Estimation
CHAPTER 8
Table of contents
2
3
Properties of
Point and Interval
Estimators
Estimation for
Single Population
(Mean and
Proportion)
1
Basic
Concepts
4
5
Independent
vs Related
Sampling
Estimation for
Two Populations
(Difference
between Means
and Proportions)
Recall
Population vs. Sample
• The POPULATION is the collection of all
elements under consideration in a study.
POPULATION
• The SAMPLE is a subset of the population.
• The PARAMETER is a summary measure
describing a specific characteristic of the
population.
• The STATISTIC is a summary measure
describing a specific characteristic of the
sample.
SAMPLE
Recall
Inferential Statistics
Includes methods by which one uses sample information to make
inferences or generalizations about a population.
Example
Problem:
There is an unknown population parameter, and we want to
make an educated guess on its value using sample data.
Example:
What is the mean income of all Filipino families using a sample
of 1000 families in Metro Manila?
Two Areas of Inferential Statistics
Estimation
inference about a parameter is made by finding a
single value or a range of values computed from the
sample data that may be used to make a statement
about the unknown value of the parameter
Hypothesis
Testing
inference about a parameter is made by assessing
whether the sample data support an assertion
made about the true value of the parameter
Two Major Types of Estimators
Point
Estimator
Interval
Estimator
Point
Estimation
Point Estimator
a single statistic whose realized value is used to estimate the true
but unknown value of the population parameter
The realized or computed value using the estimator is called a
point estimate.
An interval estimate conveys more information than a point estimate. In interval
estimation, we look at the length of the interval and the specified level of confidence.
Common Point Estimators for Parameters under
Random Sampling from an Infinite Population
Parameter
Estimator
Mean: µ
1 𝑛
Sample Mean: 𝑋ሜ = σ𝑖=1 𝑋𝑖
𝑛
Standard Error
of the Sample Mean:
σ
𝑠𝑒 𝑋ሜ =
n
Proportion: p
Standard Error of the
Sample Proportion
෠ =
𝑠𝑒(𝑃)
𝑝(1 − 𝑝)
𝑛
Estimator of Standard Error
of the Sample Mean:
𝑆
෣
ሜ =
𝑠𝑒(
𝑋)
𝑛
Sample Proportion: 𝑝Ƹ
Estimator of Standard Error
of the Sample Proportion
෣
෠ =
𝑠𝑒(𝑃)
෠ − 𝑃)
෠
𝑃(1
𝑛
Interval
Estimation
Interval Estimator
a rule that tells us how to calculate two numbers based on sample
data that will form an interval within which we expect the
population parameter to lie with a specified degree of confidence
The realized pair of numbers computed from this estimator, say
(a,b), is called an interval estimate or confidence interval
estimate.
An interval estimate conveys more information than a point estimate. In interval
estimation, we look at the length of the interval and the specified level of confidence.
(1-α) 100% Confidence Interval (CI) Estimate
The fraction (1-α) in a (1-α)100% CI estimate is called the confidence
coefficient, and the endpoints are called the lower and upper
confidence limits.
The length of the interval is defined as the difference between the
upper and lower confidence limits.
Remarks
• The researcher chooses the value of α.
• Naturally, we choose a value for 𝛼 that is close to 0 so that we can
be more confident about our inference.
• Common choices for α are 0.10, 0.05, and 0.01.
• Smaller α means higher the confidence coefficient.
• The confidence coefficient is NOT the probability that the true
value of the parameter falls in the interval estimate.
• Once a sample is drawn and the confidence interval is constructed,
the resulting interval estimate encloses the true value of the
parameter, or it doesn’t.
Interpretation
If we take repeated samples of size n and if for each
one of these samples we compute for the (1-α)100% CI,
then (1-α)100% of the resulting confidence intervals will
contain the unknown value of the parameter.
Example
Suppose we get 10 samples
of the same size and from
the same population. For
each sample, we then
compute a 90% confidence
interval estimate for the
mean.
Common Interpretation
We are (1-α)100% confident that the (population
parameter) is inside the interval (a,b).
Remarks
• A good confidence interval estimate is one that is as narrow as
possible and has a large confidence coefficient.
• For a fixed sample size n, as the confidence coefficient increases,
the length of the interval also increases.
• For a fixed n, as the confidence coefficient (1- α) increases, the
length of the interval increases.
