Download Confidence Interval Estimation of ?

Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION
10-1
Topic (10) – ESTIMATION OF PARAMETERS
OF A SINGLE POPULATION
As has been mentioned repeatedly, we use sample
information to estimate the unknown information
about the population we are interested in.
EXAMPLES
1. Rats raised in an enriched environment appear to
have larger cortexes than those raised in unenriched
environments. What is the average cortex weight, µ,
for the population of cortex weights for every rat
which could be raised in an enriched environment?
We’ll grow a sample of rats in the enriched environment
and use our sample data to estimate µ.
2. Fiddler crabs are known as such because they
have asymmetrically sized pincers. Like human
handedness, it is estimated that approximately 10%
of fiddler crabs are “left-pincer”, i.e. the left pincer is
larger than the right. What is the true proportion, π,
of left-pincered crabs on a remote island in the South
Pacific?
We’ll fly to the S. Pacific, take a random sample of crabs
and use the sample proportion p to estimate π.
Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION
10-2
Several ways we could report the results:
We could report a POINT ESTIMATE, which is a
single number from the sample. It has little context or
meaning unless additional information is provided.
For example, when reporting the sample mean it
usually is better to also report something about its
sampling variability, such as including the sample
size n and the sample standard deviation s.
So, other possibilities for reporting:
1. Report x and s.
Not useful since s is not the standard deviation of x
2. Report x and σ x
=σ
n
.
Usually can’t because we don’t know the value of σ
3. Report x and SEM =
s
n
.
• SEM is STANDARD ERROR OF THE MEAN
• SEM is an unbiased estimate of σ x (under
random sampling)
Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION
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• one common report is to list x ± SEM but it’s
not the best choice since it isn’t directly
interpretable as an interval estimate
We would like to report a range or interval of
plausible values for the population parameter we’re
estimating. Such a report is called an INTERVAL
ESTIMATE of the population parameter.
When done right, the interval estimator includes our
point estimate and an estimate of the accuracy of the
point estimator and some measure of our “comfort”
with the estimate we are providing.
That is, we would like to be somewhat confident that
the interval we report actually covers the true value
of the parameter we are trying to estimate.
Defn: A CONFIDENCE INTERVAL for a
population characteristic is an interval of plausible
values for that parameter. It is constructed so that the
true value of the parameter is captured inside the
interval with a chosen, specified level of confidence.
EXAMPLE Political polls are almost always reported as
follows: “The proportion of voters who would vote for
Gore today is 48% ± 4% with 95% confidence”. What that
Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION
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means is that based on their sample, the true proportion π
of voters who would choose Gore is somewhere inside the
interval 44% and 52%. The 95% confidence refers to the
probability that the interval captures π.
Defn: The CONFIDENCE LEVEL associated with
a confidence interval is the probability that the
interval estimate covers the true value of the
parameter. One way of thinking of it is as the success
rate of the method we are using to do the estimation!
The method includes the choice of point estimator,
the assumptions about the distribution of that
estimator, and the sampling design.
The confidence level is chosen by the researcher
doing the reporting. Common levels in the life
sciences are 90%, 95% and 99%. In social sciences
you sometimes see 85% as well.
When intervals are constructed in a way that uses
knowledge of the sampling distributions of the point
estimators, we can assign these probabilities and not
just guess what they might be.
Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION
•
10-5
π
•
•
•
•
e.g. 68% of all sample proportions are within the interval
above. If we put a confidence interval of the same width
around each estimate shown above, what percent of the
new intervals would include π ?
Important Point: the confidence level is not a
probability about a particular interval including the
true value. The interval either does or does not cover
the true value (but you’ll never know that). The
confidence level is the probability that the method
you used will give an interval that includes the true
value!!!
Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION
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1) INTERVAL ESTIMATION OF THE POPULATION PROPORTION π
How does the Gallup polling company calculate the
95% confidence interval they report? They take a
sample of size n and calculate the sample proportion
of successes p (in this case people who indicate they
would vote right now for Gore). Then they use:
A LARGE SAMPLE 95% CONFIDENCE
INTERVAL FOR THE POPULATION
PROPORTION π is
p ± 1.96
i.e.
p(1 − p )
n
⎛
p(1 − p )
⎜⎜ p − 1.96
,
n
⎝
p(1 − p ) ⎞
⎟⎟
p + 1.96
n ⎠
• Large-sample refers to a sample size that is
sufficiently large to invoke the Central Limit
Theorem.
• p is the point estimator of π so is at the center of
the interval
Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION
•
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p(1 − p )
is the estimator of σ p - we use this
n
because it is the estimate of how much sampling
variability there is in p
• 1.96 is the z-score, z*, that makes the following
statement true: 0.95 = Pr(- z* < Z < + z*). We
use this because we are using the CLT which
states that sample proportions are normally
distributed for large samples
The formula is easily adapted for other confidence
levels. Simply replace 1.96 with the appropriate
number from the table below. The z critical values
for common confidence levels are:
Confidence Level
80%
90%
95%
98%
99%
99.9%
Z critical values
1.28
1.645
1.96
2.33
2.58
3.29
Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION
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EXAMPLE Our researcher did fly to the South
Pacific and collected 150 crabs. For each crab she
recorded whether the left or right pincer was
dominant and observed that 20 crabs were leftpincered. Calculate a 90% C.I. to estimate the true
proportion of left-pincered crabs on the island.
1) Is the sample size large enough to use our method?
Need to know if nπ and n(1 − π ) ≥ 10 but we
don’t know π. Instead look to see that the
number of successes and the number of failures
in the sample both are greater than 10.
Here she saw 20 successes and 130 failures
So, yes
p = 20
2)
= 0.133,
150
.133(1 − .133 )
p(1 − p )
=
= .0277
150
n
3) 90% Confidence Î z* = 1.645
4) The 90% C.I. then is
Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION
10-9
p(1 − p )
= .133 ± 1.645(0.0277 )
n
= .133 ± .0456 = (0.0874, 0.1786 )
p ± 1.645
The true proportion of Fiddler crabs on this island
lies, with 90% confidence, between 8.4% and 17.9%.
We interpret it as follows: Since this interval includes
10% there is no evidence that the true proportion on
this island is any different than other populations.
a) What if we had calculated a 95% C.I.? Would it
be wider or shorter than the 90% C.I.?
Wider since 1.96 > 1.645, i.e.
1.96(0.0277) = 0.0543 > 1.645(0.0277) = 0.0456
b) What if she had seen 13.3% based on a sample
of 300 crabs? Would the 90% C.I. be wider or shorter
than the one based on 150 crabs?
Shorter because
1.645
.13(1 − .13 )
.13(1 − .13 )
= .0323 < 1.645
= .0456
300
150
(this interval does not include 10% Î based on a
sample of 300 this island is different!!!)
Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION
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Defn: Confidence intervals can be written in the form
point estimate ± MARGIN OF ERROR
where the margin of error (ME) is the product of the
critical value and the standard deviation of the point
estimate.
Suppose the scientist is still planning the fiddler crab
experiment and wants to calculate a 95% confidence
interval with a margin of error of no more than 2.5%.
How big a sample size should she take?
p(1 − p )
= 0.025
Margin of Error (ME) = 1.96
n
Don’t have an estimate for p yet since she hasn’t
sampled yet. She does have a guess though (10%) so
we’ll use that.
.1(1 − .1)
= 0.025 for n.
Now we need to solve 1.96
n
2
1
.
96
⎛
⎞
Get ⎜
⎟ 0.1(1 − 0.1) = n = 553.19 ≈ 554 crabs
⎝ .025 ⎠
are needed.
Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION
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General equation to estimate the needed sampled size
for a specified margin of error (ME) when estimating
a population proportion is:
⎛ z ⎞
~
~
⎟
n = π (1 − π )⎜⎜
⎟
ME
⎠
⎝
*
2
where
• π~ is a guess as to the likely true proportion in
the population (if completely unsure use 0.5)
• z* is the z-score critical value for the confidence
level desired
• ME is the desired margin of error (in decimals)
Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION
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2) INTERVAL ESTIMATION OF THE POPULATION MEAN µ
Recall that we have a CLT for the sample mean so
that if a sample is large enough the sample mean x is
Normally distributed with a mean µ x = µ
And a standard deviation σ x = σ
.
n
⇒ We should be able to construct a confidence
interval estimate of the population mean µ in a way
very similar to what we did for the population
proportion:
i.e. use point estimator ± z × std. dev. of estimator
σ
K
for a 95% C.I. for µ.
i.e. use x ± 1.96
n
Problem?
Almost never know σ either.
Since s is an unbiased estimator of σ, can we use
s
K
instead?
x ± 1.96
n
Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION
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Sort of. If n is really large (> 120, say) then yes.
Other wise we need to use a slightly different C.I.
method.
Use of z-scores in the above Confidence Intervals is
x − µx
x−µ
=
has a
based on the fact that z =
σx
σ n
standard normal distribution when x is normally
distributed.
x − µx x − µ
. In fact to
=
SEM
s n
distinguish it from z-scores, it is called a T-SCORE.
T-scores have what is known as a Student’s
T-distribution with (n-1) degrees of freedom. .
That isn’t true for t =
Defn: STUDENT’S T-DISTRIBUTION
1) is unimodal and symmetric, with equal-sized tails.
2) has a mean of 0
3) has a standard deviation that depends on the
sample size, n, used to calculate x and s.
Hence every sample size has a different t-distribution
which is denoted as t(n-1). n-1 are called the
DEGREES OF FREEDOM (df) for the distribution.
Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION
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T-distributions look like a standard normal
distribution but are more variable, i.e. slightly wider
and flatter. How wider and flatter depends on the
sample size involved. As n increases, the shape
becomes more and more like a Normal distribution.
x−µ
is more variable than z = x − µ but
s n
σ n
how much more depends on the sample size. For
large sample sizes they have almost the identical
frequency distributions!
i.e. t =
T(3 df)
T(15 df)
N(0,1)
-1.96
-2.13
-3.18
+1.96
+2.13
+3.18
Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION
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So, C.I.s for estimating the population mean µ use
the t-score for (n-1) degrees of freedom in place of
the z-score:
CONFIDENCE INTERVALS FOR THE
POPULATION MEAN µ have the general formula
⎛ s ⎞
x ± ( t critical value)⎜
⎟
⎝ n⎠
where the t critical value is based on n-1 df.
This interval is appropriate under the following:
1) sampling is random, and
2) either
a) the sample size is large so we can use the CLT or
b) the original population has a frequency
distribution that is bell-curve shaped
How do we find the t critical values that we need?
Once we do that, how do we use this equation?
Since there are different values for t-score for each
sample size we need an entire table to list them for
Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION
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some typical confidence levels. (Recall that before I
could list the 6 commonly used z values. Now I need
6 numbers for each and every sample size!). The
table is at the end of this topic.
EXAMPLE Suppose we wish to determine if the
average cortex weight for rats is larger than usual
when the rats are raised in an enriched environment.
Assign 10 randomly selected newborn rats to an
enriched environment and raise them to adult stage.
The cortexes had an average weight of x = 565 mg
and a standard deviation of s = 170 mg.
Calculate a 90% C. I. for the true mean cortex weight
of rat raised in enriched environments.
1) since n is small we should check to see if the
frequency distribution of the population sampled is
a bell curve.
• Can’t since we didn’t raise the entire population!
• Can we use the 10 sample values, i.e. do a boxplot
and make sure there are no extreme outliers or
obvious skew (hard with 10 numbers)? Since we
weren’t given the raw data we can’t do this either.
Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION
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• It is not unreasonable to assume that weight is
normally distributed or at least approximately so.
2) n=10 so n -1 = 9
90% confidence level + 9 df gives
t-score = 1.83 (note that it’s larger than z=1.645)
⎛ s ⎞
⎛ 170 ⎞
565
1
.
