Download For means - My Math Mantra

Chapter 9
Inferring Population
Means
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Central Limit Theorem, Mean Values
App1: Theory
App2: Confidence Intervals
App3: Hypothesis Testing
One-Population
Two-Population Unmatched
Two-Population Matched
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9.1
Sample Means of
Random Samples
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Central Limit Theorem, Mean Values
Theory
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Inferring a Population’s Mean Value from a
Sample’s Mean Value




In this chapter we apply the Central Limit Thm to Mean
Values in the same way did for Proportions (a.k.a. %).
Remember that Proportions are used for Categorical
Data and Means are used for Quantitative Data.
Example: Do you smoke? VS. How many cigarettes do
you smoke?
The three applications covered for Mean Values are:



Finding the mean value of a Group from within a
population, given the population’s mean.
Confidence Intervals for mean values
Hypothesis Testing for means, both single and dual pops.
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Notation: Statistics, Parameters, Means
and Proportions



Mean and Standard Deviation if the survey question
has a numerical variable.
Proportion if the survey question is Yes/No
The confidence interval and hypothesis test always
refer to the population not the sample
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Sample Proportions and the Central Limit
Theorem

Recall that all proportions (%) from all samples of
the same size from a population form a Normal Dist.
Likewise the Mean Value from all Samples,
of the same size, form a Normal Distribution
- even if the Population is Skewed!
Population
->
Individual
Samples
->
Histogram of all
Samples’ Means
->
ND Model used
->
Example: Cherry Blossom Ten Mile Run
The Race Times follow a Normal Distribution model
with a mean of 97 minutes and a standard deviation
of 17 minutes: N(97, 17).
 The population parameters are, in symbols,
 µ = 97 minutes
 σ = 17 minutes
 Now lets randomly sample 30 runners and calculate
the average of their finishing times.
 Lets repeat this many times and build a histogram of
the mean values from each sample.

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Example: Cherry Blossom Ten Mile Run

The population of all runners’s times look like this:
Dot plot of around 100 samples (remember n = 30):
 The center models the pop
 Symmetric
 Unimodal
 (Ideally would be thinner)

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Simulating Many Sample Means

As the sample size increases


Accuracy Does Not Change
Better Precision - sampling variability diminishes
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CLThm: Standard Error for Means
The standard deviation of the sampling gets smaller
with larger sample size.
 This is true for any population distribution.
 The Standard Error is the standard deviation of the
sampling distribution.
 For sample mean it is:


To complete the picture:

µ represents the mean of the population
σ represents the standard deviation of the population

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EXAMPLE 1: iTunes Library Statistics



A student’s iTunes library of mp3s has a very large
number of songs. The mean length of the songs is 243
seconds, and the standard deviation is 93 seconds. The
distribution of song lengths is right-skewed. Using his
mp3 player, this student will create a playlist that
consists of 25 randomly selected songs.
Q1: Is the mean value of 243 minutes an example of a
parameter or a statistic? Explain.
A: The mean of 243 is an example of a parameter,
because it is the mean of the population that consists of
all of the songs in the student’s library.
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EXAMPLE 1: iTunes Library Statistics
(cont)

Q2: What should the student expect the average song
length to be for his playlist?


The sample mean length can vary, but is typically the same as
the population mean: 243 seconds.
Q3: What is the standard error for the mean song length
of of all samples of 25 randomly selected songs?

The standard error is:
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Example: The mean cost per item at a grocery
store is $2.75 and the standard deviation is $1.26.
A shopper randomly puts 36 items in her cart.

Is 2.75 a parameter or a statistic?


Predict the average cost per item in the shopper’s cart.


Parameter
$2.75
Find the standard error for carts with 36 items.

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Comparing Standard Errors

The mean income for residents of the city is
$47,000 and the standard deviation is $12,000.
Find the standard error for the following sample
sizes
n=1
n=4
 n = 16
 n = 100

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Comparing Standard Errors

The mean income for residents of the city is
$47,000 and the standard deviation is $12,000.
Find the standard error for the following sample
sizes
n=1
n=4
 n = 16
 n = 100

→ $12,000
→ $6,000
→ $3,000
→ $1,200
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9.2
The Central Limit
Theorem for Sample
Means
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Application 1: Grouped Data, Means
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The Central Limit Theorem for Sample Means

The Central Limit Theorem (CLT) assures us that no matter what
the shape of the population distribution, if a sample is selected
such that the following conditions are met, then the distribution of
sample means follows an approximately Normal distribution.
 Random Sample and Independence. Each observation is
collected randomly from the population, and observations are
independent of each other.
 Normal. Either the population distribution is Normal or the
sample size is large.
 Big Population. If the sample is collected without replacement,
then the population must be at least 10 times larger than the
sample size.
 (( No “10 success / 10 failure” condition for means. It does not
make sense - no proportions involved here ))
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What is a Large Enough Sample Size?

