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Chapter 9
Inferring Population
Means
Copyright © 2014 Pearson Education, Inc. All rights reserved
Learning Objectives



9- 2
Understand when the Central Limit Theorem for
sample means applies and know how to use it to
find approximate probabilities for sample means.
Know how to test hypotheses concerning a
population mean and concerning the comparison of
two population means.
Understand how to find, interpret, and use
confidence intervals for a single population mean
and for the difference of two population means.
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Learning Objectives Continued


9- 3
Understand the meaning of the p-value and of
significance levels.
Understand how to use a confidence interval to
carry out a two tailed hypothesis test for a
population mean or for a difference of two
population means.
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9.1
Sample Means of
Random Samples
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Statistics, Parameters, Means and
Proportions



9- 5
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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Accuracy of the Sample Mean
If the sample mean is accurate, then the
average of all sample means will equal the
population mean.
 If Simple Random Sampling is used the
sample mean is accurate, also called unbiased.
 Other sampling techniques to be looked at
later produce results that are close to being
unbiased.

9- 6
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Precision and the Sample Mean
The precision of the sample mean describes
how much variability there is from one
sample mean to the next.
 If the population standard deviation is small
the sample mean will have more precision.
 If the sample size is large the sample mean
will have more precision.

9- 7
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Simulating Many Sample Means

As the sample size increases
 Better
Precision
 Accuracy Does Not Change
9- 8
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Standard Error



9- 9
The Standard Error is the standard deviation
of the sampling distribution.
x  
x 

n
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Standard Error and Sample Size
x 

n
The Standard Error is smaller for larger
sample sizes.
 Increasing the sample size by a factor of 4
decreases the standard error by a factor of 2.
 Increasing the sample size by a factor of 100
decreases the standard error by a factor of 10.

9 - 10
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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.


$2.75
Find the standard error for carts with 36 items.

9 - 11
Parameter
1.26
x 
 0.21
36
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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

9 - 12
→ $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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Conditions for the Central Limit
Theorem for Sample Means
Random Sampling Technique
 One or Both of the Following:

 Population
is Normally Distributed
 Sample Size is Large

9 - 14
Population Size is At Least 10 Times Bigger
Than the Sample Size
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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.

9 - 15
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Central Limit Theorem For Means

Central Limit Theorem: If the conditions are
met and the population has mean  and
standard deviation , then the sampling
distribution will be approximately normal.
  
N  ,

n

9 - 16
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Visualizing the Central Limit Theorem
9 - 17

Population Distribution
Skewed Right

Sampling Distribution
Approximately Normal
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The Central Limit Theorem and Shape
9 - 18
→
Sampling Distribution
Mound Shaped
→
Sampling Distribution
Unimodal
Population Distribution
Skewed Left
→
Sampling Distribution
Symmetric
Population Distribution
Not Symmetric
→
Sampling Distribution
Symmetric

Population Distribution
Uniform

Population Distribution
Bimodal


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Applying the Central Limit Theorem
The distribution of women’s pulse rates is
skewed right with  = 74 bpm,  = 13 bpm.
 If 30 women are selected, find P ( x  72)

9 - 19

13 

N  74,
  N (74, 2.1)
38 


P( x  72)  0.17
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Applying the Central Limit Theorem
The distribution of women’s pulse rates is
skewed right with  = 74 bpm,  = 13 bpm.
 If one woman is selected can you find

P(x < 72)
 No,
since the distribution is skewed right, it is not
normal. Without more information about the
distribution this probability cannot be found.
9 - 20
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Population, Sample, and Sampling
Distributions


The population distribution is the distribution of all
individuals that exist.
The distribution of the sample is the distribution of
the individuals that were surveyed.


The sampling distribution is the distribution of all
possible sample means of sample size n.

9 - 21
The mean, standard deviation, and the shape are likely to
be close to the population distribution.
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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The t-Distribution
If  is unknown, we cannot find the z-score.
 Use the sample standard deviation s instead.

x 
t
s
n

9 - 22
SEEST
s is an estimate for the

n standard error
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Facts About the t-Distribution
Bell shaped
 Tails a little bigger than Normal
 Given n there are n – 1 degrees of freedom.
 For large degrees of freedom, the distribution
is almost normal.

9 - 23
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9.3
Answering Questions
about the Mean of a
Population
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Confidence Interval for a Population
Mean
Gives a plausible range of values for the
population mean.
 Confidence level gives the percent of all
possible confidence intervals that contain the
population mean.
 Similar to confidence interval for a
population proportion, but used for a
quantitative variable.

9 - 25
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CI Example:





9 - 26
x  t  SEEST
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.
s
2

 0.30
d.f. = 44 → t = 1.68, SEEST 
n
45
Lower Bound: 9 – 1.68 x 0.30 ≈ 8.5
Upper Bound: 9 + 1.68 x 0.30 ≈ 9.5
(8.5,9.5)
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CI Interpretations: (8.5,9.5)


9 - 27
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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CI Interpretations: (8.5,9.5)


9 - 28
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,
each survey would result in a different confidence
interval. 90% of these confidence intervals will
succeed in containing the actual population mean
number of hours worked on homework and 10%
will not contain the true population mean.
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Confidence Intervals and StatCrunch
9 - 29

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.