• For a fixed confidence coefficient (1- α), as n increases, the length
of the interval decreases.
Table of contents
2
3
Properties of
Point and Interval
Estimators
Estimation for
Single Population
(Mean and
Proportion)
1
Basic
Concepts
4
5
Independent
vs Related
Sampling
Estimation for
Two Populations
(Difference
between Means
and Proportions)
Some Desirable Properties of a Point Estimator
Unbiased
Reliable
Efficient
“Good” point estimator
Unbiased Estimator
An estimator is said to be unbiased for the parameter being
estimated if the average of the estimates it produces under
repeated sampling from the same population is equal to the true
value of the parameter being estimated.
This means that the possible estimates computed from the different samples
are all centered around the true value of the parameter.
Unbiased Estimator
T(X) → θ
(estimator)
(parameter it
is estimating)
E(T) = θ
E(estimator) = true value of parameter it is estimating
Remarks
ഥ is an unbiased estimator of μ under
In general, the sample mean 𝑿
simple random sampling (whether with or without replacement) .
Also, under SRSWR, the sample variance, S2 is an unbiased estimator
of the population variance σ2.
However, the sample standard deviation, S, is a biased estimator of
the population standard deviation σ. However, the bias becomes
smaller or insignificant for large sample sizes.
ത
𝐸(𝑋)=𝜇
𝐸(𝑆 )= σ2
2
𝐸(𝑆)≠𝜎
Illustration
Population data={21,23,30,34}. Select a sample of size n=2 from this population using
SRSWR. Show that S2 is unbiased for σ2 but S is biased for σ. We have the following list of all
possible samples of size 2 and computed values of S2 and S.
Sample Data
(x1, x2)
S2
Sample Data
(x1, x2)
S
Sample #
S2
S
1
21
21
0
0
9
23
30
24.5
24.5
2
23
23
0
0
10
30
23
24.5
24.5
3
30
30
0
0
11
21
30
40.5
40.5
4
34
34
0
0
12
30
21
40.5
40.5
5
21
23
2
2
13
23
34
60.5
60.5
6
23
21
2
2
14
34
23
60.5
60.5
7
30
34
8
8
15
21
34
84.5
84.5
8
34
30
8
8
16
34
21
84.5
84.5
Sample #
Illustration
From the population data, we have
𝜎2 =
Now,
σ4𝑖=1(𝑋𝑖 −27)2
4
= 27.5 and 𝜎 = 27.5 = 5.244.
𝐸(𝑆 2 ) = (0)(4/16) + (2)(2/16) + (8)(2/16) + (24.5)(2/16)
+(40.5)(2/16) + (60.5)(2/16) + (84.5)(2/16) = 27.5
and
𝐸(𝑆) = 0 (4/16) + 2 (2/16) +
+ 40.5 (2/16) + 60.5 (2/16) +
8 (2/16) + 24.5 (2/16)
84.5 (2/16) = 4.066
Unbiased Estimator
ഥ , the sample proportion is
ෝ of X
Since p is a special case of µ, and p
an unbiased estimator of the population proportion under random
sampling from an infinite population which includes SRSWR.
It is also unbiased under SRSWOR.
Remarks
• An unbiased estimator is not necessarily the “best” estimator of a
parameter.
• A parameter may have more than one unbiased estimator.
• Among all possible unbiased estimators, we would naturally prefer
to use the one with the smallest variance or smallest standard
error.
Reliability
Our measure of reliability is the standard error of the estimator.
We want the standard error of our unbiased estimator to be small
because this implies that the estimates computed from the
different samples are close to the true value it is estimating.
Unbiasedness + Reliability
Unbiased and
Reliable
Unbiased but
Unreliable
Biased but
Reliable
Biased and
Unreliable
Note: The smallest circle at the center is the true value of the parameter of
interest.
Unbiasedness + Reliability
Note: The smallest circle at the center is the true value of the parameter of
interest.
Most Efficient Estimator
An unbiased estimator of a parameter with the smallest variance
among all the other unbiased estimators is called the most
efficient estimator.
In fact, the sample mean is the most efficient estimator when
sampling from the normal distribution.