If the population distribution is not too far from
Normal then the sample size can be small.

For most population distributions n = 25 or higher
gives sufficient accuracy.

If the population distribution is far from normal, a
larger sample size is needed.
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Example: Visualizing the Central Limit
Theorem

Distribution of annual
tuitions and fees at all twoyear colleges in the US
( 2008–2009 academic year)
>>> Skewed Right

Histogram of 30 randomly
selected samples from the
population. Each sample is
of size 30. Already looking
unimodal, not symmetric.
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Visualizing the Central Limit Theorem

Distribution of 200
randomly selected samples
from the population. Each
sample is of size 30.

Now increase sample size.
Distribution of 200
randomly selected samples
from the population. Each
sample is of size 60.
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Application 1: Grouped Data, Means
Flagship Example
EXAMPLE 2: Pulse Rates Are Not Normal

A large study in the US finds the mean resting pulse rate of adult
women is 74 beats per minute, with a standard deviation of 13
bpm. The distribution is skewed right.

QUESTION: If we take a random sample of 36 women from this
population what is the probability that the average pulse rate of
this group will be below 71 bpm?

(a) Standard Error:

(b) z-Score:

(c) Area to the Left: P( z < -1.38 ) = .0838 The probability that a
randomly selected group has a mean less than 71 bpm is 8.4%
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EXAMPLE 2: Pulse Rates - with Software

A large study in the US finds the mean resting pulse rate of adult
women is 74 beats per minute, with a standard deviation of 13
bpm. The distribution is skewed right.

Q: If 38 women are selected, find the probability that the mean of
the group is less than 72.



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Individual vs. Grouped
!
!
!
!
EXAMPLE: The EPA finds that consumer car mileage follow a
normal distribution with a mean of 24 mpg and SD of 6 mpg.
An Individual from a Normally Distributed Population
Q1: What is the probability a random car will get < 20 mpg?
A: Find the Area to the left of 20 mpg:
! z-score for 20 mpg: z = ( 20 - 24 ) / 6 = -0.67
! Z-Table: => 0.2514
or 25.1%
A Group from a Normally Distributed Population
Q2: What is the probability a randomly selected group of 8 cars
have an average < 20 mpg?
! SE = 6 / sqrt (8) = 2.121
! z = ( 20 - 24 ) / 2.121 = - 1.89
! Z-Table: ==> 0.0294 or 2.9%
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Contrasting Individual to Grouped ND




If you do not have a Normally Distributed population
you cannot find probabilities from the z-table for an
individual.
But you can find probabilities from the z-table for groups
from within the population, so long as conditions are met,
namely n > 25 or so.
If you do have a Normally Distributed population you
can find probabilities from the z-table
And you can find probabilities from the z-table for groups
from within that population, without condition, even for
small group sizes, like n = 8.
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Reviewing the different types of distributions


The population distribution is the distribution of all
individuals that exist.
The distribution within a sample is the distribution
of the individuals that were surveyed.


The mean, standard deviation, and the shape are likely to
be close to the population distribution.
The sampling distribution is the distribution of all
possible sample means of sample size n.

The mean will be the same as the population mean, but the
shape will be approximately normal and the standard
deviation will be smaller.
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Reviewing the different types of distributions
Population
Distribution
An Individual
Sample’s
Distribution
Distribution of a
bunch of
Samples’ Means
Distribution Model
for Samples’
Means
9.3
Answering Questions
about the Mean of a
Population:
Confidence Intervals
and
Hypothesis Tests
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Application 2: Confidence Intervals
One-Population, Means
Gosset’s “Student’s - t” Distribution



In the late 1800‘s William S. Gosset, an employee of the
Guinness Brewery in Dublin, Ireland, worked long and
hard to find the sampling model for Means.
The sampling model that Gosset found has been known
as Student’s t-model.
The Student’s t-models form a whole family of related
distributions that depend on a parameter known as
degrees of freedom, a value related to sample size.
 We often denote degrees of freedom as df, and a
particular instance of the model as tdf.
Introduction to the t-Distribution

The t-statistic will be:

Compare this to the z-statistic:


