.984
1.962
31.82
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
2.423
2.390
2.374
2.364
2.330
63.66
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
2.704
2.660
2.639
2.626
2.581
318.31
22.327
10.215
7.173
5.893
5.208
4.785
4.501
4.297
4.144
4.025
3.930
3.852
3.787
3.733
3.686
3.646
3.610
3.579
3.552
3.527
3.505
3.485
3.467
3.450
3.435
3.421
3.408
3.396
3.385
3.307
3.232
3.195
3.174
3.098
636.62
31.599
12.924
8.610
6.869
5.959
5.408
5.041
4.781
4.587
4.437
4.318
4.221
4.140
4.073
4.015
3.965
3.922
3.883
3.850
3.819
3.792
3.768
3.745
3.725
3.707
3.690
3.674
3.659
3.646
3.551
3.460
3.416
3.390
3.300
0.000
0.674
0.842
1.036
1.282
1.645
1.960
2.326
2.576
3.090
3.291
0%
50%
60%
70%
80%
90%
95%
Confidence Level
98%
99%
99.8%
99.9%
two-tails
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
40
60
80
100
1000
z
Park (CMUQ)
70-207
0.001
October 1, 2019
25 / 31
Confidence interval for a population mean µ
CI for µ
[ y ± Margin of Error], y is the sample mean.
CI for µ with σ known
Using a normal distribution directly from the sampling distribution of
the sample mean, y ± z α/2 pσn .
CI for µ with σ unknown
Using s =
qP
(y i −y)2
n−1
and a t distribution, y ± t α/2 psn .
When n is large, z α/2 and t α/2 are similar.
Park (CMUQ)
70-207
October 1, 2019
26 / 31
What about for a population proportion p ?
For categorical data, we record the proportions of observations
in the categories.
Ï
The proportions of employees who participated in a management
training / who didn’t.
Ï
The proportions of nations with low tariffs / moderate tariffs /
high tariffs.
Ï
The proportions of students majored in Business Administration
/ Information Systems / Computer Science.
Park (CMUQ)
70-207
October 1, 2019
27 / 31
Confidence interval for a population proportion p
CI for µ
[ y ± Margin of Error], y is the sample mean.
Recall ³that we
´ computed the margin of error based on
σ2
y ∼ N µ, n , σ y = pσn ⇒ margin of error = z α/2 × pσn .
Park (CMUQ)
70-207
October 1, 2019
28 / 31
Confidence interval for a population proportion p
CI for µ
[ y ± Margin of Error], y is the sample mean.
Recall ³that we
´ computed the margin of error based on
σ2
y ∼ N µ, n , σ y = pσn ⇒ margin of error = z α/2 × pσn .
CI for p
[p ± Margin of Error], p is the sample proportion.
r
¶
p(1 − p)
p(1 − p)
Compute the margin of error: p ∼ N p,
, σp =
n
n
r
p(1 − p)
⇒ margin of error = z α/2 ×
.
n
µ
Park (CMUQ)
70-207
October 1, 2019
28 / 31
Confidence interval for a rpopulation proportion p
In the margin of error z α/2 ×
p(1 − p)
, p is not known.
n
We substitute p for p .
CI for p
r
[p ± Margin of Error] = p ± z α/2 ×
Park (CMUQ)
70-207
p(1 − p)
n
October 1, 2019
29 / 31
Confidence interval for a rpopulation proportion p
In the margin of error z α/2 ×
p(1 − p)
, p is not known.
n
We substitute p for p .
CI for p
r
[p ± Margin of Error] = p ± z α/2 ×
p(1 − p)
n
Increased statistical error due to substituting p for p ? Should
use t α/2 instead of z α/2 ?
q
p(1−p)
s
Compare y ± t α/2 pn and p ± z α/2
.
n
Ï
Two estimates vs. one estimate.
Park (CMUQ)
70-207
October 1, 2019
29 / 31
Confidence interval for a rpopulation proportion p
In the margin of error z α/2 ×
p(1 − p)
, p is not known.
n
We substitute p for p .
CI for p
r
[p ± Margin of Error] = p ± z α/2 ×
p(1 − p)
n
Increased statistical error due to substituting p for p ? Should
use t α/2 instead of z α/2 ?
q
p(1−p)
s
Compare y ± t α/2 pn and p ± z α/2
.
n
Two estimates vs. one estimate.
Substituting p for p for the margin of error doesn’t produce an
additional error. ⇒ Always use the standard normal and z α/2 .
Ï
Park (CMUQ)
70-207
October 1, 2019
29 / 31
Exercise
Source: Rasmunssen Reports, 2012
Question: “Will today’s children be better off than their parents?”
Population: U.S. Adults
Sample size: N = 1000
Sample Data: 240 said yes.
Find a 95% confidence interval for the proportion of adults who
think that today’s children will be better off than their parents.
Interpret your result substantively.
Park (CMUQ)
70-207
October 1, 2019
30 / 31
Exercise
Point estimate: p = 0.24.
r
r
p(1 − p)
.24(1 − .24)
Margin of error : z 0.025 ×
= 1.96
= 0.026.
n
1000
Confidence interval : 0.24 ± 0.026 = [0.214, 0.266]
Interpretation
Ï
I am 95% confident that the proportion of the U.S. adults who
think that today’s children will be better off than their parents is
between 0.214 and 0.266.
Park (CMUQ)
70-207
October 1, 2019
31 / 31