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END 311E
STATISTICS
Sampling and Sampling Distributions

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Population
census

Are we able to observe all

members of the population?

Motor vehicle
registration

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Sampling
§ Sampling: The process or action of taking samples representing the population
§ Sample: a finite part of a population whose properties are studied to gain
information about the whole
§ Sample Statistics, any value of observed data, especially one used to estimate
the corresponding parameter of the population:

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Why Sample?
• Selecting a sample is less time-consuming than selecting every item in
the population.

• Selecting a sample is less costly than selecting every item in the

population.

• An analysis of a sample is less cumbersome and more practical than an

analysis of the entire population.

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Sampling Distributions
• A sampling distribution is a distribution of all of the possible values of a sample
statistic for a given size sample selected from a population.

• For example, suppose you sample 50 students from your college regarding their mean
GPA. If you obtained many different samples of 50, you will compute a different mean
for each sample. We are interested in the distribution of all potential mean GPA we
might calculate for any given sample of 50 students.

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Developing a Sampling Distribution

• Assume there is a population

D
A B C
• Population size 𝑁 = 4
• Random variable, X, is age of individuals
• Values of X: 18, 20, 22, 24 (years)

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Developing a Sampling Distribution

Summary Measures for the Population Distribution:

μ=
åX i
P(x)
N .3

18 + 20 + 22 + 24 .2
= = 21
4 .1

σ=
å (X - μ)
i
2

= 2.236
18
A
20
B
22
C
24
D
x
N
Uniform Distribution

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Developing a Sampling Distribution

Now consider all possible samples of size n=2

16 Sample Means
1st 2nd Observation
Obs
18 20 22 24
18 18,18 18,20 18,22 18,24 18 20 22 24
20 20,18 20,20 20,22 20,24 18 18 19 20 21
22 22,18 22,20 22,22 22,24 20 19 20 21 22
24 24,18 24,20 24,22 24,24 22 20 21 22 23
24 21 22 23 24
16 possible samples
(sampling with
replacement)
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Developing a Sampling Distribution

Sampling Distribution of All Sample Means

16 Sample Means Sample Means Distribution

_
P(X)
18 20 22 24
.3
18 18 19 20 21
20 19 20 21 22 .2
22 20 21 22 23
.1
24 21 22 23 24
0 _
18 19 20 21 22 23 24 X
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Developing a Sampling Distribution

Summary Measures of this Sampling Distribution:

18 + 19 + 19 + ! + 24
μX = = 21
16

(18 - 21) 2 + (19 - 21)2 + ! + (24 - 21)2

σX = = 1.58
16

Note: Here we divide by 16 because there are 16 different samples of size 2.

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Comparing the Population Distribution to the Sample Means

Distribution

Population Sample Means Distribution

N=4 n=2

μ = 21 σ = 2.236 μX = 21 σ X = 1.58
_
P(X) P(X)
.3 .3

.2 .2

.1 .1

0
18 20 22 24 X
0
18 19 20 21 22 23 24
_
X
A B C D
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Population mean = The mean of all possible samples

Sample mean is an UNBIASED ESTIMATOR for the population mean!!

Suppose we have random samples 𝑋! , 𝑋" , … , 𝑋#

#

𝐸 % 𝑋$ = 𝐸 𝑋! + 𝐸 𝑋" + ⋯ + 𝐸 𝑋# = 𝑛𝜇
$%!

𝜇 𝜇 𝜇

#
1 1
𝐸 𝑋+ = 𝐸 % 𝑋$ = 𝑛𝜇 = 𝜇
𝑛 𝑛
$%!

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Sample Mean Sampling Distribution:

Standard Error of the Mean
• Different samples of the same size from the same
population will yield different sample means
• A measure of the variability in the mean from sample to
sample is given by the Standard Error of the Mean:
(This assumes that sampling is with replacement or
sampling is without replacement from an infinite population)

σ
σX =
n
• Note that the standard error of the mean decreases as the
sample size increases
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Sample variance is a BIASED ESTIMATOR for the population variance!!

Suppose we have random samples 𝑋! , 𝑋" , … , 𝑋#

#

Var % 𝑋$ = Var 𝑋! + Var 𝑋" + ⋯ + Var 𝑋# = 𝑛𝜎 "

$%!

𝜎 𝜎 𝜎

#
1 1 𝜎"
Var 𝑋+ = Var % 𝑋$ = "
𝑛𝜎 " =
𝑛 𝑛 𝑛
$%!

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Sample Mean Sampling Distribution:

If the Population is Normal
• If a population is normal with mean μ and standard
deviation σ, the sampling distribution of X is also
normally distributed with

σ
μX = μ and σX =
n

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Z-value for Sampling Distribution of the Mean

• Z-value for the sampling distribution of : X

( X - μX ) ( X - μ)
Z= =
σX σ
n
where: X = sample mean
μ = population mean
σ = population standard deviation
n = sample size

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Sampling Distribution Properties

μx = μ
Normal Population
Distribution

μ x
(i.e. x is unbiased ) Normal Sampling
Distribution
(has the same mean)

μx
x
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Example
• A cereal firm fills thousands of boxes of cereals during a day.
• To be consistent with the package labeling, boxes should contain 368 grams of
cereal.
• Cereal weight varies from box to box.
• Given that the standard deviation of the cereal-filling process is 15 grams,
• What will the standard error be for a sample contains 25 boxes?