If σ is unknown, we cannot find the z-score.
The difference is that the SE is an estimate
Introduction to the t-Distribution






We will need to alter the z-Table we use to look up our
areas from.
This new table (distribution) is called the:
 t-distribution
This is not just a single table like the N(0,1) was.
This is a family of z-like tables, one for each sample
size, referred to as the Degree of Freedom.
In other words each dof has it’s own z-like table.
Because it is unwieldily to show a few dozen tables in
the text we will use software or only the critical values
from these tables.
Introduction to the t-Distribution

The t-distribution is broader and shorter than the
Normal distribution, thus making extremes not as
rare events.
Introduction to the t-Distribution


As sample size increases the t-distribution narrows
and it’s shape approaches the Normal distribution,
so for large n we could use the normal distribution.
For n < 10 the two are almost indistinguishable.
Introduction to the t-Distribution

For n < 40 the two are indistinguishable.
Application 2: Confidence Intervals

A confidence interval is a useful answer to the
following questions:
 “What’s the typical value for a variable in this large
group of objects or people?
 How far away from the truth might this estimate of the
typical value be?”

You should provide a confidence interval whenever
you are estimating the value of a population parameter
on the basis of a random sample from that population.
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Confidence Level
The confidence level is a measure of how well the
method used to produce the confidence interval
performs.
 The Confidence level gives the percent of
confidence intervals that contain the population
mean.
 As in proportions, the Confidence Level is
associated with a critical value, a t*-value, in
this case.

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Confidence Levels

Here are a 100 samples’ Confidence Intervals.


The 5 red ones miss the true proportion.
It appears that a 95% Confidence Level was used.
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Confidence Intervals for Means: The Steps
1. Conditions:
 Independence?
 Randomization Condition?
 10% Condition?
 Nearly Normal Condition?
2. Do the Math
 Calculate the Standard Error
 Pick the critical value (the multiplier) called t*
 Form the Margin of Error = t* x SE
 Make the Interval: (mean - ME, mean + ME)
3. Interpret the results
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Assumptions and Conditions

Independence Assumptions:
 Independence Assumption. The data values
should be independent.
 Randomization Condition: The data arise from
a random sample or suitably randomized
experiment. Randomly sampled data
(particularly from an SRS) are ideal.
 10% Condition: When a sample is drawn
without replacement, the sample should be no
more than 10% of the population.
 Nearly Normal: If the sample appears skewed
seek n > 25.
Finding the Multiplier t*

The Degree of Freedom represents the number of
‘directions’ the sample size, n, can ‘vary’.

For a single population means the df = n - 1

Once the df is calculated, we can access one family of the
t-distribution tables.

Next we utilize the selected Confidence Level.

The left and right side t-value that captures a “Confidence
Level’s worth of area” in the middle of the distribution defines the critical value t*
Recall Critical Values for z-scores

Many z-tables callout critical z* scores and the
area they contain between their + and - values.

Well, we do likewise for each of the df-families of
the t-distribution and put them in a list:
Finding t-Values By Hand

The Student’s t-model
has a different table
for each value of
degrees of freedom.

Because of this,
Statistics books
usually have one table
of only the t-model’s
critical values for
selected confidence
levels, instead of
page after page of all
t-values.
EXAMPLE 7: Finding the Multiplier t*





Suppose we collect a
sample of 15 iPads and
wish to calculate a 90%
Confidence Interval for the
mean battery life.
Q: Find the critical value, t*,
for a 90% C.I. when n = 15
SOLUTION: df = n - 1 or
here df = 15 -1 =14
A CI is a 2-tail problem,
and the complement of
90% is 10%
row 14, col 0.10 => 1.761
Application 2: Confidence Intervals
One-Population, Means
Flagship Example
EXAMPLE 8: College Tuition Costs


A random sample of 35 U.S. junior colleges had a mean
tuition of $2380, and a standard deviation of $1160. Find a
90% confidence interval for the mean tuition of all U.S. junior
colleges based on this sample.
(i) Conditions:




Independence:
Random:
Big Pop:
Nearly Normal:

(ii) Do the Math:
 t* :
 SE:

:

(iii) Interpret:
EXAMPLE 8: College Tuition Costs


A random sample of 35 U.S. junior colleges had a mean tuition of
$2380, and a standard deviation of $1160. Find a 90% confidence
interval for the mean tuition of all U.S. junior colleges based on this
sample.
(i) Conditions:





Independence: Assumed
Random: Stated
Big Pop: There are more than 350 junior colleges in the U.S.
Nearly Normal: Not mentioned but not needed since large sample
(ii) Do the Math:
 t* : df = 35 - 1 = 34,
100% - 90% = 10% =>
t*-Table = 1.691

= 2380 ± (1.691 * 196.0758) = 2380 ± 331.56
(iii) Interpret: A 90% C I for the mean tuition of all U.S. junior colleges
is ($2048, $2712).


