Anthony J Greene 1

Where We Left Off• What is the probability of randomly

selecting a sample of three individuals, all of whom have an I.Q. of 135 or more?

• Find the z-score of 135, compute the tail region and raise it to the 3rd power.

X X

So while the odds chance selection of a single person this far above the mean is not all that unlikely, the odds of a sample this far above the mean are astronomical

z = 2.19 P = 0.0143 0.01433 = 0.0000029

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Sampling Distributions

I What is a Sampling Distribution? A If all possible samples were drawn from

a populationB A distribution described with

Central Tendency µM

And dispersion σM ,the standard error

II The Central Limit Theorem

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Sampling Distributions

• What you’ve done so far is to determine the position of a given single score, x, compared to all other possible x scores

x

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Sampling Distributions

• The task now is to find the position of a group score, M, relative to all other possible sample means that could be drawn

M

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Sampling Distributions

• The reason for this is to find the probability of a random sample having the properties you observe.

M

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1. Any time you draw a sample from a population, the mean of the sample, M , it estimates the population mean μ, with an average error of:

2. We are interested in understanding the probability of drawing certain samples and we do this with our knowledge of the normal distribution applied to the distribution of samples, or Sampling Distribution

3. We will consider a normal distribution that consists of all possible samples of size n from a given population

Sampling Distributions

n

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Sampling ErrorSampling error is the error resulting from using a sample to estimate a population characteristic.

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Sampling Distribution of the Mean

For a variable x and a given sample size, the distribution of the variable M (i.e., of all possible sample means) is called the sampling distribution of the mean.

The sampling distribution is purely theoretical derived by the laws of probability.

A given score x is part of a distribution for that variable which can be used to assess probability

A given mean M is part of a sampling distribution for that variable which can be used to determine the probability of a given sample being drawn

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The Basic Concept

• Extreme events are unlikely -- single events

• For samples, the likelihood of randomly selecting an extreme sample is more unlikely

• The larger the sample size, the more unlikely it is to draw an extreme sample

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The original distribution of x: 2, 4, 6, 8

Now consider all possible samples of size n = 2

What is the distribution of sample means M

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The Sampling Distribution For n=2Notice that it’s a normal distribution with μ = 5

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Heights of the five starting players

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Possible samples and sample means for samples of size two

M

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Dotplot for the sampling distribution of the mean for samples of size two (n = 2)

M

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Possible samples and sample means for samples of size four

M

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Dotplot for the sampling distribution of the mean for samples of size four (n = 4)

M

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Sample size and sampling error illustrations for the heights of the basketball players

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Dotplots for the sampling distributions of the mean for samples of sizes one, two, three, four, and five

M

M

M

M

M

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The possible sample means cluster closer around the population mean as the sample size increases. Thus the larger the sample size, the smaller the sampling error tends to be in estimating a population mean, , by a sample mean, M.

For sampling distributions, the dispersion is called Standard Error. It works much like standard deviation.

Sample Size and Standard Error

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Standard Error of M

For samples of size n, the standard error of the variable x equals the standard deviation of x divided by the square root of the sample size:

In other words, for each sample size, the standard error of all possible sample means equals the population standard deviation divided by the square root of the sample size.

nM

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The Effect of Sample Size on Standard ErrorThe distribution of sample means for random samples of size (a) n = 1, (b) n = 4, and (c) n = 100 obtained from a

normal population with µ = 80 and σ = 20. Notice that the size of the standard error decreases as the sample size

increases.

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Mean of the Variable M

For samples of size n, the mean of the variable M equals the mean of the variable under consideration:

M .

In other words, for each sample size, the mean of all possible sample means equals the population mean.

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The standard error of M for sample sizes one, two, three, four, and five

Standard error = dispersion of MσM

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The sample means for 1000 samples of four IQs. The normal curve for x is superimposed

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Suppose a variable x of a population is normally distributed

with mean and standard deviation . Then, for samples of

size n, the sampling distribution of M is also normally

distributed and has mean

M = and standard error of

Sampling Distribution of the Mean for a Normally Distributed Variable

nM

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(a) Normal distribution for IQs(b) Sampling distribution of the mean for n = 4(c) Sampling distribution of the mean for n = 16

Samples Versus Individual Scores

Distribution of Samples

Distribution of Individual Scores

Distribution Consists of

M x

Observed Central Tendency

MM M

Theoretical Central Tendency M

Observed Dispersion

sM s

Theoretical Dispersion M

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Frequency distribution for U.S. household size

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Relative-frequency histogram for household size

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Sample means n = 3,for 1000 samples of household sizes.

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The Central Limit Theorem

For a relatively large sample size, the variable M is

approximately normally distributed, regardless

of the distribution of the underlying variable x.

The approximation becomes better and better with increasing

sample size.

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Sampling distributions fornormal, J-shaped, uniform variable

M

M M M

M M M

M MM

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APA Style: TablesThe mean self-consciousness scores for participants who were

working in front of a video camera and those who were not (controls).

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APA Style: Bar GraphsThe mean (±SE) score for treatment groups A and B.

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APA Style: Line GraphsThe mean (±SE) number of mistakes made for

groups A and B on each trial.

