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The Practice of Statistics, 5th Edition

Starnes, Tabor, Yates, Moore

Bedford Freeman Worth Publishers

Do the warm-up on your own, on a piece of paper

The Practice of Statistics, 5th Edition

Starnes, Tabor, Yates, Moore

Bedford Freeman Worth Publishers

CHAPTER 7 Sampling Distributions 7.3

Sample Means

Learning Objectives

After this section, you should be able to:

The Practice of Statistics, 5th Edition 3

FIND the mean and standard deviation of the sampling distribution

of a sample mean. CHECK the 10% condition before calculating the

standard deviation of a sample mean.

EXPLAIN how the shape of the sampling distribution of a sample

mean is affected by the shape of the population distribution and the

sample size.

If appropriate, use a Normal distribution to CALCULATE

probabilities involving sample means.

Sample Means

Why do we do sampling?

• It is usually difficult (or impossible) to do a

census. And it is also expensive.

• An unbiased sample is meant to represent

the population.

– How would we know if it was a good

representation?

Sampling Variability

How can x, based on a sample of only a few

of the 117 million American households, be

an accurate estimate of μ?

KEY CONCEPT

A second sample would likely result in a different mean . . .

The value of a statistic varies in repeated sampling . . .

It is why we have to check conditions!!

WHY DO WE CHECK

ASSUMPTIONS?

What are the assumptions we have used?

• Some form of randomization in context

• May use approximately Normal Distribution

• Independence

Eliminates bias, every member of population

has an equal chance to be drawn

Sample size must be large enough to insure

the sampling distribution is approximately

normal.

Sample size must be less than 10% of the

population . This allows us to use standard

deviation formulas.

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The Sampling Distribution of

When we record quantitative variables we are interested in other

statistics such as the median or mean or standard deviation. Sample

means are among the most common statistics.

Consider the mean household earnings for samples of size 100.

Compare the population distribution on the left with the sampling

distribution on the right. What do you notice about the shape, center,

and spread of each?

x



When we choose many SRSs from a population, the sampling

distribution of the sample mean is centered at the population mean µ

and is less spread out than the population distribution. Here are the

facts.

Sampling Distribution of a Sample Mean

as long as the 10% condition is satisfied: 10n ≤ N.

x

The standard deviation of the sampling distribution of x is

s x =s

n

The mean of the sampling distribution of x is mx = m

Suppose that x is the mean of an SRS of size n drawn from a large population

with mean m and standard deviation s. Then :

Note: These facts about the mean and standard deviation of x are true

no matter what shape the population distribution has.


Movies

Suppose the number of movies viewed in the last year by high school

students has an average of 19.3 with a standard deviation of 15.8.

Suppose we take an SRS of 100 high school students and calculate the

mean number of movies viewed by the members of the sample.

(a) What is the mean of the sampling distribution of ? x

x

(b) What is the standard deviation of the sampling distribution of ?

Remember to check that the 10% condition is met.

is an unbiased estimator of μ, μx=μ=19.3 movies

σx = 15.8/10=1.58 movies. The 10% condition is met because there are

more than 10(100) = 1000 high school students.

x


Sampling From a Normal Population

We have described the mean and standard deviation of the sampling

distribution of the sample mean x but not its shape. That's because the

shape of the distribution of x depends on the shape of the population

distribution.

In one important case, there is a simple relationship between the two

distributions. If the population distribution is Normal, then so is the

sampling distribution of x . This is true no matter what the sample size is.

Sampling Distribution of a Sample Mean from a Normal Population

Suppose that a population is Normally distributed with mean m

and standard deviation s . Then the sampling distribution of x

has the Normal distribution with mean m and standard

deviation s / n, provided that the 10% condition is met.


The Central Limit Theorem

Most population distributions are not Normal. What is the shape of the

sampling distribution of sample means when the population distribution

isn’t Normal?

It is a remarkable fact that as the sample size increases, the distribution

of sample means changes its shape: it looks less like that of the

population and more like a Normal distribution!

When the sample is large enough, the distribution of sample means is

very close to Normal, no matter what shape the population distribution

has, as long as the population has a finite standard deviation.

Uniform distribution

Sample size 1

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Sample size 2


Sample size 3


Sample size 4


Sample size 8


Sample size 16


Sample size 32

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Triangle distribution

Sample size 1


Sample size 2


Sample size 3


Sample size 4


Sample size 8


Sample size 16

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Sample size 32

Inverse distribution

Sample size 1


Sample size 2


Sample size 3


Sample size 4


Sample size 8

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Sample size 16


Sample size 32

Parabolic distribution

Sample size 1


Sample size 2


Sample size 3


Sample size 4

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Sample size 8


Sample size 16


Sample size 32



Most population distributions are not Normal. What is the shape of the

sampling distribution of sample means when the population distribution

isn’t Normal?

It is a remarkable fact that as the sample size increases, the distribution

of sample means changes its shape: it looks less like that of the

population and more like a Normal distribution!

When the sample is large enough, the distribution of sample means is

very close to Normal, no matter what shape the population distribution

has, as long as the population has a finite standard deviation.

Draw an SRS of size from any population with mean and finite

standard deviation . The says that when

is large, the sampling distribution of the sample mean is

n

n

x

central limit theorem (CLT)

approximately

Normal.