83
=
±
⎟ = 565 ± 98.4
⎟
⎜
⎝ n⎠
⎝ 10 ⎠
3) x ± t ⎜
“We are 90% confident that the average cortex
weight of rats which are raised in enriched
environments has a value within the interval 467 mg
and 663 mg.”
Suppose the researcher had been interested in
comparing enriched environments to deprived ones.
So, a sample of 10 rats were placed in a deprived
environment but otherwise treated identically. At the
end of the experiment the calculated 90% C.I. for the
deprived rats was 389 mg to 460 mg.
Is this sufficient evidence to argue that an enriched
environment increases cortex weights?
Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION
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Sort of. The comparison is valid BUT what is your
confidence level now? Not possible to know exactly
when you do this type of comparison using 2
confidence intervals. (it’s actually around 82% not
90%)
We’ll learn later a better method for comparing 2
populations which will also give us a way of
determining our confidence in the method.
EXAMPLE As part of a larger experiment on the use
of exercise to reduce stress and susceptibility to
illness, researchers gathered baseline data on the
levels of Human Beta-Endorphin (HBE) in sedentary
people. Higher levels of HBE are associated with
higher stress levels.
They randomly selected 25 people getting yearly
physicals at a nearby hospital and measured HBE
levels (pg/ml). The sample mean level was 42.5 and a
standard deviation of 16.8 pg/ml. Construct a 95%
CI of the population mean HBE level.
Population: HBE levels in all sedentary people?
Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION
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Sample: HBE levels in 11 people at a nearby hospital
Can we use the equation we learned? Don’t know the
distribution of the original population but we will
assume that it is not too skewed. Then n=25 should
be larger enough for the CLT to hold.
Calculations:
For n − 1 = 24 and 95% confidence: t = 2.06
⎛ s ⎞
⎛ 16.8 ⎞
Then x ± t ⎜
⎟ = 42.5 ± 6.8
⎟ = 42.5 ± 2.06⎜
⎝ n⎠
⎝ 25 ⎠
Conclusion: “We are 95% confident that the average
HBE level in sedentary people is 42.8±6.8 pg/ml.”
QUESTION: Suppose I had decided that I could
report an 80% confidence interval rather than a 95%
interval.
A) which interval would be wider?
B) which interval is more likely to include the true
mean that I am trying to estimate?
Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION
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Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION
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Example of Mis-Statements and Some Computations
Using a N(0,1) Random Variable
Distributions
N(0,1)
-3
-2
-1
Quantiles
100.0% maximum
99.5%
97.5%
90.0%
75.0% quartile
50.0% median
25.0% quartile
10.0%
2.5%
0.5%
0.0%
minimum
0
3.635
2.563
1.949
1.298
0.677
0.012
-0.678
-1.297
-1.976
-2.506
-3.447
1
2
3
Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION
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Take a sample of size 15 from this population and
calculate a 95% CI using the method we just learned:
-0.3272061, -2.3610111, -0.7944772, -1.6243058,
1.34490057, -0.6010952, -0.8823261, 0.50218859,
0.29919709, -0.5588271, 0.36830855, 0.46442333,
-0.3233998, 0.5258819, 0.28472816
-2
-1
Moments
Mean
Std Dev
Std Err Mean
upper 95% Mean
lower 95% Mean
n
0
1
-0.245535
0.9435046
0.2436118
0.2769608
-0.76803
15
Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION
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So, the 95% confidence interval is (-0.768, +0.277).
Correctly Stated Interpretation:
1) We are 95% confident that the population mean,
µ, has a value that is between -0.768 and +0.277.
2) There’s a 5% chance that our interval will not
cover the population mean, µ.
It’s the probability that the method results in a CI that
covers the true parameter value (µ) that is being
addressed here, not the particular sample NOR the
population values.
For example, suppose we were able to take 200
samples of 15 values from the population. For each
one we calculate x , s , and a 95% CI. Now, since we
know a N(0,1) random variable has a mean of µ = 0 ,
we can determine whether the CI covers the value of
0 or not.