Table of contents
2
3
Properties of
Point and Interval
Estimators
Estimation for
Single Population
(Mean and
Proportion)
1
Basic
Concepts
4
5
Independent
vs Related
Sampling
Estimation for
Two Populations
(Difference
between Means
and Proportions)
Common Point Estimators for Parameters under
Random Sampling from an Infinite Population
Parameter
Mean: µ
Standard Error
of the Sample Mean:
σ
ሜ
SE(X)=
n
Estimator
1 𝑛
ሜ
Sample Mean: 𝑋 = σ𝑖=1 𝑋𝑖
𝑛
Estimator of Standard Error
of the Sample Mean:
𝑆
ሜ =
𝑠𝑒(𝑋)
𝑛
Common Point Estimators for Parameters under
Random Sampling from an Infinite Population
Parameter
Proportion: p
Standard Error of the
Sample Proportion
෠ =
𝑆𝐸(𝑃)
𝑝(1 − 𝑝)
𝑛
Estimator
Sample Proportion: 𝑝Ƹ
Estimator of Standard Error
of the Sample Proportion
෠ =
𝑠𝑒(𝑃)
෠ − 𝑃)
෠
𝑃(1
𝑛
(1-α) 100% Confidence Interval (CI) Estimate for 𝝁
Suppose we have a random sample (X1, X2, …, Xn) taken from a normally distributed
population with mean, μ, and variance, σ2. This means that ALL Xi’s are independent and
normally distributed random variables with the same μ and σ2.
Cases
Case 1: σ2 is known
Case 2: σ2 is unknown and n ≤ 30
Case 3: σ2 is unknown and n > 30
Confidence Interval Estimators
𝜎
𝜎
𝑋ത − 𝑍𝛼 ∗
, 𝑋ത + 𝑍𝛼 ∗
𝑛
𝑛
2
2
𝑠
𝑠
𝑋ത − 𝑡𝛼,𝑛−1 ∗
, 𝑋ത + 𝑡𝛼,𝑛−1 ∗
𝑛
𝑛
2
2
𝑠
𝑠
𝑋ത − 𝑍𝛼 ∗
, 𝑋ത + 𝑍𝛼 ∗
𝑛
𝑛
2
2
Remarks
• The first two formulas were derived using the sampling
distributions discussed in Chapter 7.
• Formula 3 is based on the fact that as the degrees of freedom
approaches infinity, the t-distribution will approach the standard
normal distribution. This formula is useful only when the values of
t are not available for large n. In fact, most statistical software will
use formula 2 whenever 2 is unknown since these software can
determine the value of t for any degrees of freedom.
Remarks
• All 3 formulas will still provide us with good approximate (1)100% confidence intervals for  even when the parent
population is not normal (including those that are badly skewed
and those that are discrete), provided that the sample size is
sufficiently large. In many cases, n>30 works well. This result is
attributed to the Central Limit Theorem.
Example
We have the price of brown sugar per kilo based on a sample of 8
grocery stores:
20.50 21.25 19.95 22.50 20.00 22.75 23.50 21.75
Assuming that the price of brown sugar is normally distributed and
that we have a random sample, compute for a 95% confidence
interval estimate for =mean price of brown sugar.
Compute for a 99% confidence interval as well.
(1-α) 100% Confidence Interval (CI) Estimate for 𝒑
If the population proportion is not expected to be too close to 0 or 1,
and the sample size n is large, then an approximate 1 − α 100%
confidence interval estimator for the population proportion, p, is
given by
𝑃෠ − 𝑧𝛼/2
෠ − 𝑃)
෠
෠ − 𝑃)
෠
𝑃(1
𝑃(1
, 𝑃෠ + 𝑧𝛼/2
𝑛
𝑛
where 𝑧𝛼Τ2 is the 100 1 − 𝛼 Τ2 th percentile of the standard normal
distribution, 𝑃෠ is the sample proportion, and 𝑛 is the sample size.
Notes
The sample size n must be large because this formula is based on the
Central Limit Theorem, where the population proportion is viewed as
the mean of the Xis where
1 if ith element possesses characteristic of interest
𝑋𝑖 = ቐ
th
0 if i element does not possess characteristic of interest
Clearly, Xi is discrete and cannot come from a normal distribution.
As the sample size n goes to infinity, the length of the interval goes to
0.
