stats lecture 06 sampling distributions

8/3/2019 Stats Lecture 06 Sampling Distributions

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Sampling Distributions

for

Means and Proportions

Quantitative Methods for Economics

Dr. Katherine Sauer

Metropolitan State College of Denver


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Chapter Overview:

I. Sampling Distributions: Means

II. The Central Limit TheoremIII. The Normal Distribution

IV. Sampling Distributions: Proportions

V. Desirable Properties of Estimators


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We can use sample statistics to inferthings about thepopulation

parameters.

Sometimes the sample statistic (e.g. mean) will be close to thepopulation parameter, sometimes it will not.

Recall: Greek letters are used for the population, English letters are

used for the corresponding sample characteristic.


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Lets start with some review.

Suppose our population consists of five numbers.

3, 1, 5, 6, 2

Calculate the population mean, variance, and standard deviation.

= 3.4

2 = 3.44

= 1.8547


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I. Sampling Distributions: Means

Suppose we want a sample of size 2. In the table, list all possiblecombinations of samples of size 2.

Repeat for a sample of size 3.

Sample of 2 Sample of 3

3,1 3,1,5

3,5 3,1,6

3,6 3,1,2

3,2 3,5,6

1,5 3,5,2

1,6 3,6,2

1,2 1,5,65,6 1,5,2

5,2 1,6,2

6,2 5,6,2


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Now, calculate each samples mean.

Sample of 2 Mean Sample of 3 Mean

3,1 2 3,1,5 3

3,5 4 3,1,6 3.33

3,6 4.5 3,1,2 2

3,2 2.5 3,5,6 4.67

1,5 3 3,5,2 3.33

1,6 3.5 3,6,2 3.67

1,2 1.5 1,5,6 4

5,6 5.5 1,5,2 2.67

5,2 3.5 1,6,2 3

6,2 4 5,6,2 4.33

Notice that the sample means vary from the population mean.

from 1.5 to 5.5 for sample of n=2

from 2 to 4.67 for sample of n=3

Depending on the sample chosen, the sample mean could be a

good estimate of the population mean, or not.


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Lets calculate the mean of all the sample means and the standard

deviation of the sample means.

x x

For sample of size 2: For sample of size 3:

mean 3.4 3.4standard dev. 1.1358 0.7572

The mean of all the sample means is the same as the populationmean.

The standard deviation of all the sample means decreases as the

sample size increases.


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The standard deviation of all the sample means is called the

standard error of the mean.

It can be calculated directly from the samples (as we just did) or

by using the formula when the population standard deviation is

known:

1

N

nN

nx

1

N

nNis the finite population correction factor.

When N is large, this factor is approximately 1.

- if the sample size is less than 5% of the

population size, you dont need the

correction factor (finite populations)


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The difference between the population mean and its point estimate

is called the sampling error.

If point estimates are the same as the population parameters, there

is no sampling error and the standard error is zero.


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Aprobability distribution is a list of every possible outcome with the

corresponding probability.

For our example, there are 10 possible samples of size 2. The

probability of each sample being selected is 0.10.

Lets plot the probability distribution of our sample means.

Step 1: Construct a frequency distribution table.

- 3 intervals is probably appropriate

- 1.5 to less than 3, 3 to less than 4.5, 4.5 to less than 6

Interval Frequency

1.5< x < 3 3

3 < x < 4.5 5

4.5 < x < 6 2


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Step 2: Calculate the relative frequencies.

probability of particular sample x frequency

Interval Frequency Relative Frequency1.5< x < 3 3 0.3

3 < x < 4.5 5 0.5

4.5 < x < 6 2 0.2

Step 3: Plot the probability histogram.

0

0.1

0.2

0.3

0.4

0.5

0.6

1.5< x < 3 3 < x < 4.5 4.5 < x < 6

RelativeFreq

uency

Sample Mean

Distribution of Sample Means


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0

0.1

0.2

0.3

0.4

0.5

0.6

1.5< x < 3 3 < x < 4.5 4.5 < x < 6

RelativeFrequency

Sample Mean

Distribution of Sample Means

Notice, even for this very small population and sample size, the

probability distribution is tending toward the bell shape of the

Normal Distribution.


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II. The Central Limit Theorem says that the probability distribution

of the sample means

for samples of size 30 or greater

selected from any population whose mean and variance

are known

approaches a Normal distribution

with mean and standard deviation .

The distribution of sample means for sample

sizes ofn > 30.

n

x

nNx

,~


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In addition, the Central Limit Theorem applies for small samples

from Normal populations, when the population variance is

known.

for samples ofany size from a Normal

distribution with known variance.

The Central Limit Theorem allows us to calculate

- probabilities regarding sample means- the limits that contain various percentages of sample

means

( later it will also help us construct confidence intervals)

nNx

,~


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III. The Normal probability distribution

It has long been recognized that large numbers of measurements,

when sorted and plotted in a histogram, tend to look like a bell-shaped form.

