Transcript sample mean

Chapter 9
Means and
Proportions
as Random
Variables
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
9.1 Understanding Dissimilarity
Among Samples
Key: Need to understand what kind of dissimilarity
we should expect to see in various samples from the
same population.
• Suppose knew most samples were likely to provide an
answer that is within 10% of the population answer.
• Then would also know the population answer should
be within 10% of whatever our specific sample gave.
• => Have a good guess about the population value
based on just the sample value.
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Statistics and Parameters
A statistic is a numerical value computed from a
sample. Its value may differ for different samples.
e.g. sample mean x , sample standard deviation s,
and sample proportion p̂.
A parameter is a numerical value associated with
a population. Considered fixed and unchanging.
e.g. population mean m, population standard
deviation s, and population proportion p.
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Sampling Distributions
Each new sample taken =>
sample statistic will change.
The distribution of possible values of a statistic
for repeated samples of the same size from a
population is called the sampling distribution
of the statistic.
Many statistics of interest have sampling distributions
that are approximately normal distributions
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Example 9.1 Mean Hours of Sleep
for College Students
Survey of n = 190 college students.
“How many hours of sleep did you get last night?”
Sample mean = 7.1 hours.
If we repeatedly took
samples of 190 and each
time computed the sample
mean, the histogram of the
resulting sample mean
values would look like the
histogram at the right:
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9.2 Sampling Distributions
for Sample Proportions
• Suppose (unknown to us) 40% of a population
carry the gene for a disease, (p = 0.40).
• We will take a random sample of 25 people from
this population and count X = number with gene.
• Although we expect (on average) to find 10 people
(40%) with the gene, we know the number will
vary for different samples of n = 25.
• In this case, X is a binomial random variable
with n = 25 and p = 0.4.
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Many Possible Samples
Four possible random samples of 25 people:
Sample 1: X =12, proportion with gene =12/25 = 0.48 or 48%.
Sample 2: X = 9, proportion with gene = 9/25 = 0.36 or 36%.
Sample 3: X = 10, proportion with gene = 10/25 = 0.40 or 40%.
Sample 4: X = 7, proportion with gene = 7/25 = 0.28 or 28%.
Note:
• Each sample gave a different answer, which did not
always match the population value of 40%.
• Although we cannot determine whether one sample will
accurately reflect the population, statisticians have
determined what to expect for most possible samples.
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The Normal Curve Approximation
Rule for Sample Proportions
Let p = population proportion of interest
or binomial probability of success.
Let p̂ = sample proportion or proportion of successes.
If numerous random samples or repetitions of the same size n
are taken, the distribution of possible values of p̂ is
approximately a normal curve distribution with
• Mean = p
p (1  p )
• Standard deviation = s.d.( p̂ ) =
n
This approximate distribution is sampling distribution of p̂ .
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The Normal Curve Approximation
Rule for Sample Proportions
Normal Approximation Rule can be applied in two situations:
Situation 1: A random sample is taken from a population.
Situation 2: A binomial experiment is repeated numerous times.
In each situation, three conditions must be met:
Condition 1: The Physical Situation
There is an actual population or repeatable situation.
Condition 2: Data Collection
A random sample is obtained or situation repeated many times.
Condition 3: The Size of the Sample or Number of Trials
The size of the sample or number of repetitions is relatively large,
np and n(1-p) must be at least 5 and preferably at least 10.
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Examples for which Rule Applies
• Election Polls: to estimate proportion who favor a
candidate; units = all voters.
• Television Ratings: to estimate proportion of
households watching TV program; units = all households
with TV.
• Consumer Preferences: to estimate proportion of
consumers who prefer new recipe compared with old;
units = all consumers.
• Testing ESP: to estimate probability a person can
successfully guess which of 5 symbols on a hidden card;
repeatable situation = a guess.
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Example 9.2 Possible Sample Proportions
Favoring a Candidate
Suppose 40% all voters favor Candidate X. Pollsters take a
sample of n = 2400 voters. Rule states the sample proportion
who favor X will have approximately a normal distribution with
mean = p = 0.4 and s.d.( p̂ ) =
p(1  p)
n

