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Chapter 11
Inference with Means
Sampling Distribution,
Confidence Intervals, and
Hypothesis Tests
The campus of Fritz University
has a fish pond. Suppose there
are 20 fish in the pond. The
lengths of the fish (in inches)
are given below:
4.5
5.4 10.3 7.9
6.3
4.3
9.6
8.5
6.6 11.7 8.9
2.2
9.8
8.7 13.3 4.6 10.7 13.4 7.7
5.6
This is a statistic!
We caught fish with lengths 6.3 The true mean m = 8.
Let’s
catch
inches,Suppose
2.2 inches,
and
13.3
inches.
Notice
that
some
we randomly catch a sample
ofan
This
is
two
more
x = 7.27
inches
sample
means
are
example
of
3 fish from this pond and measure
their
samples
and
closer and some
2nd sample
8.5,
4.6,
and
5.6
inches.
sampling
length. What would the mean
length
of
farther
away;
some
look
at
the
x = 6.23 inches
variability
above
and
some
below
the
sample
be?
sample
means.
rd
3 sample – 10.3, 8.9, and 13.4 inches.the mean.
x = 10.87 inches
Fish Pond Continued . . .
4.5
5.4 10.3 7.9
6.3
4.3
9.6
8.5
6.6 11.7 8.9
2.2
9.8
8.7 13.3 4.6 10.7 13.4 7.7
5.6
There are 1140 (20C3) different possible samples
of size 3 from this population. If we were to
catch all those different samples and calculate
the mean length of each sample, we would have a
distribution of all possible x.
This would be the sampling distribution of x.
Sampling Distributions of x
• The distribution that would be formed
by considering the value of a sample
statistic for every possible sample of a
given size from a population.
In this case, the sample
statistic is the sample
mean x.
Fish Pond Revisited . . .
Suppose there are only 5 fish
in the pond. The lengths of
the fish (in inches) are given
below:
6.6 11.7 8.9
2.2
What is the mean
mand
7.84
x = standard
deviation of this
sxpopulation?
= 3.262
9.8
We will keep the
population size
small so that we
can find ALL the
possible samples.
Fish Pond Revisited . . .
6.6 11.7 8.9
2.2
9.8
mx = 7.84 and sx = 3.262
Pairs
6.6 &
11.7
6.6 &
8.9
6.6 &
2.2
6.6 &
9.8
x
9.15
7.75
4.4
8.2
11.7 &
8.9
11.7 &
2.2
11.7 &
9.8
8.9 &
2.2
8.9 &
9.8
the9.35
10.3Let’s
6.95 find
10.75all5.55
samples of size 2.
2.2 &
9.8
6
These values determine the How
doisthese
What
the mean
mx = 7.84
sampling
distribution of x for
valuesand
compare
to
standard
samples of size 2.
the
population
deviation
of these
sx = 1.998
meansample
and standard
means?
deviation?
Fish Pond Revisited . . .
6.6 11.7 8.9
2.2
9.8
mx = 7.84 and sx = 3.262
11.7,
Now6.6,let’s11.7,
find 11.7,
all the
2.2,
8.9,
8.9,
2.2,
samples
of
size
3.
9.8
2.2
9.8
9.8
Triples
6.6,
11.7,
8.9
6.6,
11.7,
2.2
6.6,
11.7,
9.8
6.6,
8.9,
2.2
6.6,
8.9,
9.8
x
9.067
6.833
9.367
5.9
8.433
6.2
7.6
10.133
7.9
What is the mean
mThese
values
values determine How
the do
x = 7.84
andthese
standard
compare to
sampling distribution of x for
deviation
ofthe
these
sx =samples
1.332of size 3. population
and
samplemean
means?
standard deviation?
8.9,
2.2,
9.8
6.967
What do you notice?
• The mean of the sampling distribution
EQUALS the mean of the population.
mx = m
• As the sample size increases, the
standard deviation of the sampling
distribution decreases.
as n
sx
General Properties of Sampling
Distributions of x
Rule 1:
Rule 2:
mx  m
sx 
s
n
Note that in the
previous fish pond
examples this standard
deviation formula was
not correct because
the sample sizes were
more than 10% of the
population.
This rule is exact if the population is infinite, and is
approximately correct if the population is finite and
no more than 10% of the population is included in
the sample
Sampling Distribution Activity – Day 1
General Properties Continued . . .
Rule 3:
When the population distribution is
normal, the sampling distribution of x is
also normal for any sample size n.
The paper “Mean Platelet Volume in Patients with
Metabolic Syndrome and Its Relationship with Coronary
Artery Disease” (Thrombosis Research, 2007) includes
data that suggests that the distribution of platelet
volume
of use
patients
whotodogenerate
not haverandom
metabolic
syndrome
We can
Minitab
samples
from
population.normal
We will
generate
random
samples
is this
approximately
with
mean m500
= 8.25
and standard
of n s= =5 0.75.
and compute the sample mean for each.
deviation
Platelets Continued . . .
Similarly, we will generate 500 random samples of n = 10,
n = 20, and n = 30. The density histograms below display
the resulting 500 x for each of the given sample sizes.