EXAMPLE 8: College Tuition Costs

Statdisk:
 Analysis -> Confidence Interval -> Mean - One Sample
Another CI Example



45 randomly selected college students worked on
homework for an average of 9 hours per week. Their
standard deviation was 2 hours. Find a 90% confidence
interval for the population mean.
d.f. = 45 - 1 = 44 → t* = 1.68
Standard Error:

Interval:
Lower Bound: 9 – 1.68 x 0.30 ≈ 8.5
Upper Bound: 9 + 1.68 x 0.30 ≈ 9.5

Next Slide


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Example (cont.) - CI Interpretations

45 randomly selected college students worked on
homework for an average of 9 hours per week. Their
standard deviation was 2 hours. Find a 90% confidence
interval for the population mean.

Interpretation of Confidence Interval: We are 90%
confident that the population mean number of hours
worked on homework for all college students is between
8.5 and 9.5 hours.
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Example (Cont.) - CI Interpretations

45 randomly selected college students worked on
homework for an average of 9 hours per week.
Their standard deviation was 2 hours. Find a 90%
confidence interval for the population mean.

Interpretation of Confidence Level: If many groups
of 45 randomly selected students were surveyed,
90% of these confidence intervals will succeed in
containing the actual population mean number of
hours worked on homework, but, 10% will not
contain the true population mean.
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Sample Size

To find the sample size needed for a particular
confidence level with a particular margin of error (ME),
solve this equation for n:

The problem with using the equation above is that we
don’t know most of the values. There a two ways to
overcome this:
 We can use s from a small pilot study.
 We can use z* in place of the necessary t value.
Application 3: Hypothesis Tests,
One-Population
Part II: Hypothesis Test for a Population Mean

The same four steps apply for a hypothesis test for a population mean:
1.
Hypothesize.
 State your hypotheses about the population parameter.
2.
Prepare.
 Get ready to test: Choose and state a significance level.
 Choose a test statistic appropriate for the hypotheses.
 Check conditions and state any assumptions that must be
made.
3.
Compute to compare.
 Compute the observed value of the test statistic in order to
compare the null hypothesis value to our observed value.
 Find the p-value to measure your level of surprise.
4.
Interpret.
 Do you reject or fail to reject the null hypothesis?
Copyright
What
does
this
mean?
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Pearson
Education,
Inc.. All rights reserved.
Test Statistic for One-population Means

Compare the observed value of the sample mean,
x-bar, to the value claimed by the null hypothesis, µ.

Called the one-sample t-test, is very similar in
structure to the test for one proportion
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Conditions
“Anyone can make a decision, but only a statistician
can measure the probability that the decision is right
or wrong.”
 We need to know the sampling distribution of our test
statistic.
 The sampling distribution follows the t-distribution
under these conditions:


Condition 1: Random Sample and Independence. The data must
be a random sample from a population, and observations must
be independent of one another.

Condition 2: Normal. The population distribution must be
Normal or the sample size must be large. For most situations, 25
is large enough.
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Application 3: Hypothesis Test,
One-Population
Flagship Example
Example: Word Length

Textbook authors must be careful that the reading level of their
book is appropriate for the target audience. Some methods of
assessing reading level require estimating the average word
length. We have randomly chosen 20 words from a randomly
selected page in Stats: Modeling the World and counted the
number of letters in each word:
5, 5, 2, 11, 1, 5, 3, 8, 5, 4, 7, 2, 9, 4, 8, 10, 4, 5, 6, 6

Suppose that our editor was hoping that the book would have a
mean word length of 6.5 letters. Does this sample indicate that
the authors failed to meet this goal? Test an appropriate
hypothesis and state your conclusion.
Example: Word Length (cont.)

Step 1: Hypothesis

Let µ represents the population mean of word lengths. Then




H0: µ = 6.5, The Claim
Ha: µ ≠ 6.5 We will thus use a Two-Tail t-test
The null hypothesis says that the mean is targeting the intended
audience at 6.5 letters per word.
The alternative is that the mean is not targeting the intended
audience.
 The mean word length is either too low or too high
Example: Word Length (cont.)