Using JMP, I simulated exactly the repeated
sampling and got 9 out of the 200 (4.5%) did not
cover µ = 0 . The next table shows some of the
results.
Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION
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Sample
Mean
Sample
Standard
Deviation
95% CI
Lower
Bound
95% CI
Upper
Bound
CI Cover
µ=0?
-1.1049107
0.84677211
-2.1310767
1.24204895
1.79642982
2.10221624
-1.1471177
1.08150832
-1.1533823
-0.3213116
-1.5549539
0.36021018
1.25636375
-1.9770196
0.54086785
-1.5792089
2.1220965
-0.8529584
0.41356674
0.55320446
2.16646888
1.35168506
-0.7047083
-0.0917601
0.32999897
0.51051004
0.21348212
-1.3225274
-0.4963301
0.25010729
-1.2455287
-0.5720726
2.26894349
0.31171105
-0.2931029
0.25012456
-0.7918701
-0.0735042
0.85730542
0.5527533
0.42753442
0.33801653
0.78304642
0.3464639
0.75434897
0.16443923
0.2669313
0.36706328
0.35510663
1.24317913
2.0860707
0.99486637
1.03681304
1.36476692
1.02076565
1.17371463
1.13639355
0.91286459
2.56664539
1.26726012
1.23548621
1.2013229
0.97791792
1.59662907
1.29138324
0.41676529
1.01401071
1.28280408
1.29482828
0.97027273
2.53712976
0.98406263
1.47509377
2.02997874
0.46674714
1.75337152
1.30972571
1.94757715
1.16635039
0.57662726
-2.0218808
0.12179875
-3.8105442
0.49895779
0.17851219
1.74952917
-1.7196284
0.29423588
-1.9150103
-2.9876656
-6.0291305
-1.773566
-0.9673791
-4.9041535
-1.6484567
-4.0965765
-0.3152253
-2.8108583
-5.0913401
-2.1647982
-0.4833855
-1.2248963
-2.8021336
-3.5161889
-2.4397426
-0.3833626
-1.9613545
-4.0738685
-3.2734606
-1.8309207
-6.6871308
-2.682677
-0.894818
-4.0421603
-1.294176
-3.5104833
-3.6009524
-4.2506417
-1.6442674
-0.6839892
-0.1879405
1.57174548
-0.4516092
1.98514011
3.41434745
2.4549033
-0.574607
1.86878077
-0.3917544
2.34504248
2.91922279
2.49398633
3.48010655
0.95011428
2.73019243
0.93815856
4.55941826
1.10494137
5.91847361
3.2712071
4.81632325
3.92826642
1.39271708
3.33266868
3.09974055
1.40438269
2.38831878
1.42881374
2.2808003
2.33113532
4.19607344
1.53853185
5.43270497
4.66558242
0.70797014
4.01073245
2.01721219
4.10363339
3.35887822
1.78949576
No
No
No
No
No
No
No
No
No
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION
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0.54496347
0.26360623
-1.7795686
1.76889744
1.36987175
0.15408404
0.61575948
1.75964955
-0.032808
0.29596459
0.71309248
0.14647825
-1.5130959
0.25016094
0.34062799
0.38727695
-0.6162605
0.7944431
0.00054833
-1.2182523
-0.3770921
-0.3548987
0.18431648
0.17146672
2.26729026
0.87556293
1.11354895
-1.2177139
0.24642509
0.2851893
1.26836219
-1.0806759
-1.0910837
-0.7211335
0.19407641
0.28889596
-0.1802052
-0.6870411
-0.2942918
-0.7288637
-1.3148595
1.29699205
-0.5270651
0.09414593
2.34976751
1.48932661
1.00724524
1.39430358
2.02759735
1.26877604
0.85684775
1.81143194
1.0357546
0.75204033
0.69665007
0.7872054
1.04700543
1.02961537
1.21204694
1.05993052
2.27295095
0.77982765
1.61345228
2.40089814
0.66754035
1.14986562
2.13065507
1.01970509
1.15701845
1.69744134
1.98171209
1.61473751
1.06505164
1.53116947
1.20047293