What do
do you
you notice
notice
What
do
you
notice
What
about the
themeans
standard
about
the
shape
of
about
deviation
of these
these
histograms?
these
histograms?
histograms?
General Properties Continued . . .
Rule 4: Central Limit Theorem
When n is sufficiently large, the sampling
distribution of x is well approximated by
a normal curve, even when the population
distribution is not itself normal.
How large is “sufficiently large”
CLT can safely be anyway?
applied if n exceeds 30.
The paper “Is the Overtime Period in an NHL Game Long
Enough?” (American Statistician, 2008) gave data on the
time (in minutes) from the start of the game to the
first goal scored for the 281 regular season games from
the 2005-2006 season that went into overtime. The
density histogram for the data is shown below.
Let’s consider these 281 values as
a population. The distribution is
strongly positively skewed with
mean m = 13 minutes and with a
median of 10Using
minutes.
Minitab, we will
generate 500 samples of
the following sample sizes
from this distribution:
n = 5, n = 10, n = 20, n = 30.
What
do
you
notice
These
Are
these
are
histograms
What
dothe
youdensity
notice
about
the
standard
histograms
centered
for
the
at 500
about
the
shape
of
deviations
of
these
samples
approximately
m = 13?
these
histograms?
histograms?
A hot dog manufacturer asserts that one of its
brands of hot dogs has a average fat content of 18
grams per hot dog with standard deviation of 1
gram. Consumers of this brand would probably not
be disturbed if the mean was less than 18 grams,
but would be unhappy if it exceeded 18 grams. An
independent testing organization is asked to
analyze a random sample of 36 hot dogs. Suppose
the resulting sample mean is 18.4 grams. Does this
Since the sample size is
result suggest that the manufacturer’s claim is
greater than 30, the
incorrect?
Central Limit Theorem
applies.
So the distribution of x is
approximately normal with
mx  18 and sx 
1
 .1667
36
Hot Dogs Continued . . .
mx  18 and sx 
1
 .1667
36
Suppose the resulting sample mean is 18.4 grams.
What do we
Does this result suggest that the manufacturer’s
call this?
claim is incorrect?
P(x > 18.4) =
Yes, the probability of getting a sample mean
of 18.4 is small enough to cause us to doubt
that the manufacturer’s claim is correct.
Now let’s look at confidence
intervals to estimate the mean
m of a population.
Cosmic radiation levels rise with increasing
altitude, promoting researchers to consider
how pilots and flight crews might be affected
by increased
exposure
cosmic
A study
The use
of theto
value
of sradiation.
introduces
reportedextra
a mean
annual cosmic
radiation we
dose
of 219
variability.
Therefore,
will
mrems with a standard deviation of 35 mrems for a
use the distribution of t values since it
sample of flight personnel of Xinjiang Airlines.
a standard
Supposehas
thismore
mean variability
is based on than
a random
sample of
normal curve.
100 flight crew members.
Remember that:
sx 
s
Are these values
parameters or statistics?
n
What value would you use
for s?
What are t-distributions?
• Created by William Gosset
Let’s look at some graphs of
t-distributions
Important Properties of t
Distributions
t distributions
are described
The t distribution
corresponding
to any by
particular
degreesofoffreedom
freedom
(df).shaped and
number of degrees
is bell
centered at zero (just like the standard normal (z)
distribution).
2) Each t distribution is more spread out than the
standard normal distribution.
1)
z curve
t curve for 2 df
0
Why is the z
curve taller
than the t curve
for 2 df?
Important Properties of t
Distributions Continued . . .
3) As the number of degrees of freedom increases,
the spread of the corresponding t distribution
decreases.
For what df would the
4) As the number of degrees oftfreedom
increases,
distribution
be
the corresponding sequence approximately
of t distributions
the
approaches the standard normal
samedistribution.
as a standard
normal
z curve
distribution?
t curve for 2 df
t curve for 5 df
0
Confidence intervals for m when s
is unknown
Confidence intervals for m when s
is unknown
The general formula for a confidence interval for a
population mean m based on a sample of size n is. . .
But this is unknown – so what
What
is this called?
What What
is thisiscalled?
this
called?
do
we use?
Standard error
s 
x  (t critical value)