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Summary

• We already knew how to determine the position of an individual score in a normal distribution

• Now we know how to determine the position of a sample of scores within the sampling distribution

• By the Central Limit Theorem, all sampling distributions are normal with

nM

of dispersion and mean

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Sample Problem 1

• Given a distribution with μ = 32 and σ = 12 what is the probability of drawing a sample of size 36 where M > 48

82

3248

236

12

M

M

Mz

Does it seem likely that M is just a chance difference?

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Sample Problem 2

• In a distribution with µ = 45 and σ = 45 what is the probability of drawing a sample of 25 with M >50?

55.09

4550

925

45

z

M

Anthony J Greene 3912.84 OR 98.55

903*96.1 OR 903*96.1

336

18 96.1

MM

MM

MM

M

Sample problem 3

• In a distribution with µ = 90 and σ = 18, for a sample of n = 36, what sample mean M would constitute the boundary of the most extreme 5% of scores?

zcrit = ± 1.96

z -1.96 +1.96M 84.12 95.88

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Sample Problem 4

• In a distribution with µ = 90 and σ = 18, what is the probability of drawing a sample whose mean M > 93?

63

18

9

18

nM

What information are we missing?n = 9

3085.0 50.06

9093

P

Mz

M

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Sample Problem 5

• In a distribution with µ = 90 and σ = 18, what is the probability of drawing a sample whose mean M > 93?

5.44

18

16

18

nM

n = 16

2514.0 67.05.4

9093

P

Mz

M

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Sample Problem 6

• In a distribution with µ = 90 and σ = 18, what is the probability of drawing a sample whose mean M > 93?

6.35

18

25

18

nM

n = 25

2033.0 83.06.3

9093

P

Mz

M

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Sample Problem 7

• In a distribution with µ = 90 and σ = 18, what is the probability of drawing a sample whose mean M > 93?

0.36

18

36

18

nM

n = 36

1587.0 00.10.3

9093

P

Mz

M

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Sample Problem 8

• In a distribution with µ = 90 and σ = 18, what is the probability of drawing a sample whose mean M > 93?

0.29

18

81

18

nM

n = 81

0668.0 50.10.2

9093

P

Mz

M

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Sample Problem 9

• In a distribution with µ = 90 and σ = 18, what is the probability of drawing a sample whose mean M > 93?

38.113

18

169

18

nM

n = 169

0146.0 17.238.1

9093

P

Mz

M

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Sample Problem 10

• In a distribution with µ = 90 and σ = 18, what is the probability of drawing a sample whose mean M > 93?

72.025

18

625

18

nM

n = 625

00001.0 17.472.0

9093

P

Mz

M

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Sample Problem 10

• In a distribution with µ = 90 and σ = 18, what is the probability of drawing a sample whose mean M > 93?

0.181

18

1

18

nM

n = 1

5675.0 17.00.18

9093

P

Mz

M

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0.00

0.10

0.20

0.30

0.40

0.50

0.60

0 100 200 300 400 500 600 700

Sample Size

P-v

alue

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Sample Problem 11

• In a distribution with µ = 200 and σ = 20, what sample mean M corresponds to the most extreme 1% ?

0.201

20

1

20

nM

n = 1

4.148 ,6.251

20020*58.2 20020*58.2

*z

M

MM

MM

z MM

z = ±2.58

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Sample Problem 12

• In a distribution with µ = 200 and σ = 20, what sample mean M corresponds to the most extreme 1% ?

0.102

20

4

20

nM

n = 4

74.21 ,8.225

20010*58.2 20010*58.2

*z

M

MM

MM

z MM

z = ±2.58

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Sample Problem 13

• In a distribution with µ = 200 and σ = 20, what sample mean M corresponds to the most extreme 1% ?

0.54

20

16

20

nM

n = 16

1.781 ,9.212

2005*58.2 2005*58.2

*z

M

MM

MM

z MM

z = ±2.58

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Sample Problem 14

• In a distribution with µ = 200 and σ = 20, what sample mean M corresponds to the most extreme 1% ?

5.28

20

64

20

nM

n = 64

6.193 ,4.206

2005.2*58.2 2005.2*58.2

*z

M

MM

MM

z MM

z = ±2.58

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Sample Problem 15

• In a distribution with µ = 200 and σ = 20, what sample mean M corresponds to the most extreme 1% ?

25.116

20

256

20

nM

n = 258

8.196 ,2.203

20025.1*58.2 20025.1*58.2

*z

M

MM

MM

z MM

z = ±2.58

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150

170

190

210

230

250

270

0 50 100 150 200 250 300

Sample Size

Scor

e

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n = 1

z -2.58 +2.58M 148 .4 251.6

150 175 200 225 250

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n = 4

z -2.58 +2.58M 174.2 225.8

150 175 200 225 250

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n = 16

z -2.58 +2.58M 187.1 212.9

150 175 200 225 250

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n = 64

z -2.58 +2.58M 193.6 206.4

150 175 200 225 250

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n = 258

z -2.58 +2.58M 196.8 203.2

150 175 200 225 250

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n = ∞

150 175 200 225 250