Stat →T statistics → One sample →with summary
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Confidence Intervals Given Data


9 - 30
Data was collected on recovery time for
the newest flu virus. Find a 95%
confidence interval for the population
mean recovery time.
Stat →T statistics → One sample →with data
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Confidence Intervals Given Data



9 - 31
Data was collected on recovery time for the newest
flu virus. A 95% confidence interval for the
population mean recovery time was (5.9,7.5).
We are 95% confident that the mean recovery time
for all people who get the flu is between 5.9 and 7.5
days.
If many random groups are observed then a different
confidence interval would result from each. 95% of
these confidence intervals will contain the true
population mean recovery time while 5% will not
contain the true population mean recovery time.
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Hypothesis Test for a Population Mean

The same four steps apply for a hypothesis
test for a population mean:
1.
2.
3.
4.
9 - 32
Hypothesize: State H0 and Ha.
Prepare: Choose a, check conditions and
assumptions and determine the test statistic to
use.
Compute to Compare: Compute the test
statistic and the p-value and compare p with a.
Interpret: Reject or fail to Reject H0? Write
down the conclusion in the context of the study.
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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
 H0:
2.
Prepare

9 - 33
 = 40, Ha:  ≠ 40
Choose a = 0.05, Use t-statistic: random and
large sample
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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.
39.5  40
t
 2.11
1.4
35
p  value  0.04
Interpret
= 0.04 < a = 0.05
 Reject H0. Accept Ha. There is statistically
significant evidence to conclude that your Focus
does not get 40 mpg on average.
 p-value
9 - 34
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Hypothesis Test with Data Using
StatCrunch

1.
150 minutes per week of exercise is
recommended. Do college students exercise
more than 150 minutes per week? 42
randomly selected college tracked their
weekly exercise.
Hypothesize
 H0:
2.
Prepare
 Use
9 - 35
 = 150, Ha:  > 150
a = 0.05 and find a t-statistic, large sample.
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150 minutes per week of exercise is recommended.
Do college students exercise more than 150 minutes
per week? 42 randomly selected college tracked
their weekly exercise.
3. Compute to Compare

9 - 36
Stat →T statistics → One sample →with data
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150 minutes per week of exercise is recommended.
Do college students exercise more than 150 minutes
per week? 42 randomly selected college tracked
their weekly exercise.
4.
Interpret



9 - 37
P-value = 0.081 > 0.05 = a
Fail to reject H0
There is statistically insufficient evidence to conclude
that the average college student exercises more than the
recommended 150 hours.
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9.4
Comparing Two
Population Means
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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

9 - 39
If there is no pairing then the samples are
independent.
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Independent (Ind) or Dependent (Dep)?




9 - 40
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.
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→ Ind
→ Dep
→ Dep
→ Ind
Independent Samples Standard Error and
Margin of Error

SEEST
2
1
2
s
s2


n1 n2
Margin of Error  t  SEEST
 Degrees of Freedom is approximately the
smaller of n1 – 1 and n2 – 1.
 Use a computer or calculator for better
accuracy.

9 - 41
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Requirement for Independent Samples
Both samples are randomly taken and each
observation is independent of any other.
 The two samples are independent of each
other (not paired).
 Either both populations are Normally
distributed or each sample size is greater than
25.

9 - 42
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Example: Independent Samples



9 - 43
38 randomly selected engineer majors and 42
randomly selected psychology majors were
observed to estimate the difference in how long it
takes to graduate. xE  5.1, sE  0.4, xP  5.6, sP  0.5
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.
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38 randomly selected engineer majors and 42
randomly selected psychology majors were observed
to estimate the difference in how long it takes to
graduate. xE  5.1, sE  0.4, xP  5.6, sP  0.5
Find a 95% confidence interval for the difference.
 Stat → T Statistics → Two sample → with summary
9 - 44
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38 randomly selected engineer majors and 42
randomly selected psychology majors were observed
to estimate the difference in how long it takes to
graduate. xE  5.1, sE  0.4, xP  5.6, sP  0.5
Find a 95% confidence interval for the difference.