 n
estimate
Margin of error
Where the t critical value is based on df = n - 1.
How to find t*
• Use Table B for t distributions
• Look up confidence level at bottom & df
on the sides
• df = n – 1
Can also use invT on the calculator!
Find these t*
90% confidence when n = 5
95% confidence when n = 15
t* = 2.132
t* = 2.145
Steps for a Confidence Interval
We typically use either a boxplot or a normal
• Verify assumptions
probability plot to assess normality.
– Random sample
– Distribution is (approximately) normal AND why
• Given
• Large sample size
• Graph data
You MUST show the graph
• Calculate interval
• Write conclusion in context
We are ____% confident that the true mean ______
is between ___ and ___.
Cosmic radiation levels rise with increasing
altitude, promoting researchers to consider
how pilots and flight crews might be affected
by increased exposure to cosmic radiation. A study
reported a mean annual cosmic radiation dose of 219
mrems and a standard deviation of 35 mrems for a
sample of flight personnel of Xinjiang Airlines.
Suppose this mean is based on a random sample of
100 flight crew members.
Calculate and interpret a 95% confidence interval
for the actual mean annual cosmic radiation exposure
for Xinjiang flight crew members.
1)Data is from a random sample of crew members
First, verify that the
2)Sample size n is large
(n > 30)
conditions
are met.
Cosmic Radiation Continued . . .
Let
What would happen to the width of
x = 219
mrems
this
interval if the confidence
level
wascrew
90%members
instead of 95%?
n = 100
flight
s = 35 mrems.
Calculate and interpret a 95% confidence interval
for the actual mean annual cosmic radiation exposure
for Xinjiang flight crew members.
 35 
219  1.984
  (212.06, 225.94)
 100 
What does this mean
in context?
We are 95% confident that the actual mean annual
cosmic radiation exposure for Xinjiang flight crew
members is between 212.06 mrems and 225.94 mrems.
The article “Chimps Aren’t Charitable” (Newsday,
November 2, 2005) summarized the results of a research
study published in the journal Nature. In this study,
chimpanzees learned to use an apparatus that dispersed
food when either of two ropes was pulled. When one of
the ropes was pulled, only the chimp controlling the
Firstthe
verify
apparatus received food. When
otherthat
rope was
conditions
forcontrolling
a
pulled, food was dispensedthe
both
to the chimp
t-interval
are met.cage. The
the apparatus and also a chimp
in the adjoining
accompanying data represent the number of times out of
36 trials that each of seven chimps chose the option that
would provide food to both chimps (charitable response).
23
22
21
24
19
20
20
Compute a 99% confidence interval for the
mean number of charitable responses for
the population of all chimps.
Chimps Continued . . .
23
Normal Scores
2
1
-1
-2
22
21
24
19
20
20
The plot is
straight,
Let’s suppose itreasonable
is reasonable
to
Since n is small,
we this
needsample
toitverify
ifplausible
it is
so
seems
regard
of seven
plausible
that this
sample
is
from
a the
that
the
population
chimps
as
representative
of
20
22
24
Number
of
Charitable
Responses
population that ischimp
approximately
normal.
distribution
of
population.
number of charitable
responses is
Let’s use a normal probability plot.
approximately
normal.
Chimps Continued . . .
23
22
21
24
19
20
x = 21.29 and s = 1.80
20
df = 7 – 1 = 6
 s 
x  (t critical value)

