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Chapter-6: Point
Estimation and
Confidence Intervals
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Introductory Statistics-I
Confidence Intervals
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Section 6.1
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POINT ESTIMATE
✗ An estimate of a population parameter given by a single number is
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called a Point Estimate for that parameter.
✗ A point estimate of a population parameter is an estimate of the
parameter using a single number.
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𝑥ҧ 𝑖𝑠 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 𝑒𝑠𝑡𝑖𝑚𝑎𝑡𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑟𝑢𝑒 𝑚𝑒𝑎𝑛 𝜇.
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𝑝Ƹ 𝑖𝑠 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 𝑒𝑠𝑡𝑖𝑚𝑎𝑡𝑒 𝑜𝑓 𝑡𝑟𝑢𝑒 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝑝.
✗ The Margin of Error is the magnitude of the difference between
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the sample point estimate and the true proportion parameter
value.
Confidence Intervals
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ESTIMATING POPULATION MEAN, 𝜇:𝜎 –KNOWN
We have a simple random sample of size n drawn
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✗
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from a population.
The population standard deviation σ is known.
✗
If the x distribution is normal, then our methods
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✗
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work for any sample size n.
Confidence Intervals
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Z-CONFIDENCE VALUES FOR 𝜎-KNOWN
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✗ 𝐴 𝑐𝑜𝑛𝑓𝑖𝑑𝑒𝑛𝑐𝑒 𝑙𝑒𝑣𝑒𝑙, 𝑐, 𝑖𝑠 𝑎𝑛𝑦 𝑣𝑎𝑙𝑢𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 0
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𝑎𝑛𝑑 1 (𝑜𝑟 0% 𝑡𝑜 100%) 𝑡ℎ𝑎𝑡 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑠 𝑡𝑜 𝑡ℎ𝑒
𝑎𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑡ℎ𝑒 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑛𝑜𝑟𝑚𝑎𝑙 𝑐𝑢𝑟𝑣𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛
−𝑧𝑐 𝑎𝑛𝑑 + 𝑧𝑐
𝐿𝑒𝑣𝑒𝑙 𝑜𝑓 𝐶𝑜𝑛𝑓𝑖𝑑𝑒𝑛𝑐𝑒, 𝑐
𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑉𝑎𝑙𝑢𝑒, z𝑐
0.90 or 90%
1.645
0.95 or 95%
1.96
0.99 or 99%
2.58
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NOTE: Usually c is equal to a number such as 0.90, 0.95 or 0.99
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This table is given below the Standard Normal Z-table provided
Confidence Intervals
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z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
0.0
.5000
.5040
.5080
.5120
.5160
.5199
.5239
.5279
.5319
.5359
0.1
.5398
.5438
.5478
.5517
.5557
.5596
.5636
.5675
.5714
.5753
0.2
.5793
.5832
.5871
.5910
.5948
.5987
.6026
.6064
.6103
.6141
0.3
.6179
.6217
.6255
.6293
.6331
.6368
.6406
.6443
.6480
.6517
α
0.08
=
= 0.04
2
2
α
1 − = 1 − 0.04 = 0.96
2
0.4
.6554
.6591
.6628
.6664
.6700
.6736
.6772
.6808
.6844
.6879
0.5
.6915
.6950
.6985
.7019
.7054
.7088
.7123
.7157
.7190
.7224
0.6
.7257
.7291
.7324
.7357
.7389
.7422
.7454
.7486
.7517
.7549
4. Find the area “0.96” in
the body of our z-table
chart and hence find the
corresponding z-score
value covering this area.
0.7
.7580
.7611
.7642
.7673
.7704
.7734
.7764
.7794
.7823
.7852
0.8
.7881
.7910
.7939
.7967
.7995
.8023
.8051
.8078
.8106
.8133
0.9
.8159
.8186
.8212
.8238
.8264
.8289
.8315
.8340
.8365
.8389
1.0
.8413
.8438
.8461
.8485
.8508
.8531
.8554
.8577
.8599
.8621
1.1
.8643
.8665
.8686
.8708
.8729
.8749
.8770
.8790
.8810
.8830
1.2
.8849
.8869
.8888
.8907
.8925
.8944
.8962
.8980
.8997
.9015
1.3
.9032
.9049
.9066
.9082
.9099
.9115
.9131
.9147
.9162
.9177
1.4
.9192
.9207
.9222
.9236
.9251
.9265
.9279
.9292
.9306
.9319
1.5
.9332
.9345
.9357
.9370
.9382
.9394
.9406
.9418
.9429
.9441
1.6
.9452
.9463
.9474
.9484
.9495
.9505
.9515
.9525
.9535
.9545
1.7
.9554
.9564
.9573
.9582
.9591
.9599
.9608
.9616
.9625
.9633
1.8
.9641
.9649
.9656
.9664
.9671
.9678
.9686
.9693
.9699
.9706
1.9
.9713
.9719
.9726
.9732
.9738
.9744
.9750
.9756
.9761
.9767
2.0
.9772
.9778
.9783
.9788
.9793
.9798
.9803
.9808
.9812
.9817
Z-CRITICAL VALUE FOR α = 5% AND 8% FOR
CI AND TWO TAILED HYPOTHESIS TESTING
1. α = 5% = 0.05
5. z = 1.75 for confidence
interval and z = ±1.75
for a Two-Tailed
Hypothesis Testing
Problem using z
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5. z = 1.96 for confidence
interval and z = ±1.96 for
a Two-Tailed Hypothesis
Testing Problem using z
3.
