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Lecture 7 Part II .pdf



Nom original: Lecture 7 - Part II.pdf
Titre: Lecture7
Auteur: Giuliana Cortese

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As sample size (n) grows…
n=1

n=64

n=4

…standard error shrinks!
…shape of sampling distribution gets closer to “normal”!

Children
n=16
1.75
.405

Sampling distributions
(sample means)

1.75
.81

1.75
.2025

If all the sample means are very close to the population
mean, then the standard error of the mean would be
small.
If the sample means vary considerably, then te standard
error of the mean would be large.

Sampling distributions
have a “normal” shape

• Properties of the “normal” distribution:

– In 90% of all samples, the sample mean is

within 1.64

within 1.96

standard errors from the population mean (10% of samples

have means that are further away).

– In 95% of all samples, the sample mean is

within 2.58

standard errors from the population mean (5% of samples

have means that are further away)

– In 99% of all samples, the sample mean is

standard errors from the population mean (1% of samples

have means that are further away)

Most values within 2 SD’s from the sample mean.
Sample mean usually within 2 SE’s from pop. mean

Example 1
Population: “US adults”;
Variable Y: “How many children have you ever had?”
We know that µY=1.75, σY=1.62.
Consider samples of size n =16.
95% of all sample means Y are within Z =1.96 standard errors from
pop. mean µY
95%

Example 2
Variable Y: “How many children have you ever had?”
We know µY=1.75, σY=1.62. Samples of size n = 45:
99% of all sample means are within 2.58 standard errors from the
population mean

Example 3: Weights of doctors

• Experimental question: Are practicing doctors
setting a good example for their patients in their
weights?
• Experiment: Take a sample of practicing doctors
and measure their weights
• Sample statistic: mean weight for the sample

IF weight is normally distributed in doctors with a
mean of 150 lbs and standard deviation of 15,
how much would you expect the sample average to
vary if you could repeat the experiment over and
over?

Example 3: Weights of doctors

• take 1000 samples of 100 doctors and calculate
their average weight….
• We are almost sure that 95% of sample means is
between 147 and 153

error

“Margin of error”

error
MARGIN OF ERROR:

Z = 1.64, 1.96, 2.58

The value of 1.96 is based on the fact that
95% of the area of a standard normal
distribution is within 1.96 standard
deviations from the mean;

90 - (1.96)(12) = 90 – 23.52 = 66.48
90 + (1.96)(12) = 90 + 23.52 = 113.52

These limits were computed by adding and
subtracting 1.96 standard deviations
to/from the mean of 90 as follows:

The shaded area represents the middle
95% of the distribution and stretches from
66.48 to 113.52.

Mean = 90
Standard error of the mean = 12

Margin of error: Interpretation
n=9

Figure 1 shows that 95% of the means
are no more than 23.52 units (1.96
standard deviations) from the mean of
90

The Central Limit Theorem







N = 2 and N = 10.
The parent population was a
uniform distribution.
The distribution for N =2 is far
from a normal distribution. Scores
are denser in the middle than in the
tails.
For N = 10 the distribution is quite
close to a normal distribution.
The means of the two distributions
are the same, but the spread of the
distribution for N = 10 is smaller.

The Central Limit Theorem

If all possible random samples, each of size n, are taken
from any population with a mean µ and a standard
deviation σ, the sampling distribution of the sample
means will:
have mean:

)

be approximately a normal distribution, regardless of the
shape of the parent population (for n large enough)

The Central Limit Theorem
• The parent population is very
non-normal.
• n=5: The sampling distribution
of the sample means has a
slight positive skew.
• n=25: The larger the sample
size n is, the closer the
sampling distribution is to a
normal distribution.

average of 100

average of 2

Uniform on [0,1]

average of 1

average of 5

average of 1

~Exp(1)

~Bin(40, .05)

average of 5

average of 1

average of 5

average of 2

average of 100

average of 2

average of 100

Central Limit Theorem (CLT)
The sampling distribution of Y :
As the sample size n gets larger,
– the standard error gets smaller
– and the sampling distribution gets closer to “normal.”
Usually good approximation from n=30!

Examples/Exercises



Figure 2: sampling distribution
of the mean.