 n
 1.80 
21.29  3.71
  (18.77, 23.81)
 7 
We are 99% confident that the
mean number of charitable
responses for the population of all
chimps is between 18.77 and 23.81.
Choosing
Sample
Size
Thisarequires
s to be
known – since it is rarely
known,
MUST
assume
The margin
ofwe
error
associated
with a
it is some
specified
confidence
interval
is value.
 s 
m  z *

 n
Solve this for n:
We can use
this to find the
necessary
sample size for
a particular
margin of
error.
The financial aid office wishes to
estimate the mean cost of textbooks per
quarter for students at a particular
university. For the estimate to be useful,
it should be within $20 of the true
population mean. How large a sample
should be used to be 95% confident of
achieving this level of accuracy?
Assume s = $100
Always round
sample size up
to the next
whole number!
 100 
20  1.96

 n 
n  96.04 
n  97
Hypothesis Tests
for a Population Mean
Let’s review the assumptions for a
confidence interval for a population mean
1) x is the sample mean from a random sample,
2) The distribution is (approximately) normal
because sample size n is large (n > 30) or the
graph indicates it’s plausible, and
3) s, the population standard deviation, is
unknownThe assumptions are the same for a
large-sample hypothesis test for a
population mean.
The One-Sample t-test for a
Population Mean
Null hypothesis:
Alternative Hypothesis:
Test Statistic:
H0: m = hypothesized value
Ha: m > hypothesized value
Ha: m < hypothesized value
Ha: m ≠ hypothesized value
x m
t 
s
n
How to find a p-value
To find a p-value:
• Use tcdf(LB, UB, df)
What would you do if this
were a two-sided test?
Find the p-value for a one-sided test with
a test statistic t = -1.5 when n = 15.
tcdf(-∞,-1.5,14) = .0779
A study conducted by researchers at Pennsylvania
State University investigated whether time
perception, an indication of a person’s ability to
concentrate, is impaired during nicotine withdrawal.
After a 24-hour smoking abstinence, 20 smokers were
asked to estimate how much time had elapsed during a
45-second period. Researchers wanted to see whether
smoking abstinence had a negative impact on time
perception, causing elapsed time to be overestimated.
Suppose the resulting data on perceived elapsed time
(in seconds) were as follows:
69 55 72 73 59 55 39 52 57 57
56 50 70 47 56 45 70 54 57 53
What is the mean and standard
x = 57.30 s = 9.30
n
=
20
deviation of the sample?
Smoking Abstinence Continued . . .
69 55 72 73 59 55 39 52 57 57
56 50 70 47 56 45 70 54 57 53
x = 57.30 s = 9.3 n = 20
Where m is the true mean perceived elapsed
H0: m = 45
time for smokers who haveState
abstained
from
the
Ha: m > 45
smoking for 24-hours
Since the boxplot is approximately hypotheses.
Assumptions:
symmetrical, it is plausible that the
1) It is reasonable
todistribution
believe thatisthe sampleVerify
of smokers
population
To do this, we need to
graph the
assumptions.
is representative
of
all
smokers.
approximately normal.
data using a boxplot or normal
2) Since the sample size is probability
not
plot
at least 30, we must
determine if it is plausible
that the population
distribution is approximately
40 50 60 70
normal.
Smoking Abstinence Continued . . .
69 55 72 73 59 55 39 52 57 57
56 50 70 47 56 45 70 54 57 53
x = 57.30 s = 9.3 n = 20
H0: m = 45
Ha: m > 45
Where m is the true mean perceived elapsed
time for smokers whoCompute
have abstained
from
the test
smoking for 24-hours
Test statistic:
P-value ≈ 0
statistic and P-value.
57.30  45
t 
 5.92
9.30
20
a = .05
Since P-value < a, we reject H0. There is convincing
evidence that the mean perceived elapsed time is
greater than the actual elapsed time of 45 seconds.
Smoking Abstinence Continued . . .
69 55 72 73 59 55 39 52 57 57
56 50 70 47 56 45 70 54 57 53
x = 57.30 s = 9.3 n = 20
Compute
the appropriate
Where
m
is
the
true
mean
perceived elapsed
H0: m = 45
confidence
time for smokers
who have interval.
abstained from
Ha: m > 45
smoking for 24-hours
Since P-value < a, we reject H0.
Notice that the
a = .05

hypothesized value 57.30  1.729 9.30



of 45 is NOT in the
20 

90%
confidence
Since
this is a one-tailed
test,
a )
(
53
.
705
,
60
.
895
interval
wetail. .05 goes in
goes inand
thethat
upper
“rejected”
H0! leaving .90 in the
the lower tail,
middle.
A growing concern of employers is time spent in activities
like surfing the Internet and emailing friends during work
hours. The San Luis Obispo Tribune summarized the
findings of a large survey of workers in an article that
ran under the headline “Who Goofs Off More than 2
Hours a Day? Most Workers, Survey Says” (August 3,
2006). Suppose that the CEO of a large company wants
to determine whether the average amount of wasted time
during an 8-hour day for employees of her company is less
than the reported 120 minutes. Each person in a random
sample of 10 employees was contrasted and asked about
daily wasted time at work. The resulting data are the
following:
108 112 117 130 111
131 113 113 105 128
What is the mean and standard
x = 116.80 s = 9.45 n = 10
deviation of the sample?
Surfing Internet Continued . . .
108 112 117 130 111 131 113 113 105 128
x = 116.80 s = 9.45 n = 10
H0: m = 120
Ha: m < 120
Where m is the true mean daily wasted
time for employees of this company
The boxplot reveals some skewness,
State
the
Verify
the
but
there
is
no
outliers.
It
is
plausible
Assumptions:
hypotheses.
assumptions.
that
the
population
distribution
is
1) The given sample was a random sample of employees
approximately normal.
2) Since the sample size is not
at least 30, we must
determine if it is plausible
that the population
110
120
130
distribution is approximately
normal.
Surfing Internet Continued . . .
108 112 117 130 111 131 113 113 105 128
x = 116.80 s = 9.45 n = 10
H0: m = 120
Ha: m < 120
Where m is the true mean daily wasted
time for employees of this company
the
testwe
116potential
.80 Compute
120 error
What
could

 and
1.07
Test Statistic: thave
statistic
TypeP-value.
II
9made?
.45
P-value =.150
10
a = .05
Since p-value > a, we fail to reject H0. There is not
sufficient evidence to conclude that the mean daily
wasted time for employees of this company is less than
120 minutes.