This bell-shaped curve is the Normal probability distribution

curve.

Formula:

This formula would trace out a bell-curve, symmetrical around

the mean of .

The area under the curve sums to 1.

- true of any probability distribution

2

2

1

2

1)(

x

exf


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The probability that a random variable,X, has a value betweenx = a

andx = b is given by the area under the curve betweenx = a and

x = b.

2

21

2

1)(

x

exf

However, we actually dont need to do the integration because the

Normal curve has some special characteristics that let us find the

area from a single table.


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Special properties of the Normal distribution:

1. Total area under the curve is one. (true of any probability

distribution)

2. The curve is symmetrical about the mean.

- the area to the left of the mean is 0.5- the area to the right of the mean is 0.5

3. The area under the curve between the mean and any point

depends on the number of standard deviations between the pointand the mean.

- theZ-score is the number of standard deviations

between the point and the mean


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The area between the mean and a point which is one standard

deviation from the mean is 0.3413.

68.26% of the total area is within one standard deviation

The area between the mean and a point which is two standard

deviations from the mean is 0.4772.95.44% of the total area is within two standard deviations

The area between the mean and a point which is three standarddeviations from the mean is 0.4986.

99.72% of the total area is within three standard deviations


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0.3413 = 34%0.3413

0.135 = 13.5%

0.0235 = 2.35%

0.0015 = 0.15%

0.135

0.0235

0.0015


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The Z-score is calculated as

Z is the number of standard deviations between the point (x) and

the mean.

Calculate Z to two decimal points.

Once you have Z, use a Normal probability distribution table tofind the area under the curve.

xZ


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Here is an excerpt from the table.

Ex: Z = 1.00

Area in upper

tail = 0.1587

Area between

and +

= 0.50.1587

= 0.3413


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Example: Suppose the time it takes to process an email inquiry is

normally distributed with a mean time of 500 seconds and a

standard deviation of 10 seconds. What is the probability that a

selected email will be processed in more than 505 seconds?

Step 1: Sketch the curve and indicate relevant information.

Step 2: Calculate Z.

Z = 505500 = 0.510

Step 3: Look up in table.


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When Z = 0.5, the area in the upper tail is 0.3085.


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The probability that an email will take more than 505 seconds to

process is 0.3085.


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What if instead we wanted to know the probability that processing

an email will take less than 485 seconds?



Z = 485500 = -1.5

10

Step 3: Look up in table.

For Z = 1.5 area in tail is 0.0668

The probability that processing an email will take less than 485

seconds to process is 0.0668.

h if i d d k h b bili h i


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What if instead we wanted to know the probability that processing

an email will take between 485 and 505 seconds?



Z1 = 485500 = -1.5 Z2 = 505500 = 0.5

10 10

Step 3: Look up in the table.

For Z = 1.5, area in tail is 0.0668

For Z = 0.5, area in tail is 0.3085

Subtract from 1. 10.06680.3085 = 0.6247


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Example: An importer of Herbs and Spices claims that the average

weight of packets of saffron is 20 grams. However, packets are

actually filled to an average weight of 19.5 grams with a standard

deviation of 1.8 grams. A random sample of 36 packets is selected.

Find the probability that the average weight is 20 grams or more.

In this example we are dealing with a sample of size n > 30. Well

apply the CLT and calculate the mean and standard error of thedistribution of means.

For our sample, and

nx

x

5.19 x 3.036

8.1

nx


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Step 2: Calculate Z (using the calculated mean and std error).

Z = 2019.5 = 1.67

0.3

Step 3: Look up Z in the table.

For Z = 1.67, the area in the tail is 0.0475

This is the probability that the average weight is 20 grams or more.


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Instead, lets find the lower and upper limit within which 95% of all

packets weigh.

In this case, we are dealing with the population, not the sample. Use

the population mean and standard deviation.


Step 2: Look up the Z that corresponds to a tail area of 0.025.

=19.5

Area in tail above

line = 0.025

Area in tail below

line = 0.025


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When the area of the upper tail is 0.025, Z = 1.96.


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=19.5

Area in tail above

line = 0.025

Area in tail below

line = 0.025

15.972

Step 3: Find the upper and lower limits.

Z x = number of units from the mean

1.96 x 1.8 = 3.528 grams

19.5 + 3.528 = 23.028 grams is the upper limit

19.53.528 = 15.972 grams is the lower limit

23.028

95% of the packets of saffron are between 15.972 and 23.028

grams in weight.


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Instead, lets calculate the two limits within which 95% of all

average weights fall.

Now we are dealing with the sample of n = 36.

The methodology is the same as when we use the entire population,

except well use the standard error of the means instead of the

standard deviation for the population.


Area in tail above

line = 0.025

Area in tail below

line = 0.025

5.19 x

3.036

8.1

nx


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Step 2: Look up the Z that corresponds to a tail area of 0.025.

Z = 1.96

Step 3: Find the upper and lower limits.