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4. Find the area “0.975” in
the body of our z-table
chart and hence find the
corresponding z-score
value covering this area.
2.
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3.
α
0.05
=
= 0.025
2
2
α
1 − = 1 − 0.025 = 0.975
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2.
1. α = 8% = 0.08
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CONFIDENCE INTERVALS
✗ For a confidence level c, the critical value zc is the number such
𝜎
𝑛
< 𝑥ҧ − 𝜇 < 𝑧𝑐
𝜎
𝑛
=𝑐
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i.e., 𝑃 −𝑧𝑐
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that the area under the standard normal curve between
– 𝑧𝑐 𝑎𝑛𝑑 𝑧𝑐 𝑒𝑞𝑢𝑎𝑙𝑠 𝑐.
✗ The area under the normal curve is the probability that the
standard normal variable z in that interval.
‘c’ is the probability that the sample mean will
𝜎
𝑛
𝑧𝑐 = 𝑧𝛼ൗ , 𝑤ℎ𝑒𝑟𝑒 𝛼 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑡ℎ𝑒 𝑙𝑒𝑣𝑒𝑙 𝑜𝑓 𝑠𝑖𝑔𝑛𝑖𝑓𝑖𝑐𝑎𝑛𝑐𝑒.
2
𝑊𝑒 𝑤𝑖𝑙𝑙 𝑢𝑠𝑒 𝑡ℎ𝑖𝑠 𝑖𝑛 𝑛𝑒𝑥𝑡 𝑐ℎ𝑎𝑝𝑡𝑒𝑟
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Note:
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differ from the population mean by at most ±𝑧𝑐
Confidence Intervals
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(MAXIMAL) MARGIN OF ERROR
► Since μ is the unknown, the margin of error |𝑥ҧ − μ| is also
𝐸 = 𝑧𝑐
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unknown.
✗ Using confidence level, c, we can say 𝑥ҧ differs from μ by
at most:
𝜎
𝑛
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Here E is called Maximal Margin of error and is also denoted
by M.E.
Finally, confidence interval for μ is an interval computed from
sample data in such a way that c is the probability of generating
an interval containing the actual value of μ.
𝑃 𝑥ҧ − 𝐸 < 𝜇 < 𝑥ҧ + 𝐸 = 𝑐
Confidence Intervals
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CONFIDENCE INTERVAL FOR MEAN, 𝜇: 𝜎-KNOWN
𝐸 = 𝑧𝑐
i.e., CI for μ when σ known is
𝜎
𝜎
(𝑥ҧ − 𝑧𝑐
, 𝑥ҧ + 𝑧𝑐 )
𝑛
𝑛
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✗
𝜎
𝑛
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Then the confidence interval (CI) for mean μ
when σ is known is as follows: (𝑥ҧ − 𝐸, 𝑥ҧ + 𝐸);
where
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✗
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c = confidence level (0 ≤ c ≤ 1)
zc = critical value for confidence level c based on the
standard normal distribution (see table given in slide 5)
Confidence Intervals
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Confidence Interval of µ is 𝑥ҧ − 𝐸, 𝑥ҧ + 𝐸
II.
Given (Point Estimate of Mean µ is 𝑥)ҧ , 𝑥ҧ = 31.39, 𝑛 =
✗ Consider a normally
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82, 𝑐 = 95% = 0.95 𝑎𝑛𝑑 𝜎 = 0.8
{𝜎 − 𝑘𝑛𝑜𝑤𝑛 ⇒ 𝑧 − 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛}
III.
zc = z0.95 = 1.96 (𝐹𝑟𝑜𝑚 𝑧 𝑡𝑎𝑏𝑙𝑒 −
𝑏 , 𝑏𝑒𝑙𝑜𝑤 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡𝑎𝑏𝑙𝑒)
IV.