Example 1

n

Figure 1 shows three pool balls, each with a number
on it. Two of the balls are selected randomly (with
replacement) and the average of their numbers is
computed.

n

It is the sampling distribution of
the mean for a sample size of 2
(n = 2).

In this example, the distribution
of pool balls and the sampling
distribution are both discrete
distributions.

Example 2
The mean expenditure per customer at a
tire store is $85.00, with a standard
deviation of $9.00.
If a random sample of 40 customers is
taken, what is the probability that the
sample average expenditure per
customer for this sample will be $87.00
or more?

Because the sample size is greater than 30, the central limit

σ =9

n = 40

theorem says the sample means are normally distributed.

µ= 85,

distribution with parameters µ and σ, then we transform to a

standard normal Z ...

• The question is
• This is

• Therefore, 7.9% of the times, a random sample of 40
customers from this population will yield a mean
expenditure of $87.00 or more.
OR
From any random sample of 40 customers, 7.9% of
them will spend on average $87.00 or more.

Example 4

Suppose that during any hour in a
large department store, the average
number of shoppers is 448, with a
standard deviation of 21 shoppers.

What is the probability that a
random sample of 49 different
shopping hours will yield a sample
mean between 441 and 446
shoppers?

Example 5
• Assume that the weights of 10-year old children
are normally distributed with a mean of 90 and a
standard deviation of 36. What is the sampling
distribution of the mean for a sample size of 9?
• Answer: the sampling distribution of the mean has
a mean of 90 and a standard deviation of 36/3 =
12. The standard deviation of a sampling
distribution is its standard error.

Sampling distribution
of a sample proportion

Overview

• We know about sampling distribution for a
sample mean
• We will learn the same for a proportion
• Connection:

– A proportion is the mean of a dummy variable.

0

Dummy variable (Population)
Y=
1
Distribution function:

Expected value:
Variance:
Standard Deviation:

Dummy variables (sample)
• Let Y be a dummy variable
• In a sample,

– where p is the proportion of the sample with Y =1

• In the population,

– where π is the proportion of the population with Y=1

Sampling distribution of the mean

• Mean

• Across all possible samples
Y has a normal distribution with
• and standard error

Sampling distribution of the proportion p

• Across all possible samples
p has a normal distribution with
• Mean

• and standard error (of the sample proportions)

Example: 1936 election
• Let Y be a dummy variable
(1=Roosevelt, 0=not).

• Literary Digest poll

– unrepresentative sample of 10 million voters
– wrong: called election for Landon

• Gallup poll

– “quota sample” of 50,000 voters
– right: called election for Roosevelt

• Sociology 549 poll

– simple random sample of 100 voters
– right or wrong?

Example
Sampling distribution of a sample proportion
• π =0.61, or 61%, of voters will vote for Roosevelt
– But we don’t know this.

• We sample n=100 voters
– In sample, p will vote for Roosevelt

• Standard error of p


Sampling distribution of a sample proportion

.65

95% of samples

p!
.60

.70

.75

p

Roosevelt

 

• Distribution of p across all possible samples

.55

– when π =0.61 and n=100
Probability
0.1
0.08
0.06
0.04
0.02

.50

Drawing a sample

!
.60

.65

.70

.75

p

Roosevelt

 

• We ask n=100 voters if they’ll vote for Roosevelt

.55

– With π =0.61, one likely result is p=0.63
Probability
0.08
0.06
0.04
0.02

.50

Drawing another sample

!
p
.60

.65

.70

.75

p

Roosevelt

 

• We ask n=100 voters if they’ll vote for Roosevelt

.55

– With π =0.61, one likely result is p=0.59
Probability
0.08
0.06
0.04
0.02

.50

.75

p

Roosevelt

 
Chapter 11 – 78

Sampling distribution of a sample proportion

.65

95% of samples

p!

.60

.70

• Distribution of p across all possible samples

.55

– when π =0.61 and n=100
Probability
0.1
0.08
0.06
0.04
0.02

.50

Margin of error
• Poll’s “margin of error” is typically ~2 SE’s
• Using normal distribution,
– In 95% of all samples the sample proportion p
are within 1.96 standard errors of population
proportion

• Here, in 95% of samples, we’ll get
0.61 +/- 1.96 (0.05) = 0.51 to 0.71, or 51% to
71%
voting for Roosevelt
Chapter 11 – 79


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