Step 2: Conditions
 We will test using a 5% significance level.
Conditions
 Random Sample? Stated.
 Independence?
Assumed.
 10% Condition?
The textbook has more than 200 words.
 Normal? A histogram of the observed word lengths looks
‘roughly’ unimodal and symmetric, so the population of all
word lengths may be approximately normal.
It is appropriate to use a one sample t-test.
Example: Word Length (cont.)


Step 3: Do The Math
Enter the word length into a calculator: x-bar = 5.50, s = 2.685



where
together we get:
2.685
t*-Table: Referring to the two-tail header (since we have a two-tail
alternative hypothesis) we see on row 19 that 1.67 is between
combined tail areas of 0.10 to 0.20 (technology yields 0.11).
This means that our p-value, between 0.10 and 0.20, is larger
than the significance value of 0.05 so we fail to reject the null.
Example: Word Length (cont.)

Step 4: Interpret

Statistician's statement:
 The p-value is higher than 0.05, and thus we fail to reject the
null hypothesis!

LA Times:
 So we conclude that this sample does not provide evidence
that the average word length differs from the goal of 6.5
letters.
Example: Word Length (cont.)


Statdisk
Analysis -> Hypothesis Testing -> Mean - One Sample
EXAMPLE 9: Dieting




Is the Weight Watchers diet effective? researchers
examined 40 subjects who were randomly assigned to
this diet. Researchers recorded the change in weight after
12 months.
Only 26 of the 40 subjects stayed with the diet for that
long, so we have data on only these 26 people.
Test the hypothesis that people on the Weight Watchers
diet tend to lose weight. (A negative weight change
means the person lost weight.)
Data: After a year, the average change in weight of the
26 people who stayed on the diet was negative 4.6
kilograms (about 10 pounds), with a standard deviation
of 5.4 kg.
EXAMPLE 9: Dieting (cont.)






Step 1: Hypothesis
Let µ represent the mean weight change of the
population.
H 0: µ = 0
Ha: µ < 0 => Left-Tail t-test
The null hypothesis says that the mean is 0,
because the neutral position here is that no
change occurs, on average. This is the same as
saying that the diet is ineffective.
The alternative is that the mean change is
negative - people lost weight!
EXAMPLE 9: Dieting (cont.)




Step 2: Conditions
We will test using a 5% significance level.
Condition 1: Random Sample and Independence? The
subjects in this study were not selected randomly from the
population of all dieters. Independence assumed.
Condition 2: Normal? The distribution of the sample does
not look Normal, so we suspect the population distribution is
not Normal. But because the sample size is larger than 25,
this condition is satisfied.
EXAMPLE 9: Dieting (cont.)




Step 3: Do The Math
= ( -4.5 - 0) / 1.0590 = -4.34
We use a t-distribution with n - 1 = 25 degrees of freedom.
On row 25 we see that 4.34 is to the left of 0.005, that is our
p-value is less than 0.005 (technology yields 0.0001)
EXAMPLE 9: Dieting (cont.)

Step 4: Interpret

The p-value is much smaller than 0.05, and thus
we reject the null hypothesis!

So we conclude that the mean weight change is
in fact negative, meaning that people do tend to
lose weight after one year on this diet.
EXAMPLE 9: Dieting (cont.)


Statdisk
Analysis -> Hypothesis Testing -> Mean - One Sample
Hypothesis Test Example (by Formula)

1.
Ford claims that its 2012 Focus gets 40 mpg
on the highway. Does your Focus’ mpg differ from
40 mpg? You chart your Focus over 35 randomly
selected highway trips and find it got 39.5 mpg
with a standard deviation of 1.4 mpg.
Hypothesize

2.
H0: µ = 40, Ha: µ ≠ 40
Prepare

Choose α = 0.05, Use t-statistic: random and large sample
Copyright © 2013 Pearson Education, Inc.. All rights reserved.
Ford claims that it’s 2012 Focus gets 40 mpg
on the highway. Does your Focus’ mpg differ from 40
mpg? You chart your Focus over 35 randomly
selected highway trips and find it got 39.2 mpg with a
standard deviation of 1.4 mpg.
3.
Compute to Prepare

4.
Interpret


p-value = 0.04 < α = 0.05
Reject H0. Accept Ha. There is statistically significant
evidence to conclude that your Focus does not get 40 mpg on
average.
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9.4
Comparing Two
Population Means
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Two-Populations Means
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Comparing Two-Populations, Means

We finish this chapter by comparing two
populations to each other.
 We first create a confidence interval for the
differences between two groups
 We
next test samples to challenge a claim between
two groups (a hypothesis test)

We will do this for two types of data sets:
Independent and Dependent (aka unmatched and
matched, also, unpaired and paired)
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Independent vs. Dependent (Paired)

Two samples are dependent or paired if each
observation from one group is coupled with a
particular observation from the other group.