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Example

The standard error is 3.

𝜎 15
= =3 The variation in the sample means for samples
𝑛 25 of n = 25 is much less than the variation in the
individual boxes of cereal

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Example
• What is the probability that the mean of a sample (n=25) being less than 365 gr?

( X - μX ) ( X - μ)
Z= =
σX σ
n

#$%&#$'
𝑍= #
= −1
P{𝑍 < −1} = 0.1587

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Example 𝑃{𝑍 < −1}

𝑃{𝑍 > 1}

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Example

𝑃{𝑍 < 1}

1 − 𝑃{𝑍 < 1}

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Example
• Find the probability that the mean being less than 365 gr using the population
parameters.
#$%&#$'
𝑍= 𝟏𝟓
= −0.2
P{𝑍 < −0.2} = 0.4207

many more individual boxes than

sample means are below 365 grams

the chance that the sample mean of

25 boxes is far away from
the population mean is less than the
chance that a single box is far away!
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Example
taking a larger sample results in
less variability in the sample means from
sample to sample!

• What is the probability that the mean of a sample (n=100) being less than 365

15
𝜎-, = = 1.5
100
#$%&#$'
𝑍= = −2
*.%

P{𝑍 < −2} = 0.0228

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Sampling Distribution Properties

As n increases, Larger sample

size
σ x decreases

Smaller sample
size

μ x
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Determining An Interval Including A Fixed Proportion of the

Sample Means
Find a symmetrically distributed interval around µ
that will include 95% of the sample means when µ
= 368, σ = 15, and n = 25.

• Since the interval contains 95% of the sample

means 5% of the sample means will be outside
the interval
• Since the interval is symmetric 2.5% will be above
the upper limit and 2.5% will be below the lower
limit.
• From the standardized normal table, the Z score
with 2.5% (0.0250) below it is -1.96 and the Z
score with 2.5% (0.0250) above it is 1.96.

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Determining An Interval Including A Fixed Proportion of the

Sample Means
• Calculating the lower limit of the interval
σ 15
XL = μ + Z = 368 + (-1.96) = 362.12
n 25
• Calculating the upper limit of the interval
σ 15
XU = μ + Z
= 368 + (1.96) = 373.88
n 25
• 95% of all sample means of sample size 25 are between 362.12 and
373.88

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Sample Mean Sampling Distribution:

If the Population is not Normal
• We can apply the Central Limit Theorem:
• Even if the population is not normal,
• …sample means from the population will be
approximately normal as long as the sample size is large
enough.

Properties of the sampling distribution:

σ
μx = μ and σx =
n
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𝐸 𝑆! = 𝜎!

∑%"#$ 𝑋" − 𝑋, !
𝐸 = 𝜎 ! ⟹ 𝑘 =?
𝑘
%
1
𝐸[9 𝑋" − 𝑋, ! ] = 𝜎 !
𝑘
"#$

𝑋" − 𝜇 − (𝑋, − 𝜇) 𝑋" − 𝑋,

%
1 !
𝐸[9 𝑋" − 𝜇 − (𝑋, − 𝜇) ] = 𝜎 !
𝑘
"#$

%
1
𝐸[9 𝑋" − 𝜇 !
− 2 𝑋" − 𝜇 𝑋, − 𝜇 + (𝑋, − 𝜇)! ] = 𝜎 !
𝑘
"#$

%
1
9(𝐸[ 𝑋" − 𝜇 ! ] − 2𝐸[ 𝑋" − 𝜇 𝑋, − 𝜇 ] + 𝐸[(𝑋, − 𝜇)! ]) = 𝜎 !
𝑘
"#$

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%
1
9(𝐸[ 𝑋" − 𝜇 ! ] − 2𝐸[ 𝑋" − 𝜇 𝑋, − 𝜇 ] + 𝐸[(𝑋, − 𝜇)! ]) = 𝜎 !
𝑘
"#$

𝜎! The square of the

&!
standard error:
%

%
1 𝜎! 𝒏 constant
(𝑛𝜎 ! + 𝑛 − 2 𝐸[9 𝑋" − 𝜇 𝑋, − 𝜇 ]) = 𝜎 !
𝑘 𝑛 𝒏
"#$

%
1 𝒏
(𝑛𝜎 ! + 𝜎 ! − 2 𝐸[ 𝑋, − 𝜇 9 𝑋" − 𝜇 ]) = 𝜎 !
𝑘 𝒏
"#$

%
1 𝑋" − 𝜇
(𝑛𝜎 ! + 𝜎 ! − 2𝒏𝐸[ 𝑋, − 𝜇 9 ]) = 𝜎 !
𝑘 𝒏
"#$

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%
1 𝑋" − 𝜇
(𝑛𝜎 ! + 𝜎 ! − 2𝒏𝐸[ 𝑋, − 𝜇 9 ]) = 𝜎 !
𝑘 𝒏
"#$