E = zc
σ
⟹ E = 1.96
n
0.8
82
= 0.17
95% CI for mean µ is,
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distributed dataset with a
standard deviation of 0.8.
Given that a random
sample of size 82 from
that dataset has a sample
mean of 31.39. Construct a
95% confidence interval
for the true mean of the
dataset and interpret it.
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I.
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EXAMPLE-1
V.
31.39 − 0.17, 31.39 + 0.17 = 31.22, 31.56
We are 95% confident that the true mean is
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VI.
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between 31.22 and 31.56
Confidence Intervals
308
II.
✗ A sample of 36 randomly
III.
IV.
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{𝜎 − 𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝐵𝑈𝑇 𝑛 > 30
⇒ 𝑢𝑠𝑒 𝑧 − 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑢𝑠𝑖𝑛𝑔 𝐶𝐿𝑇 𝑡𝑟𝑒𝑎𝑡𝑖𝑛𝑔
𝑠 ≅ 𝜎}
zc = z0.9 = 1. 645 & z0.95 =
1.96 {𝐹𝑟𝑜𝑚 𝑧 𝑡𝑎𝑏𝑙𝑒 − 𝑏 , 𝑏𝑒𝑙𝑜𝑤 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡𝑎𝑏𝑙𝑒)
E = zc
σ
⟹ E = 1.645
n
𝐸 = 1.96
6.7
36
= 1.84 and
6.7
= 2.19
36
V.
90% CI for mean µ is, 23 − 1.84, 23 + 1.84 =
21.16, 24.84 𝑎𝑛𝑑
95% CI for mean µ is, 23 − 2.19, 23 + 2.19 =
20.81, 25.19
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selected candy boxes
recorded a mean weight of
23 lbs and a standard
deviation of 6.7 lbs.
Construct 90% and 95%
confidence intervals of
true mean weight of candy
boxes and comment about
it.
Confidence Interval of µ is 𝑥ҧ − 𝐸, 𝑥ҧ + 𝐸
Given (Point Estimate of Mean µ is 𝑥)ҧ , 𝑥ҧ = 23; 𝑛 = 36, 𝑠 =
6.7 𝑐 = 90% = 0.9 𝑎𝑛𝑑 𝑐 = 95% = 0.95
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EXAMPLE-2
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I.
VI.
We are 90% confident that the true mean is between
21.16 and 24.84.
We are 95% confident that the true mean is between 20.81 and
25.19
Confidence Intervals
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Given (Point Estimate of Mean µ is 𝑥)ҧ , 𝑥ҧ = 150, 𝑛 =
20, 𝜎 = 15.5, 𝑎𝑛𝑑 𝑐 = 99% = 0.99
{𝜎 − 𝑘𝑛𝑜𝑤𝑛 ⇒ 𝑧 − 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛}
III.
zc = z0.99 = 2. 58 (𝐹𝑟𝑜𝑚 𝑧 𝑡𝑎𝑏𝑙𝑒 −
(𝑏), 𝑏𝑒𝑙𝑜𝑤 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡𝑎𝑏𝑙𝑒)
IV.
E = zc
σ
⟹ 2.58
n
15.5
20
= 8.94
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scores out of 200 of a
statistics class are
considered for study. The
average score for these 20
students is 150. Based on
previous studies, standard
deviation of these scores is
15.50. Construct and
interpret a 99% confidence
interval for true averages
of final scores.
II.
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✗ A sample of 20 final exam
Confidence Interval of µ is 𝑥ҧ − 𝐸, 𝑥ҧ + 𝐸
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EXAMPLE-3
I.
V.
99% CI for mean µ is, 150 − 8.94 150 + 8.94 =
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141.06, 158.94
VI.
We are 99% confident that the true mean is
between 141.06 and 158.94.
Confidence Intervals
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EXAMPLE-4
A survey of 120 smokers resulted
that they smoke 13 packs of
that they smoke 13 packs of
cigarettes per week and with a
cigarettes per week and with a
standard deviation of 3 packs per
week. If the population data of
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week. If the population data of
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standard deviation of 3 packs per
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A survey of 120 smokers resulted
smokers does not follow normal
Find and interpret a 90% confidence
distribution, Find and interpret a 90%
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smokers follow a normal distribution,
confidence interval for true average
smoked per week.
of packs smoked per week.