Before and After
Identical Twins
Husband and Wife
Older Sibling and Younger Sibling
If there is no pairing then the samples are
independent.
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Quiz: Independent or Dependent?




Do women perform better on average than men on
their statistics final? 60 women and 40 men were
surveyed.
40 people’s blood pressure was measured before
and after giving a public speech. Does blood
pressure change on average?
Is the average tip percent greater for dinner than
lunch? 35 wait staff who worked both lunch and
dinner looked at their receipts.
Are Americans more stressed out on average
compared to the French? 50 from each country
were given a stress test.
Copyright © 2013 Pearson Education, Inc.. All rights reserved.
Test: Independent or Dependent?




Do women perform better on average than men
on their statistics final? 60 women and 40 men →
were surveyed.
40 people’s blood pressure was measured before
and after giving a public speech. Does blood →
pressure change on average?
Is the average tip percent greater for dinner than
lunch? 35 wait staff who worked both lunch and→
dinner looked at their receipts.
Are Americans more stressed out on average
compared to the French? 50 from each country →
were given a stress test.
Copyright © 2013 Pearson Education, Inc.. All rights reserved.
Ind
Dep
Dep
Ind
Comparing Two Means

First, when comparing
populations, examine
the side-by-side
boxplots

Our parameter of
interest : µ1 – µ2.
Comparing Two Means (cont.)




Recall that Variance = SD2
For independent random quantities, variances add
2
2
 SD2
total = SD 1 + SD 2
So, the standard deviation of the sum or difference of
two populations is the square root of the sum of
variances:
We still don’t know the true standard deviations of the
two groups, so we need to estimate and use the
standard error
Comparing Two Means (cont.)

Because we are working with means and estimating
the standard error of their difference using data, we
use the t-distribution model.

The confidence interval we build is called a
two-sample t-interval.

The hypothesis test is called a two-sample ttest.
Sampling Distribution for the Difference
Between Two Means


When the conditions are met, the standardized sample
difference between the means of two independent groups is
.
The standard error for estimates will be
Assumptions and Conditions

Conditions (Each condition needs to be checked for both
groups.
 Independence Assumption: Is each member of a
sample independent from each other?

Randomization Condition: Were the data collected with
suitable randomization (representative random
samples or a randomized experiment)?

10% Condition: We don’t usually check this condition
for differences of means. We will check it for means
only if we have a very small population or an
extremely large sample.
Assumptions and Conditions (cont.)

Normal Population Assumption:
 Nearly Normal Condition: This must be checked for
both groups. A violation by either one violates the
condition.

Independent Groups Assumption: The two groups we are
comparing must be independent of each other.

See the Next Part (matched/paired) if the groups are not
independent of one another
App2: Two-Population Confidence Interval
When the conditions are met, we are ready to find the confidence
interval for the difference between means of two independent groups.
The confidence interval is
where the standard error of the difference of the means is
The critical value depends on the particular confidence level, C, that you
specify and on the number of degrees of freedom, which we get from the
sample sizes and a special formula.
App2: Two-Population Confidence Interval

The special formula for the degrees of freedom (or
row #) for our t*- value is:

We will let technology calculate degrees of
freedom for us!

((For Exams, we will use the smaller of each
populations degree of freedom: n1 - 1 and n2 - 1 ))
Application 2: Confidence
Interval, Two-Populations
Flagship Example
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EXAMPLE 11 Comparing Men’s and Women’s
Senses of Smell