0.4(1  0.4)
2400
 0.01
Histogram at right
shows sample
proportions resulting
from simulating this
situation 400 times.
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Estimating the Population Proportion
from a Single Sample Proportion
In practice, we don’t know the true population proportion p,
so we cannot compute the standard deviation of p̂ ,
s.d.( p̂ ) =
p (1  p )
n
.
In practice, we only take one random sample, so we only have
one sample proportion p̂ . Replacing p with p̂ in the standard
deviation expression gives us an estimate that is called the
standard error of p̂ .
s.e.( p̂ ) =
pˆ (1  pˆ )
n
.
If p̂ = 0.39 and n = 2400, then the standard error is 0.01. So
the true proportion who support the candidate is almost surely
between 0.39 – 3(0.01) = 0.36 and 0.39 + 3(0.01) = 0.42.
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9.3 What to Expect of
Sample Means
• Suppose we want to estimate the mean weight loss
for all who attend clinic for 10 weeks. Suppose
(unknown to us) the distribution of weight loss is
approximately N(8 pounds, 5 pounds).
• We will take a random sample of 25 people from
this population and record for each X = weight loss.
• We know the value of the sample mean will vary
for different samples of n = 25.
• What do we expect those means to be?
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Many Possible Samples
Four possible random samples of 25 people:
Sample 1: Mean = 8.32 pounds, standard deviation = 4.74 pounds.
Sample 2: Mean = 6.76 pounds, standard deviation = 4.73 pounds.
Sample 3: Mean = 8.48 pounds, standard deviation = 5.27 pounds.
Sample 4: Mean = 7.16 pounds, standard deviation = 5.93 pounds.
Note:
• Each sample gave a different answer, which did not always
match the population mean of 8 pounds.
• Although we cannot determine whether one sample mean will
accurately reflect the population mean, statisticians have
determined what to expect for most possible sample means.
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The Normal Curve Approximation
Rule for Sample Means
Let m = mean for population of interest.
Let s = standard deviation for population of interest.
Let x = sample mean.
If numerous random samples of the same size n are taken, the
distribution of possible values of x is approximately a normal
curve distribution with
• Mean = m
s
• Standard deviation = s.d.( x ) =
n
This approximate distribution is sampling distribution of x .
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The Normal Curve Approximation
Rule for Sample Means
Normal Approximation Rule can be applied in two situations:
Situation 1: The population of measurements of interest is
bell-shaped and a random sample of any size is measured.
Situation 2: The population of measurements of interest is not
bell-shaped but a large random sample is measured.
Note: Difficult to get a Random Sample? Researchers usually
willing to use Rule as long as they have a representative sample
with no obvious sources of confounding or bias.
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Examples for which Rule Applies
• Average Weight Loss: to estimate average weight
loss; weight assumed bell-shaped; population = all
current and potential clients.
• Average Age At Death: to estimate average age at
which left-handed adults (over 50) die; ages at death not
bell-shaped so need n  30; population = all left-handed
people who live to be at least 50.
• Average Student Income: to estimate mean monthly
income of students at university who work; incomes not
bell-shaped and outliers likely, so need large random
sample of students; population = all students at university
who work.
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Example 9.4 Hypothetical Mean
Weight Loss
Suppose the distribution of weight loss is approximately
N(8 pounds, 5 pounds) and we will take a random sample
of n = 25 clients. Rule states the sample mean weight loss
will have a normal distribution with
s
5

 1 pound
mean = m = 8 pounds and s.d.( x ) =
n
25
Histogram at right shows
sample means resulting
from simulating this
situation 400 times.
Empirical Rule:
It is almost certain that
the sample mean will be
between 5 and 11 pounds.
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Standard Error of the Mean
In practice, the population standard deviation s is rarely
known, so we cannot compute the standard deviation of x ,
s
s.d.( x ) =
.
n
In practice, we only take one random sample, so we only have
the sample mean x and the sample standard deviation s.
Replacing s with s in the standard deviation expression gives
us an estimate that is called the standard error of x .
s
s.e.( x ) =
.
n
For a sample of n = 25 weight losses,
the standard deviation is s = 4.74 pounds.
So the standard error of the mean is 0.948 pounds.
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Increasing the Size of the Sample
Suppose we take n = 100 people instead of just 25.
The standard deviation of the mean would be
s
5