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interval for true average of packs
Confidence Intervals
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SAMPLE SIZE ESTIMATION FOR MEANS
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✗ When designing a statistical study, it is good practice to decide in
advance:
The confidence level (𝑐)
-
The maximal margin of error (𝐸 𝑜𝑟 𝑀. 𝐸. )
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-
✗ Using these we can then calculate the required (minimum)
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sample size,𝑛, to meet these goals.
Confidence Intervals
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𝐸 = 𝑧𝑐
𝜎
𝑛
𝑤ℎ𝑒𝑛 𝜎 − 𝑘𝑛𝑜𝑤𝑛.
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𝑧𝑐 𝜎 2
𝑛=
𝐸
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𝐼𝑛 𝑡ℎ𝑖𝑠 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑖𝑓 𝑤𝑒 𝑘𝑛𝑜𝑤 𝐸, 𝑐, 𝜎,
𝑤𝑒 𝑐𝑎𝑛 𝑐𝑜𝑚𝑝𝑢𝑡𝑒 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒 𝑛 𝑎𝑠 𝑓𝑜𝑙𝑙𝑜𝑤𝑠:
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𝑇ℎ𝑒 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑚𝑎𝑟𝑔𝑖𝑛 𝑜𝑓 𝑒𝑟𝑟𝑜𝑟 𝑀. 𝐸. 𝑜𝑟 𝐸
𝑓𝑜𝑟 𝑚𝑒𝑎𝑛𝑠 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦
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CHOOSING SAMPLE SIZE FOR MEANS
PROOF
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Always round n up to the next
integer!!
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Note:
if the calculated n value is 10.75, round up to n=11
if the calculated n value is 10.23, round up to n=11
if the calculated n value is 10, round up to n=11
Confidence Intervals
313
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EXAMPLE-1
𝐺𝑖𝑣𝑒𝑛 𝑑𝑒𝑡𝑎𝑖𝑙𝑠 𝑎𝑟𝑒, 𝑐 = 90%;
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𝜎 = 10.
𝑇𝑜 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑡ℎ𝑒 𝑆𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒, 𝑛.
𝐸 = 5, 𝜎 = 10, 𝑧𝑐 = 𝑧0.9 = 1.645.
𝐹𝑜𝑟 𝑆𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒 𝑒𝑠𝑡𝑖𝑚𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑀𝑒𝑎𝑛𝑠, 𝑤𝑒 ℎ𝑎𝑣𝑒
𝑧𝑐 𝜎 2
⟹𝑛=
𝐸
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estimate the mean age at
which a child learns to talk.
Find the sample size
necessary for a 90%
confidence level to ensure
that the sample mean is
within 5 weeks for the mean
age at which a child learns to
talk. Assume σ =10 weeks.
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✗ A child psychologist wants to
𝑀𝑎𝑥𝑖𝑚𝑎𝑙 𝑀𝑎𝑟𝑔𝑖𝑛 𝑜𝑓 𝑒𝑟𝑟𝑜𝑟, 𝐸 = 5; 𝑎𝑛𝑑
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𝑛=
1.645×10 2
5
= 10.82 ≈ 11
Remember: Always round n up to the next integer!!
Confidence Intervals
314
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EXAMPLE-2
𝐶𝑜𝑛𝑓𝑖𝑑𝑒𝑛𝑐𝑒 𝑙𝑒𝑣𝑒𝑙, 𝑐 = 99%;
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𝑀𝑎𝑥𝑖𝑚𝑎𝑙 𝑀𝑎𝑟𝑔𝑖𝑛 𝑜𝑓 𝑒𝑟𝑟𝑜𝑟, 𝐸 = 2; 𝑎𝑛𝑑
𝜎 = 1.4.
𝑇𝑜 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑡ℎ𝑒 𝑆𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒, 𝑛.
𝐸 = 2, 𝜎 = 1.4, 𝑧𝑐 = 𝑧0.99 = 2.58
𝐹𝑜𝑟 𝑆𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒 𝑒𝑠𝑡𝑖𝑚𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑀𝑒𝑎𝑛𝑠, 𝑤𝑒 ℎ𝑎𝑣𝑒
at
data is normally distributed
with standard deviation of
1.4. Determine the
minimum sample size
required when you want to
be 99% confident that the
sample mean is within two
units from the mean.
M
✗ Assume that a population
Ve
nk
𝑛=
𝑧𝑐 𝜎 2
⟹𝑛=
𝐸
2.58×1.4 2
2
= 3.26 ≈ 4
Remember: Always round n up to the next integer!!
Confidence Intervals
315