We compare men and women whose index of smell was
measured while they were lying down. The summary statistics are:
 Men: x-bar = 10.0694, s = 3.3583, n = 18
 Women: x-bar = 11.1250, s = 2.7295, n = 18
The box-plots give us a visual comparison:
Q: Find a 95% confidence interval for the means difference in
smelling ability between men and women.
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EXAMPLE 11 Comparing Men’s and Women’s
Senses of Smell
!
(i) Conditions
! Independence Assumption
!
Randomization Condition:
!
10% Condition:
!
Nearly Normal Condition:
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EXAMPLE 11 Comparing Men’s and Women’s
Senses of Smell
!
(i) Conditions
! Independence Assumption These data consist of two
independent samples: 18 men and 18 women.
!
Randomization Condition: Not Mentioned
!
10% Condition: Not needed if large population.
!
Nearly Normal Condition: The box plots show Q2 and Q3 as
being symmetric, but long tails exist. Proceed with caution since
both sample sizes > 25.
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EXAMPLE 11 Comparing Men’s and Women’s
Senses of Smell
!
!
(ii) Do the Math
! df
! critical value:
!
ME =
!
Interval:
(iii) Conclusion
!
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EXAMPLE 11 Comparing Men’s and Women’s
Senses of Smell
!
!
(ii) Do the Math
! df = min of (18-1 and 18 - 1) = 17 Technology yields = 33
! critical value: t-table row 17, col 0.05, two-tail => t* = 2.110
!
ME =
= 2.11x1.020 = 2.1522
!
Interval: -1.0556 +/- 2.1522, or about (-3.2, 1.1)
(iii) Conclusion
! Because the interval contains zero, we cannot rule out the possibility
that the mean difference in the population is 0. This suggests that men
and women may not differ in their ability to smell.
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EXAMPLE 11 Comparing Men’s and Women’s
Senses of Smell

We compare men and women whose index of smell ...
Men: x-bar = 10.0694, s = 3.3583, n = 18
 Women: x-bar = 11.1250, s = 2.7295, n = 18


Statdisk: Analysis -> Confidence Interval -> Mean 2-Ind Samples
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Conf Int 2 -Pop Indep: Engr &Psych

38 randomly selected engineer majors and 42 randomly
selected psychology majors were observed to estimate the
difference in how long it takes to graduate. Use this
data:

Q: Find a 95% confidence interval for the difference.

The two population are independent since there is no
pairing between each engineer major and each
psychology major.
The students were selected randomly, independently, and
the sample sizes are both greater than 25.

Copyright © 2013 Pearson Education, Inc.. All rights reserved.
Conf Int 2 -Pop Indep: Engr &Psych
! Statdisk: Analysis
-> Conf Intvl -> Mean 2-Ind Samples
Conf Int 2 -Pop Indep: Engr &Psych

38 randomly selected engineer majors and 42 randomly selected
psychology majors were observed to estimate the difference in how
long it takes to graduate. Use this data:

We are 95% confident that the average time it takes to
graduate is between 0.3 and 0.7 years longer for
psychology majors than for engineer majors.
Copyright © 2013 Pearson Education, Inc.. All rights reserved.
Application 3: Hypothesis
Tests, Two-Populations
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II. Hypothesis Test for the Difference Between
Two Means
We test the hypothesis H0: µ1 – µ2 = Δ0, where the
hypothesized difference, Δ0, is almost always 0, using
the statistic
The standard error is
When the conditions are met and the null hypothesis is
true, this statistic can be closely modeled by a
Student’s t-model with a number of dof given by a
special formula. We use that model to obtain a P-value.
Back Into the Pool

Remember that when we know a proportion, we
know its standard deviation.
 Thus, when testing the null hypothesis that
two proportions were equal, we could
assume their variances were equal as well.
 This led us to pool our data for the hypothesis
test.
Back Into the Pool (cont.)

For means, there is also a pooled t-test.
 Like the two-proportions z-test, this test
assumes that the variances in the two groups
are equal.
 But, be careful, there is no link between a
mean and its standard deviation…

We will avoid pooled situations in Means 2-pop
HT
What Can Go Wrong?


Watch out for paired data.
 The Independent Groups Assumption
deserves special attention.
 If the samples are not independent, you can’t
use two-sample methods. Regression?
Look at the boxplots.
 Check for outliers and non-normal
distributions by making and examining
boxplots.
HT 2-pop Paired: Chocolate & Memory

Does eating chocolate improve memory. 12 people
were give a memory test before and after eating
chocolate. The data for the number of words recalled
out of 50 are shown below. Assume Normality.

1.
16
33
9
42
38
27
30
41
After
20
29
11
42
39
25
34
44
26
Hypothesize

2.
Before 24
H0: µdiff = 0,
Ha: µdiff ≠ 0
Prepare

α = 0.05, T-Statistic, large sample
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HT 2-pop Paired: Chocolate & Memory
3.
Compute to Compare

4.