Illustration
Suppose we have 10 students. Define 𝑋𝑖 as follows:
1, if the ith student passed Stat 101 on his or her first take
Xi = ቊ
0, if the ith student did not pass Stat 101 on his or her first take
We have the following dataset
𝑖
1
2
3
4
5
𝑋𝑖 𝑖 𝑋𝑖
1 6 1
1 7 0
0 8 1
1 9 1
1 10 1
σ10
𝑋𝑖
8
𝑖=1
𝑋ത = 𝑃෠ =
=
= 0.8 = 80%
10
10
Notes
𝑛
σ
• if we define 𝑌 = 𝑖=1 𝑋𝑖 , where the 𝑋𝑖 s are independent, then
𝑌~𝐵𝑖𝑛𝑜𝑚𝑖𝑎𝑙(𝑛, 𝑝)
• By the Central Limit Theorem,
𝑝 1−𝑝
𝑃෠ ≈ 𝑁𝑜𝑟𝑚𝑎𝑙(𝑝,
) if 𝑛𝑝 ≥ 5 and 𝑛(1 − 𝑝) ≥ 5
𝑛
Example
According to a 1984 American study, about one in three individuals feels
shopping is an unpleasant experience (Journal of Marketing Research,
February/March 1984). Suppose we take a national sample of 4,100 Filipino
male and female adults, and we determine each respondent’s opinion on the
pleasantness of shopping. The survey produced the following results:
Sample Size
Number who think shopping is an unpleasant
experience
Males
2,015
850
Females
2,085
570
Compute a 90% confidence interval for the proportion of males in the
sample who think shopping is an unpleasant experience.
Example
We are 90% confident that the interval 40.4% to 44.0% contains the
population proportion of males who find shopping an unpleasant
experience.
Can we say that less than half of the men find shopping unpleasant?
Margin of Error
The margin of error, denoted by e, is the upper bound on the
absolute difference between the estimator and the parameter called
the error of estimation, though there is an associated risk  of
selecting a sample that yields an estimate whose error of estimation
is greater than this upper bound, e.
Margin of Error
• By definition, when we use 𝑋ሜ to estimate , the margin of error, e,
satisfies the condition: 𝑃(|𝑋ሜ − 𝜇| > 𝑒) = 𝛼, or equivalently, 𝑃(|𝑋ሜ −
𝜇| ≤ 𝑒) = 1 − 𝛼 . Likewise, when we use 𝑃෠ to estimate p, the
margin of error, e, satisfies the condition: 𝑃(|𝑃෠ − 𝑝| > 𝑒) = 𝛼
•
𝜎
𝜎
𝜎
𝑃(𝑋ሜ − 𝑧𝛼/2 < 𝜇 < 𝑋ሜ + 𝑧𝛼/2 ) = 1 − 𝛼 means that 𝑒 = 𝑧𝛼/2
𝑛
𝑛
𝑛
• 𝑃 𝑃෠ − 𝑧𝛼
2
𝑝 1−𝑝
𝑛
2
𝑝 1−𝑝
𝑛
𝑧𝛼/2
𝑝(1−𝑝)
𝑛
< 𝑝 < 𝑃෠ + 𝑧𝛼
= 1 − 𝛼 means that 𝑒 =
Remarks
• We can use the margin of error to describe the “goodness” of the
point estimate. The smaller the margin of error, the better.
• The margin of error usually involves unknown parameters. In this
case, we use sample data to estimate these unknown parameters.
• In estimating the mean or proportion, the margin of error is the
term that we add/subtract to the point estimate to compute
the (1-α)100% confidence interval estimate. It is a function of 2
factors: a) standard error and b) the risk 𝛼.
Example
Based on a multistage probability sample of 1,800 likely voters 18 years old
and above, Pulse Asia’s nationwide survey has a ± 2.3% margin of error at
95% confidence.
Poe is “statistically tied” with
Cayetano. Cayetano is “statistically
tied” with Go and Poe. That is, their
respective CIs overlap. In other
words, their respective differences
are within the margin of error.
Table of contents
2
3
Properties of
Point and Interval
Estimators
Estimation for
Single Population
(Mean and
Proportion)
1
Basic
Concepts
4
5
Independent
vs Related
Sampling
Estimation for
Two Populations
(Difference
between Means
and Proportions)
Independent vs Related Sampling
Independent
Sampling
Related or
Paired
Sampling
Independent Sampling
If the selection of the sample from the first population is
independent of the selection of the sample from the second
population, then we are taking independent samples from the two
populations.