Consider the strange population distribution

from the Rice University sampling distribution

applet.

Describe the shape of the sampling

distributions as n increases. What do you

notice?



Normal/Large Condition for Sample Means

The central limit theorem allows us to use Normal probability

calculations to answer questions about sample means from many

observations even when the population distribution is not Normal.

As the previous example illustrates, even when the population

distribution is very non-Normal, the sampling distribution of the sample

mean often looks approximately Normal even with small sample sizes.

If the population distribution is normal, then so is the sampling

distribution of . This is true no matter what the sample size (n)

is.

If the population distribution is not Normal (aka unknown,

skewed, etc.), the Central Limit Theorem (“CLT”) tells us that the

sampling distribution of will be approximately Normal in most

cases, if n ≥ 30. x

x

http://onlinestatbook.com/stat_sim/sampling_dist/

http://onlinestatbook.com/stat_sim/sampling_dist/

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How do you know if you can use an approximately

normal distribution?

1. It’s stated in the problem stem. If so, say so in your assumptions.

2. n is greater than or equal to 30, then you have a large enough

sample size to approximate the sampling distribution with a normal

curve.

3. Graph the actual data, draw your graph (boxplot) in the

assumptions, and state if there are no extreme outliers or severe

skewedness that we may use an approximately normal distribution

Point of clarification:

If we can’t prove a sampling distribution is approximately normal; don’t

speculate what it is . . . We only know we can not assume it is . . .



x

A statistic used to estimate a

parameter is unbiased if the

mean of its sampling

distribution is equal to the

true value of the parameter

being estimated.

Example: At the P. Nutty Peanut Company, dry-roasted, shelled

peanuts are placed in jars by a machine. The distribution of weights in

the jars is approximately Normal, with a mean of 16.1 ounces and a

standard deviation of 0.15 ounces.

(a) Without doing any calculations, explain which outcome is more

likely:

1) randomly selecting a single jar and finding that the contents

weigh less than 16 ounces or

2) randomly selecting 10 jars and finding that the average

contents weigh less than 16 ounces.

(b) Find the probability of each event described above, you may

assume all the assumptions have been met..

It’s more likely to have a

single jar weigh less,

than a group of 10

(they’d be closer to the

mean).





(b) Find the probability of each event described above, you may assume

all the assumptions have been met..

1) randomly selecting a single jar and finding that the contents

weigh less than 16 ounces or

State the distribution and values of interest.

Let µ = the true average weight of a jar of peanuts. N(16.1, 0.15)

x

Perform the calculations.

2514.0)67.0(

67.015.0

1.1616

zP

z OR Normalcdf(-9999,16,16.1,0.15) = 0.2525

Answer the question.

There is a 25% chance that a single jar will contain less than 16 oz of peanuts.





(b) Find the probability of each event described.

2) randomly selecting 10 jars and finding that the average

contents weigh less than 16 ounces.


Let µ = the true average weight of 10 randomly selected jars of peanuts.

Thus has roughly N(16.1, 0.047)


0166.0)13.2(

13.2047.0

1.1616

zP

z OR Normalcdf(-9999,16,6.1,0.047) = 0.0167


There is a 2% chance that a random sample of 10 jars will contain less than 16 oz

of peanuts.

047.010

15.0

nx

x

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Example: Suppose that the number of texts sent during a typical day by a

randomly selected high school student follows a right-skewed distribution with

a mean of 45 and a standard deviation of 35.

Assuming that students at your school are typical texters, how likely is it that a

random sample of 50 students will have sent more than a total of 2500 texts in

the last 24 hours?


Let = true average number of texts sent by 50 randomly selected students.

Thus has roughly N(45, 4.95)

Check assumptions!

Problem stated randomly selected high school students

10%, the sample (50 students) is less than 10% of all students

95.450

35

nx

Since n = 50 and 50 > 30, we may use an approximately

normal distribution for the sampling distribution of x

Example: Suppose that the number of texts sent during a typical day by a

randomly selected high school student follows a right-skewed distribution with

a mean of 45 and a standard deviation of 35.

Assuming that students at your school are typical texters, how likely is it that a

random sample of 50 students will have sent more than a total of 2500 texts in

the last 24 hours?


Let = the true average number of texts sent by 50 randomly selected students.

Thus has roughly N(45, 4.95)


1562.0)01.1(

01.195.4

4550

zP

z OR Normalcdf(50,1E99,45,4.95) = 0.1562


There is a 16% chance that a random sample of 50 students will have sent over

2500 texts in the last 24 hours.

95.450

35

nx

What is the shape of the sampling distribution of

a mean when the sample is not taken from a

normally distributed population. Does the sample

size matter. Does the concept have a name.

Depends on either the size being large

enough (n>30) or on graphing the actual data

in a boxplot . . .

Try Practice problems on page 11

Review other practice problem in the packet

10-11

12 Internet exercise

13- more practice . . . Section Summary

In this section, we learned how to…


FIND the mean and standard deviation of the sampling distribution of

a sample mean. CHECK the 10% condition before calculating the

standard deviation of a sample mean.

EXPLAIN how the shape of the sampling distribution of a sample

mean is affected by the shape of the population distribution and the

sample size.

If appropriate, use a Normal distribution to CALCULATE probabilities

involving sample means.

Sample Means

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