Stat → T Statistics → Paired
Interpret
P-value = 0.13 > 0.05 = α
Fail to Reject H0
Conclusion:
There is insufficient evidence to make a conclusion about the
mean number of words memorized increasing after eating
chocolate.
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HT 2-pop Indp Samples: Hot & Cold Batteries

1.
Do batteries last longer in colder climates than in
warmer ones? The table shows some randomly selected
battery lives in months.
Florida
19
22
25
21
18
19
27
25
Montreal
37
49
22
26
47
41
38
37
Hypothesize
 H0: µF = µM
 Ha:
µF < µM
2. Prepare
= 0.05
 Independent Samples,
 Assume Normal Distributions
α
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28
15
HT 2-pop Indp Samples: Hot & Cold Batteries
3.
Compute to Compare
 Stat → T Statistics → Two sample → with data
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HT 2-pop Indp Samples: Hot & Cold Batteries
Florida
19
22
25
21
18
19
27
25
Montreal
37
49
22
26
47
41
38
37
4.
28
15
Interpret

P-value = 0.0009 < 0.05 = α
Reject H0

Accept Ha

Conclusion: There is statistically significance evidence to
support the claim that on average batteries last longer in
Montreal than in Florida.

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Chapter 9
Guided Exercise 1
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HT 1-pop: Is the Mean Body Temperature
really 98.6?


1.
A random sample of 10 independent healthy people showed
body temperatures (in degrees Fahrenheit) as follows:
 98.5, 98.2, 99.0, 96.3, 98.3,
98.7, 97.2, 99.1, 98.7, 97.2
Use α = 0.05.
Hypothesize
 H0: µ = 98.6

Ha: µ ≠ 98.6
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HT 1-pop: Is the Mean Body Temperature
really 98.6?
2. Prepare

Not far from normal.

Sample collected randomly.

Use the t-statistic.
Copyright © 2013 Pearson Education, Inc.. All rights reserved.
HT 1-pop: Is the Mean Body Temperature
really 98.6?
3. Do the Math



t ≈ -1.65
p-value ≈ 0.13
p-value ≈ 0.13 > 0.05 = α
4. Interpret

We cannot reject 98.6 as the population mean body
temperature from these data at the 0.05 level.
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Chapter 9
Guided Exercise 2
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HT 2-pop Indep: TV Watching and Wealth

A two-sample t-test for the number of televisions
owned in households of random samples of students
at two different community colleges. Assume
independence. One of the schools is in a wealthy
community (MC), and the other (OC) is in a less
wealthy community.
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HT 2-pop Indep: TV Watching and Wealth
1. Hypothesize

Let µoc be the population mean number of televisions owned
by families of students in the less wealthy community (OC),
and let µmc be the population mean number of televisions
owned by families of students at in the wealthy community
(MC).
 H0: µoc = µm

Ha: µoc ≠ µm
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HT 2-pop Indep: TV Watching and Wealth
2. Prepare




Choose an appropriate t-test. Because the sample sizes are 30,
the Normality condition of the t-test is satisfied. State the
other conditions, indicate whether they hold, and state the
significance level that will be used.
Use a t-test with two independent samples.
The households were chosen randomly and independently.
The population of all households of each type is more than 10
times the sample sizes.
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HT 2-pop Indep: TV Watching and Wealth
3. Do the Math


t = 0.95
p-value = 0.345
4. Interpret

Since the p-value = 0.345 is very large, we fail to reject H0.

At the 5% significance level, we cannot reject the hypothesis that the mean
number of televisions of all students in the wealthier community is the same
as the mean number of televisions of all students in the less wealthy
community.
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Chapter 9
Guided Exercise 3
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HT 2-pop Paired: Pulse Before and After Fright

Test the hypothesis that the
mean of college women’s pulse
rates is higher after a fright,
using α = 0.05.

1. Hypothesize
 H0: µbefore = µafter

Ha: µbefore > µafter
2. Prepare
 Choose a test: Should it be a paired t-test or a
two-sample t-test? Why? Assume that the
sample was random and that the distribution
of differences is sufficiently Normal.
Mention the level of significance.
 Paired t-test since before and after.
 Level of Significance: α = 0.05.
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HT 2-pop Paired: Pulse Before and After Fright
3. Compute to Compare

t ≈ 4.9

p-value = 0.002

0.002 < 0.05
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HT 2-pop Paired: Pulse Before and After Fright
4. Interpret

Reject or do not reject H0. Then write a sentence that includes
“significant” or “significantly” in it. Report the sample mean pulse rate
before the scream and the sample mean pulse rate after the scream.
 Reject H0. There is statistically significant evidence to support the
claim that mean blood pressure is higher after a fright.
 µbefore ≈ 74.8

µafter ≈ 83.7
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