2.21306374
2.19968934
1.28751128
0.88624608
1.25641273
1.35738017
1.78051269
1.42903957
1.34804592
0.7970715
1.37794775
1.48434652
1.73039274
-4.4947866
-2.9306817
-3.9398947
-1.2215863
-2.9788921
-2.5671699
-1.2219962
-2.1254856
-2.2542807
-1.3170015
-0.7810733
-1.5419094
-3.7586992
-1.9581444
-2.2589541
-1.8860479
-5.4912554
-0.8781209
-3.4599626
-6.3676666
-1.8088238
-2.8211151
-4.3854841
-2.0155832
-0.2142675
-2.7650867
-3.1368008
-4.6809814
-2.0378835
-2.9988426
-1.3063962
-5.8272255
-5.8089481
-3.4825706
-1.7067324
-2.4058413
-3.0914961
-4.505861
-3.3592768
-3.6201346
-3.0244079
-1.6584119
-3.7106718
-3.6171774
5.58471354
3.45789412
0.38075764
4.75938119
5.71863555
2.875338
2.45351513
5.64478466
2.18866463
1.90893068
2.20725829
1.83486592
0.73250742
2.45846629
2.94021013
2.66060182
4.25873447
2.46700707
3.46105932
3.93116209
1.05463954
2.1113178
4.7541171
2.35851664
4.74884803
4.51621252
5.36389866
2.24555357
2.53073368
3.56922119
3.84312054
3.66587375
3.62678073
2.04030352
2.0948852
2.98363327
2.73108572
3.13177879
2.77069327
2.16240729
0.39468881
4.25239604
2.65654154
3.80546925
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
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Yes
10-
Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION
26
1.14444656
…
…
1.65451125
-0.1351413
-1.8632532
-0.5569952
-2.2413172
-0.0982971
0.36542458
-0.3203501
-1.2338743
0.40171205
-0.8315755
0.97514472
0.16588811
-0.5426113
0.00646725
1.18488525
1.10724653
1.30641932
2.00831787
0.88297557
1.75641268
1.60158032
2.2890512
2.11831252
1.29986472
1.42429477
0.66079362
0.72570411
0.81747025
0.75995975
1.27561648
0.72405208
0.80901499
-1.3968796
-3.4763009
-3.2327707
-2.6529022
-2.0289355
-5.6303838
-3.9920433
-7.1508438
-4.6416256
-2.422508
-3.3751586
-2.6511356
-1.1547685
-2.5848748
-0.6548068
-2.5700371
-2.0955485
-1.7286973
3.68577266
1.27331435
2.37121085
5.96192468
1.75865294
1.90387729
2.878053
2.66820932
4.44503137
3.15335712
2.73445833
0.18338709
1.95819256
0.92172379
2.60509628
2.90181336
1.01032599
1.74163183
10-
Yes
Yes
Yes
Yes
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Yes
Yes
Must be very careful that you state the interpretation
correctly.
Mis-statement #1: we are confident that 95% of the
data fall between -0.768 and +0.277.
Here, only 4 out of 15 (26.67%) fall within that
interval!
-0.3272061, -2.3610111, -0.7944772, -1.6243058,
1.34490057, -0.6010952, -0.8823261, 0.50218859,
0.29919709, -0.5588271, 0.36830855, 0.46442333,
-0.3233998, 0.5258819, 0.28472816
Topic (10) – ESTIMATION OF PARAMETERS OF A SINGLE POPULATION
27
10-
Mis-statement #2: we are confident that 95% of the
population falls between -0.768 and +0.277.
Let’s calculate the Pr(-0.768 < Z < +0.277), i.e. the
probability that a randomly selected value falls in the
interval (or equivalently the proportion of the
population values falling within the interval).
Pr(-0.768 < Z < +0.277)
= Pr(Z < +0.277) - Pr(Z < -0.768)
= 0.6091 - 0.2212
= 0.3879
Here, only 39% of the population values fall between
-0.768 and +0.277.