Population 1
𝜇𝑋 , 𝜎𝑋2
Sample 1 of size n1
(X1 , X2 , ..., Xn1 )
ത 𝑠𝑋2
𝑋,
Use these
samples
to infer on
μX –μY
Population 2
𝜇𝑌 , 𝜎𝑌2
Sample 2 of size n2
(X1 , X2 , ..., Xn2 )
ത 𝑠𝑌2
𝑌,
Method 1:
random samples are taken
separately from two
populations and the same
response variable is
recorded for everyone
Example: We want to know whether boys
are taller than girls on the average.
We randomly selected 10 boys and 10
girls from an elementary school, and we
get the height of the 20 participants.
Experimental unit: child
Response variable: height of a child
1 Treatment: measuring the height of the
children
Xi = height of the girls
Yi = height of the boys
Parameter of interest: X - Y
Extraneous variables: age, year level
Method 2:
participants are randomly
assigned to one of two
treatment conditions and
the same response variable
is recorded for each
individual unit
Example: A researcher wants to determine the
effect of music in the growth of plants by
subjecting a sample of plants to music twice a
day and another sample receives no music.
Experimental unit: plant
Response variable: height of the plant
2 Treatments: playing music twice a day
Xi = height of the plants that received music
twice a day
Yi = height of the plants that did not receive
music
Parameter of interest: X - Y
Extraneous variables: type of soil, amount of
sunlight
Related / Paired Sampling
If the selection of the sample from the first population is related (in
any of its forms) to the selection of the sample from the second
population, then we are taking related or paired samples from the
two populations.
Population 1
𝜇𝑋 , 𝜎𝑋2
X1
X2
…
Xn
Sample
D1 = X1 – Y1
D2 = X2 – Y2
...
Dn = Xn – Yn
Population 2
𝜇𝑌 , 𝜎𝑌2
Y1
Y2
…
Yn
Methods of Generating Paired Data
Paired data: {(X1,Y1), (X2,Y2), …, (Xn,Yn)}
Forming paired data will be beneficial when the two measures in
the ith pair, Xi and Yi, exhibit strong direct relationship so that when
Xi is high then so is Yi as a result of sharing the same values on the
extraneous variable/s.
Method 1:
all experimental units in
the sample receive both
treatments
Example: Two formulations of a new whitening soap
are to be compared as to their whitening effect. A
random sample of 40 potential users of the soap is
selected.
Each person uses a randomization
mechanism to determine which formulation is
applied on the left arm, so that the other formulation
is applied on the right arm. After two weeks, they
measured the effect of each formulation.
Experimental unit: person
Response variable: degree of fairness of a person
2 Treatments: Formulation A and Formulation B of
whitening soap
Xi = degree of fairness of arm of ith person where
Formulation A was applied
Yi = degree of fairness of arm of ith person where
Formulation B was applied
Parameter of interest: X - Y
Extraneous variables: original degree of fairness of
person, biological characteristics that affect person’s
reaction to any treatment
Method 2:
taking measurements
before and after the
treatment is applied to
the experimental units
(can be viewed as a
special case of Method 1)
Example: A police department wants to assess
the effects of an obvious radar trap on the
speeds of cars. Ten cars are randomly selected
on a highway, and their speeds are measured
just before a radar trap comes into view and
right after they pass the obvious radar trap.
Experimental unit: car
Response variable: speed of car
2 Treatments: visible radar trap
Xi = speed of ith car before seeing radar trap
Yi = speed of ith car after seeing radar trap
Parameter of interest: X - Y
Extraneous variables: driver, type of car, age of
car, etc.
Method 3:
use naturally occurring
pairs such as twins, or
husbands and wives, or
siblings, etc.
Example: A gym instructor measures the effect
of an exercise in lowering blood pressure
among twins by making one twin exercise four
times a week while the other twin has no
exercise.
Experimental unit: person
Response variable: blood pressure
2 Treatments: exercise
Xi = blood pressure of ith person with exercise
Yi = blood pressure of ith person without
exercise
Parameter of interest: X - Y
Method 4:
form pairs of
experimental units that
have the same values or
levels of the extraneous
variable
Example: A science teacher has developed new teaching
materials and wants to evaluate the effectiveness of these
materials in improving the students’ comprehension. Prior
to sampling, the teacher formed pairs of students so that
students belonging in the same pair received about the
same final grade in science the previous term. The
teacher then selected a sample of pairs of students. The
teacher randomly selects which student in each pair will
be taught using the new materials so that the other one
will be taught using the old materials. At the end of the
term, all the students in the sample were given a
standardized test.