1
(𝑛𝜎 ! + 𝜎 ! − 2𝒏𝐸[ 𝑋, − 𝜇 ! ]) = 𝜎 !
𝑘

𝜎!
𝑛

1 𝜎! 1
(𝑛𝜎 ! + 𝜎 ! − 2𝑛 ) = 𝜎 ! ⟹ (𝑛𝜎 ! + 𝜎 ! − 2𝜎 ! ) = 𝜎 !
𝑘 𝑛 𝑘

1 1 ∑%
"#$(𝑋" − 𝑋)
!
(𝑛𝜎 ! − 𝜎 ! ) = 𝜎 ! ⟹ (𝑛 − 1)𝜎 ! = 𝜎 ! ⟹ 𝑘 = (𝑛 − 1) ⟹ 𝑆=
𝑘 𝑘 𝑛−1

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Central Limit Theorem

the sampling
As the n↑
distribution of the
sample sample mean becomes
size gets almost normal
large regardless of shape of
enough… population

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Sample Mean Sampling Distribution:

If the Population is not Normal

Population Distribution
Sampling distribution
properties:

Central Tendency

μx = μ
μ x
Variation Sampling Distribution
σ (becomes normal as n increases)
σx = Larger
n Smaller sample
size
sample
size

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Example
• Suppose a population has mean μ = 8 and standard
deviation σ = 3. Suppose a random sample of size n
= 36 is selected.

• What is the probability that the sample mean is

between 7.8 and 8.2?

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Example
Solution:
• Even if the population is not normally distributed, the
central limit theorem can be used (n is relatively large)
• … so the sampling distribution of x is approximately
normal
• … with μx = 8
σ 3
• …and σx = = = 0.5
n 36

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Example
Solution (continued):
æ ö
ç 7.8 - 8 X -μ 8.2 - 8 ÷
P(7.8 < X < 8.2) = Pç < < ÷
ç 3 σ 3 ÷
è 36 n 36 ø
= P(-0.4 < Z < 0.4) = 0.6554 - 0.3446 = 0.3108

Population Sampling Standard Normal

Distribution Distribution Distribution
???
? ??
? ?
? ? ? Sample Standardize
?
7.8 8.2 -0.4 0.4
Z
μ=8 X
μX = 8
x μz = 0
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Population Proportions
π = the proportion of the population having
some characteristic
• Sample proportion (p) provides an estimate of π:
X number of items in the sample having the characteristic of interest
p= =
n sample size
• 0≤p≤1
• p is approximately distributed as a normal distribution when n is large
(assuming sampling with replacement from a finite population or without replacement from an infinite
population)

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Sampling Distribution of p
• Approximated by a
Sampling Distribution
normal distribution if: P(ps)
.3
•
nπ ³ 5 .2
.1
and 0
0 .2 .4 .6 8 1 p

n(1 - π ) ³ 5
where
π(1- π )
μp = π and σp =
n
(where π = population proportion)
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Standardize p to a Z value with the formula:

p -p p -p
Z= =
σp p (1- p )
n
Suppose q is the probability of success and (1-q) is the probability of failure where
2
𝑞 = 3.
𝐸 𝑋 = 𝑛𝑞
Var 𝑋 = 𝑛𝑞(1 − 𝑞)

𝑋 𝑛𝑞
𝐸 𝑝 =𝐸 = =𝑞
𝑛 𝑛

𝑋 1 𝑞(1 − 𝑞)
Var 𝑝 = Var = 4 𝑛𝑞 1 − 𝑞 =
𝑛 𝑛 𝑛
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Example

• If the true proportion of voters who support

Proposition A is π = 0.4, what is the probability that
a sample of size 200 yields a sample proportion
between 0.40 and 0.45?

i.e.: if π = 0.4 and n = 200, what is

P(0.40 ≤ p ≤ 0.45) ?

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Example
if 𝝅 = 𝟎. 𝟒 and 𝒏 = 𝟐𝟎𝟎, what is
𝑷(𝟎. 𝟒𝟎 ≤ 𝒑 ≤ 𝟎. 𝟒𝟓) ?

Find σ p: p (1- p ) 0.4(1- 0.4)

σp = = = 0.03464
n 200

Convert to æ 0.40 - 0.40 0.45 - 0.40 ö

P(0.40 £ p £ 0.45) = Pç £Z£ ÷
standardized normal: è 0.03464 0.03464 ø
= P(0 £ Z £ 1.44)

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Example
if 𝝅 = 𝟎. 𝟒 and 𝒏 = 𝟐𝟎𝟎, what is
𝑷(𝟎. 𝟒𝟎 ≤ 𝒑 ≤ 𝟎. 𝟒𝟓) ?
Utilize the cumulative normal table:
P(0 ≤ Z ≤ 1.44) = 0.9251 – 0.5000 = 0.4251

Standardized
Sampling Distribution Normal Distribution
0.4251

Standardize

0.40 0.45 0 1.44

p Z
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