9 - 45
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.
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Hypothesis Test: Paired Samples


1.
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.
Before 24
16
33
9
42
38
27
30
41
After
20
29
11
42
39
25
34
44
Hypothesize
 H0:
9 - 46
26
diff = 0,
Ha: diff ≠ 0
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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.
2.
Prepare

3.
a = 0.05, T-Statistic, large sample
Compute to Compare
 Stat
9 - 47
→ T Statistics → Paired
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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.
4.
Interpret
= 0.13 > 0.05 = a
 Fail to Reject H0
 Conclusion:
There is insufficient evidence to make a
conclusion about the mean number of words
increasing after eating chocolate.
 P-value
9 - 48
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Hypothesis Test: Independent Samples

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
1.
Hypothesize
F = M
 Ha: F < M
 H0:
9 - 49
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28
15
Do batteries last longer in colder
climates than in warmer ones?
 Prepare
a
= 0.05
 Independent Samples,
 Assume Normal Distributions
9 - 50
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Do batteries last longer in colder
climates than in warmer ones?
3.
9 - 51
Compute to Compare
 Stat → T Statistics → Two sample → with data
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Do batteries last longer in colder
climates than in warmer ones?
Florida
19
22
25
21
18
19
27
25
Montreal
37
49
22
26
47
41
38
37
4.
28
15
Interpret
= 0.0009 < 0.05 = a
 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.
 P-value
9 - 52
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9.5
Overview of Analyzing
Means
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General Formulas

Hypothesis Test Statistic
Test statistic

estimated value    null hypothesis value 


SE
Confidence Interval
CI : estimated value   multiplier  SEEST
9 - 54
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Finding the p-value Given the Test
Statistic

Left Tailed Hypothesis:
 Find
the probability that a value is less than the
test statistic .

Right Tailed Hypothesis:
 Find
the probability that a value is greater than
the test statistic .

Two Tailed Hypothesis:
 Make
the test statistic negative. Then find the
probability that a value is less than the test
statistic. Finally multiply by 2.
9 - 55
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Comparing CI and Hypothesis Tests

It can be concluded at the 5% level that the
value is not the mean, proportion, or
difference if
a
value falls outside the 95% confidence interval
 the p-value is less than 0.05

9 - 56
A 95% (90%, 99%) confidence interval is
equivalent to a two-tailed test with a = 0.05
(0.1, 0.01) when it comes to rejecting or
failing to reject H0.
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Hypothesis Tests and CI Example

Suppose that a hypothesis test:
H0:  = 80
Ha:  ≠ 80
was done for the average height of male
college basketball players. If p-value = 0.02
can the 95% confidence interval contain 80?
 No.
Since the p-value < 0.05, H0 is rejected. 80
cannot be in the confidence interval.
9 - 57
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Hypothesis Test or Confidence Interval:
Which Should be Used?
For one-tailed testing: hypothesis test
 For two tailed testing: either can be used
 Confidence Intervals give more than
hypothesis tests.

 CI
gives a plausible range for the population
value.

9 - 58
The hypothesis test addresses the question of
whether H0 is false
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Chapter 9
Case Study
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Epilepsy, Drugs, and Giving Birth
Four drugs are taken for epilepsy:
carbamazepine, lamotrigine, phenytoin, and
valproate.
 Three years after pregnant mothers took the
medicine, their children were given a IQ test.
 The New England Journal of Medicine
reported that taking valproate increased the
risk of impaired cognitive development.

9 - 60
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95% Confidence Intervals
These give us a visual comparison.
 The valporate CI does not overlap with the
lamotrigine CI.
 For better comparisons, use confidence
intervals for the difference between means.

9 - 61
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Confidence Intervals for Differences


9 - 62
None contain 0. A hypothesis test for a difference
between the means will reject H0.
There is statistically significant evidence to
conclude that the mean IQ for children born to
mothers taking valproate is different than for any of
the other drugs.
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Chapter 9
Guided Exercise 1
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Is the Mean Body Temperature really
98.6?

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

1.
Use a = 0.05.
Hypothesize
 = 98.6
 Ha:  ≠ 98.6
 H0:
9 - 64
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2. Prepare
Not far from normal.
 Sample collected randomly.
 Use the t-statistic.

9 - 65
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3. Compute to Compare
t ≈ -1.65
 p-value ≈ 0.13
 p-value ≈ 0.13 > 0.05 = a

9 - 66
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4. Interpret

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
p-value = 0.13 > 0.05 = a
 We cannot reject 98.6 as the population mean
body temperature from these data at the 0.05
level.

9 - 67
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Chapter 9
Guided Exercise 2
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
9 - 69
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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1. Hypothesize



9 - 70
Let oc be the population mean number of televisions
owned by families of students in the less wealthy
community (OC), and let mc be the population mean
number of televisions owned by families of students
at in the wealthy community (MC).
H0: oc = m
Ha: oc ≠ m
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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.

9 - 71
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3. Compute to Compare
t = 0.95
 p-value = 0.345

9 - 72
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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.

9 - 73
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Chapter 9
Guided Exercise 3
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Pulse Before and After Fright
Test the hypothesis that the
mean of college women’s pulse
rates is higher after a fright,
using a = 0.05.
 1. Hypothesize

before = after
 Ha: before > after
 H0:
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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: a = 0.05.
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3. Compute to Compare
t ≈ 4.9
 p-value = 0.002
 0.002 < 0.05

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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.



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Reject H0. There is statistically significant evidence to
support the claim that mean blood pressure is higher after
a fright.
before ≈ 74.8
after ≈ 83.7
Copyright © 2014 Pearson Education, Inc. All rights reserved