Experimental unit: student
Response variable: score in standardized test
2 Treatments: old and new teaching materials
Xi = score of ith student using the old teaching materials
Yi = score of ith student using the new teaching
materials
Parameter of interest: X - Y
Extraneous variables: aptitude in science
Example #1
Identify whether the samples are independent or related:
The principal of a school wishes to determine if the Grade 6 boys are
better in mathematics than the Grade 6 girls. A random sample of
boys were selected. Then a random sample of girls were selected. All
of the students in the two random samples were asked to take a
standardized test in mathematics and their scores were determined.
Example #2
Identify whether the samples are independent or related:
A tire manufacturer is testing two new tread designs in terms of
stopping distance. To do this, the company uses two test cars driving
side by side at the same speed. Both cars have automatic braking
systems so that both sets of brakes engage simultaneously on a
signal. Then the stopping distances for both cars are measured after
repeating the experiment 10 times.
Example #3
Identify whether the samples are independent or related:
A company which does a large volume of business by mail decides to
test whether there is a difference in mail delivery between those
items brought to a post office as compared to those put in a corner
mailbox. A random sample of 100 customers from the same city were
selected and their letters were mailed from the post office. Another
random sample of 100 customers were then selected but their letters
were sent from the corner mailbox.
Table of contents
2
3
Properties of
Point and Interval
Estimators
Estimation for
Single Population
(Mean and
Proportion)
1
Basic
Concepts
4
5
Independent
vs Related
Sampling
Estimation for
Two Populations
(Difference
between Means
and Proportions)
Notations
1st Population (X) 2nd Population (Y)
Population mean
𝜇𝑋
𝜇𝑌
Population Standard
Deviation
𝜎𝑋
𝜎𝑌
Sample mean
𝑋ത
𝑌ത
Sample Standard
Deviation
𝑆𝑋
𝑆𝑌
Proportion
𝑝1
𝑝2
Common Point Estimators for the Difference of Two
Parameters under Random Sampling from an Infinite
Population
Parameter
µX - µY
P1 – P2
Estimator
𝑋ሜ − 𝑌ሜ
𝑝Ƹ1 − 𝑝Ƹ 2
(1-α) 100% CI Estimate for 𝝁𝑿 − 𝝁𝒀 for Two Independent Samples
Let 𝑋1 , 𝑋2 , … , 𝑋𝑛1 be a random sample from a Normal population with mean 𝜇𝑋 and variance 𝜎𝑋2
Let 𝑌1 , 𝑌2 , … , 𝑌𝑛2 be a random sample from a Normal population with mean 𝜇𝑌 and variance 𝜎𝑌2
ഥ and 𝒀
ഥ denote the sample mean and 𝑆𝑋2 and 𝑆𝑌2 denote the sample variance of the two random samples, respectively.
Let 𝑿
Cases
2
2
Case 1: 𝜎𝑋 and 𝜎𝑌 are
known
2
2
Case 2: 𝜎𝑋 and 𝜎𝑌 are
2
2
unknown but 𝜎𝑋 = 𝜎𝑌
Confidence Interval Estimators
𝑋ത − 𝑌ത − 𝑍𝛼
2
𝑋ത − 𝑌ത − 𝑡𝛼,𝑛 +𝑛 −2
2 1
2
𝑆𝑝2
𝜎𝑋2 𝜎𝑌2
𝜎𝑋2 𝜎𝑌2
+
, 𝑋ത − 𝑌ത + 𝑍𝛼
+
𝑛1 𝑛2
𝑛2
2 𝑛1
1
1
1
1
2
+
, 𝑋ത − 𝑌ത + 𝑡𝛼,𝑛 +𝑛 −2 𝑆𝑝
+
𝑛1 𝑛2
𝑛1 𝑛2
2 1 2
2
2
𝑛
−
1
𝑆
+
𝑛
−
1
𝑆
1
2
𝑋
𝑌
2
𝑆𝑝 =
𝑛1 + 𝑛2 − 2
(1-α) 100% CI Estimate for 𝝁𝑿 − 𝝁𝒀 for Two Independent Samples
Let 𝑋1 , 𝑋2 , … , 𝑋𝑛1 be a random sample from a Normal population with mean 𝜇𝑋 and variance 𝜎𝑋2
Let 𝑌1 , 𝑌2 , … , 𝑌𝑛2 be a random sample from a Normal population with mean 𝜇𝑌 and variance 𝜎𝑌2
ഥ and 𝒀
ഥ denote the sample mean and 𝑆𝑋2 and 𝑆𝑌2 denote the sample variance of the two random samples, respectively.