Z x x = number of units from the mean

1.96 x 0.3 = 0.588 grams

19.5 + 0.588 = 20.088 grams is the upper limit

19.50.588 = 18.912 grams is the lower limit

95% of the samples average weights are between 18.912 and 20.088

grams.


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IV. Sampling Distributions: Proportions

Aproportion is the number of elements with a given characteristic

divided by the total number of elements in the group.ex: The proportion of people who vote in an election is

the number who vote divided by the number eligible to

vote.

X or x are the number of elements with a given characteristic.

Often times proportions are quoted as percentages.

The sample proportion is a point estimate of the population

proportion.

N

X

n

xp


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Example: Suppose we have the following population of data.3, 1, 5, 6, 2

Calculate the population proportion of even numbers.

= 2 = 0.4

5

Referring back to our samples of size 2 and 3, calculate the sample

proportion of even numbers.


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3,1 0 3,1,5 0

3,5 0 3,1,6 1/3 = 0.33

3,6 1/2 = 0.5 3,1,2 1/3 = 0.33

3,2 1/2 = 0.5 3,5,6 1/3 = 0.33

1,5 0 3,5,2 1/3 = 0.33

1,6 1/2 = 0.5 3,6,2 2/3 = 0.67

1,2 1/2 = 0.5 1,5,6 1/3 = 0.33

5,6 1/2 = 0.5 1,5,2 1/3 = 0.33

5,2 1/2 = 0.5 1,6,2 2/3 = 0.67

6,2 2/2 = 1 5,6,2 2/3 = 0.67

Sample

Proportion

Sample

ProportionSample of 3Sample of 2

Calculate the mean of all sample proportions for each sample size.

The mean of all the sample proportions is the same as thepopulation proportion.

For samples of size 2: p = 0.4

For samples of size 3: p = 0.4


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The standard deviation of all the sample proportions decreases as

the sample size increases.

The standard error of all sample proportions is given by

(when N is large, we can omit the finite population correction

factor)

For samples of size 2:

= (0.3464)(0.8660)

= 0.29998

= 0.3

1

)1(

N

nN

np

15

25

2

)4.01(4.0

p


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For samples of size 3:

= (0.2828)(0.7071)= 0.199967

= 0.2

15

35

3

)4.01(4.0

p


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The list of every possible sample proportion with its probability

is called the sampling distribution of proportions.

Lets plot the probability distribution of our proportions for the

samples of size 2.

Step 1: Construct a frequency distribution table.

- we only have 3 values forp (0, 0.5, 1)

p Frequency

0 30.5 6

1 1


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Step 2: Calculate the relative frequency distribution. (probability

distribution)

probability of particular sample x frequency

0.10

Step 3: Plot the probability histogram.

p Frequency Relative Frequency

0 3 0.3

0.5 6 0.6

1 1 0.1


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Notice, even for this very small population and sample size, the

probability distribution is tending toward the bell shape of the

Normal Distribution.

For samples of size 30 or greater the distribution of

sample proportions is approximately Normal with

mean and standard deviation

The distribution of sample proportions

for sample sizes ofn > 30.

p

np

)1(

n

Np)1(

,~


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Example: In a certain neighborhood, it is known that 12% of

people age 16 to 24 are unemployed. If a random sample of 150

people age 16 to 24 is selected, what is the probability that the

sample contains at most 10% unemployed?

Step 1: Calculatep and p .

In this case, n = 150.

If 12% of the population is unemployed then = 0.12.

p = = 0.12

= 0.0265150

)12.01(12.0

p

Step 2: Sketch the curve and indicate relevant information


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Step 3: Calculate Z

= 0.100.12 = -0.7547 = -0.75

0.0265

Step 4: Look up Z in the table.

Area in tail is 0.2266.

The probability that at most 10% of the sample is unemployed is

0.2266.

= 0.12

0.10

p

pZ


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Instead, lets calculate the probability that the sample contains at

most 25 unemployed people.

Step 1: Convert the number into a proportion.

25 / 150 = 0.16667

Step 2: Calculatep and p .

p = = 0.12 p = 0.0265

S 3 Sk h h d i di l i f i


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Pr(p


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Many times the value of the population proportion is unknown.

We can approximate the mean and standard error of the

proportions by :

pp n

ppsp

)1(


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V. Some desirable properties of estimators

1. Estimators should be unbiased.

accurate

An estimator is unbiased if the average value of all the pointestimates is equal to the population parameter being estimated.

To prove that x is an unbiased estimator of we would need to

show that the expected value of the sample mean is equal to the

population mean.

E(x) =.


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2. The values of sample statistics vary around the population

parameter. It is desirable to keep this variance at a minimum

minimum variance

precise

An estimator is precise when the values of the estimates are

close.


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Concepts:

- Central Limit Theorem- Normal distribution

- desirable properties of estimators

Skills:For both means and proportions:

- calculate the mean of all the sample means (proportions) and the

standard deviation of the sample means (proportions)

- construct a probability distribution table

- calculate the probability of an event

stats lecture 06 sampling distributions

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