Let 𝑿
Cases
Confidence Interval Estimators
𝑋ത − 𝑌ത − 𝑡𝛼,𝑣
2
2
2
Case 3: 𝜎𝑋 and 𝜎𝑌 are
2
2
unknown but 𝜎𝑋 ≠ 𝜎𝑌
𝑠𝑋2 𝑠𝑌2
𝑠𝑋2 𝑠𝑌2
+ , 𝑋ത − 𝑌ത + 𝑡𝛼,𝑣
+
𝑛1 𝑛2
𝑛1 𝑛2
2
𝑠𝑋2ൗ
𝑣=
𝑛1
𝑠𝑋2ൗ
𝑛2
2
𝑠𝑌2ൗ
𝑛1
Case 4: 𝜎𝑋2 and 𝜎𝑌2 are
unknown but 𝑛1 >
30 𝑎𝑛𝑑 𝑛2 > 30
𝑛1 − 1
𝑋ത − 𝑌ത − 𝑍𝛼
2
2
𝑠𝑌2ൗ
+
2
𝑛2
+
𝑛2 − 1
𝑠𝑋2 𝑠𝑌2
𝑠𝑋2 𝑠𝑌2
+ , 𝑋ത − 𝑌ത + 𝑍𝛼
+
𝑛1 𝑛2
𝑛2
2 𝑛1
(1-α) 100% CI Estimate
for 𝝁𝑿 − 𝝁𝒀 for Two
Independent Samples
Remarks
• The formulas hold for random samples from a normal
distribution. However, they provide good approximate (1−𝛼)100%
confidence intervals even when the distributions are not that
different from the normal distribution so long as the sample size is
sufficiently large.
• By Central Limit Theorem, you can also use the Case 4 when the
sample sizes are sufficiently large.
• Even if the population variances are considerably different, Case 2
will still provide a good estimate provided that n1=n2 and both
populations are normal. Therefore, in a planned experiment, one
should make every effort to equalize the size of the samples.
Example
Estimate the mean difference of wages between those who work in
an urban setting and those who work in a rural area. Use a 95%
confidence level. Do not assume equal variances between the two
populations.
Example
We are 95% confident that the mean difference of wages between
those who work in an urban setting and those who work in a rural
area is between $92.6 and $188.
Can we say that those who work in an urban area earn more?
Interpretations
(−, +)
(+, +)
(−, −)
If the computed interval estimate contains 0, then we
do not have sufficient evidence to conclude that the
two means are different from each other.
Sometimes positive, sometimes negative…
If the computed interval estimate contains positive
values only then we can conclude with (1-α)100%
confidence that μX is greater than μY.
𝜇𝑋 − 𝜇𝑌 > 0 → 𝜇𝑋 > 𝜇𝑌
If the computed interval estimate contains negative
values only then we can conclude with (1-α)100%
confidence that μX is less than μY
𝜇𝑋 − 𝜇𝑌 < 0 → 𝜇𝑋 < 𝜇𝑌
Example
Suppose that a random sample of 12 plots planted with pechay were
selected from each of two types of plants in a farm in Nueva Ecija. The yields,
in kilograms, of pechay are measured from each of the plots. The
measurements generated are given to be:
Type I
Type II
10.100
8.000
8.700
7.700
7.850
5.400
9.500
3.300
4.900
7.550
9.200
8.300
5.705
7.250
6.450
7.800
5.625
5.750
10.350
6.150
7.450
4.575
8.100
5.000
Obtain a 99% confidence interval estimate for the difference in mean yield of
pechay between the two types of plots. Assume equal variances and the
normality of yield for both types of plots.
Example
We are 99% confident that the mean difference between Type I and
Type II plots is in the interval -1.23 and 2.96.
Does one Type I yield more than the Type II plot?
(1-α) 100% CI Estimate for 𝝁𝑿 − 𝝁𝒀 for Two Paired Samples
Let {(𝑋1 , 𝑌1 ), (𝑋2 , 𝑌2 ), … , (𝑋𝑛 , 𝑌𝑛 )} be your sample data. Denote 𝐷𝑖 = 𝑋𝑖 − 𝑌𝑖 for
i=1, 2, …, n
𝑛
1
ഥ = ෍ 𝐷𝑖 = 𝑋ത − 𝑌ത
𝐷
𝑛
𝑖=1
𝑆𝐷 =
ഥ )2
σ𝑛𝑖=1(𝐷𝑖 − 𝐷
𝑛−1
A 100(1-α)% Confidence Interval Estimator for the Mean of the Differences
𝑆𝐷
𝑆𝐷
ഥ − 𝑡𝛼
ഥ + 𝑡𝛼
𝐷
,
𝐷
,𝑛−1 𝑛
,𝑛−1 𝑛
2
2
Example
Twenty first year college students were divided into 10 pairs, each member of the
pair having approximately the same IQ. One of each pair was selected at random
and assigned to a mathematics section using programmed materials only. The
other member of each pair was assigned to a section in which the professor
lectured. At the end of the semester each group was given the same examination
and the following results were recorded.
Pair
Programmed
Materials
Lectures
𝑑𝑖
1
2
3
4
5
6
7
8
9
10
76
60
85
58
91
75
82
64
79
88
81
-5
52
8
87
-2
70
-12
86
5
77
-2
90
-8
63
1
85
-6
83
5
Find the 98% confidence interval estimate for the true difference in the two
learning procedures. Assume normality.
Example
We are 98% confident that the mean difference between the grades
is in the interval -4.09 and 7.29.
Is one method more effective than the other?
Approximate (1-α) 100% CI Estimate for 𝑷𝟏 − 𝑷𝟐
Given two independent random samples of sizes n1 and n2,
a point estimator for 𝑃1 – 𝑃2 is 𝑝Ƹ1 − 𝑝Ƹ 2
(𝑝Ƹ1 −𝑝Ƹ2 ) − 𝑍𝛼
2
𝑝Ƹ1 1 − 𝑝Ƹ1
𝑝Ƹ 2 1 − 𝑝Ƹ 2
𝑝Ƹ1 (1 − 𝑝Ƹ1 ) 𝑝Ƹ 2 (1 − 𝑝Ƹ 2 )
+
, (𝑝Ƹ1 −𝑝Ƹ 2 ) + 𝑍𝛼
+
𝑛1
𝑛2
𝑛1
𝑛2
2
This approximation will only hold when the sample sizes are large.
Thus, we require the sample sizes n1 ≥ 30 and n2 ≥ 30.
Furthermore, we have the condition that both
P1 and P2 are not expected to be too close to 0 or 1.
Example
In a random sample of 80 females, 40 sleeps in Stat 101. In another
random sample of 120 males, 108 sleeps in Stat 101. Construct a 95%
confidence interval estimate for PM – PF where PF and PM are the
true proportions of females and males, respectively, who sleeps in
Stat 101.
Who sleeps more in class, males or females?
Exercises
Lengths of Children’s Animated Films
The lengths (in minutes) of a random selection of popular children’s
animated films are listed below. Estimate the true mean length of all
children’s animated films with 95% confidence
93 83 76 92 77 81 78 78 76 75 100
Exercises
Television Set Ownership
A survey of 90 families showed that 40 owned at least one television
set. Find the 95% confidence interval of the true proportion of
families who own at least one television set.
Exercises
Length of Life of Light Bulbs
Suppose an electrical firm manufactures light bulbs that have a
length of life that is normally distributed, with a standard deviation of
55 hours.
If a random sample of 27 bulbs has a mean length of life of 1,320
hours, find a 99% confidence interval for the population mean length
of life of all bulbs produced by this firm. Interpret the interval.
Exercises
Running Time of Films
The following data represent the running time of a random sample of
films produced by two motion picture companies:
A
B
103
97
94
82
Time (in minutes)
110
87
98
123
92
175
88
118
Compute a 90% confidence interval for the difference between the
mean running time of films produced by the two companies. Assume
that the running times for each of the companies are approximately
normally distributed with equal variances.
End of Chapter 8
Next: Hypothesis Testing