 0.5 pounds.
s.d.( x ) =
n
100
• For samples of n = 25,
sample means are likely to
range between 8 ± 3 pounds
=> 5 to 11 pounds.
• For samples of n = 100,
sample means are likely to
range only between 8 ± 1.5
pounds => 6.5 to 9.5 pounds.
Larger samples tend to result in more accurate estimates
of population values than smaller samples.
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Sampling for a Long, Long Time:
The Law of Large Numbers
LLN: the sample mean x will eventually get
“close” to the population mean m no matter how
small a difference you use to define “close.”
LLN = peace of mind to casinos, insurance companies.
• Eventually, after enough gamblers or customers,
the mean net profit will be close to the theoretical mean.
• Price to pay = must have enough $ on hand to pay the
occasional winner or claimant.
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9.4 What to Expect in Other
Situations: CLT
The Central Limit Theorem states that if n is
sufficiently large, the sample means of random
samples from a population with mean m and
finite standard deviation s are approximately
normally distributed with mean m and
standard deviation s n .
Technical Note:
The mean and standard deviation given in the CLT hold
for any sample size; it is only the “approximately normal”
shape that requires n to be sufficiently large.
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Example 9.5 California Decco Winnings
California Decco lottery game: mean amount lost per
ticket over millions of tickets sold is m = $0.35; standard
deviation s = $29.67 => large variability in possible
amounts won/lost, from net win of $4999 to net loss of $1.
Suppose store sells 100,000 tickets in a year. CLT =>
distribution of possible sample mean loss per ticket
is approximately normal with …
mean (loss) = m = $0.35and s.d.( x) =
s
$29.67

 $0.09
n
100000
Empirical Rule: The mean loss is almost surely between
$0.08 and $0.62 => total loss for the 100,000 tickets is
likely between $8,000 to $62,000!
There are better ways to invest $100,000.
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9.5 Sampling Distribution
for Any Statistic
Every statistic has a sampling distribution,
but the appropriate distribution may not always
be normal, or even approximately bell-shaped.
Construct an approximate sampling distribution
for a statistic by actually taking repeated samples
of the same size from a population and constructing
a relative frequency histogram for the values of the
statistic over the many samples.
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Example 9.6 Winning the Lottery
by Betting on Birthdays
Pennsylvania Cash 5 lottery game:
Select 5 numbers from integers 1 to 39.
Grand prize won if match all 5 numbers.
One strategy = 5 numbers bet correspond to birth
days of month for 5 family members => no chance
to win if highest number drawn is 32 to 39.
What is the probability of this?
Statistic of interest = H = highest of five integers
randomly drawn without replacement from 1 to 39.
e.g. if numbers selected are 3, 12, 22, 36, 37 then H = 37.
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Example 9.6 Winning the Lottery
by Betting on Birthdays (cont)
Summarized below: value of H for 1560 games.
Highest number over 31 occurred in 72% of the games.
Most common value of H = 39 in 13.5% of games.
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9.6 Standardized Statistics
If conditions are met, these standardized
statistics have, approximately, a standard
normal distribution N(0,1).
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Example 9.7 Unpopular TV Shows
Networks cancel shows with low ratings. Ratings based
on random sample of households, using the sample
proportion p̂ watching show as estimate of population
proportion p. If p < 0.20, show will be cancelled.
Suppose in a random sample of 1600 households, 288 are
watching (for proportion of 288/1600 = 0.18). Is it likely
to see p̂ = 0.18 even if p were 0.20 (or higher)?
z
pˆ  p
p( 1 p)
n

0.18 0.20
0.20( 1 0.20 )
1600
 2.00
The sample proportion of 0.18 is about
2 standard deviations below the mean of 0.20.
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9.7 Student’s t-Distribution:
Replacing s with s
Dilemma: we generally don’t know s. Using s we have:
xm
xm
t


s.e.( x ) s / n
n (x  m)
s
If the sample size n is small,
this standardized statistic will
not have a N(0,1) distribution
but rather a t-distribution with
n – 1 degrees of freedom (df).
More on t-distributions and tables of probability areas in Chapters 12-13.
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Example 9.8 Standardized Mean Weights
Claim: mean weight loss is m = 8 pounds.
Sample of n =25 people gave a sample mean
weight loss of x = 8.32 pounds and a sample
standard deviation of s = 4.74 pounds.
Is the sample mean of 8.32 pounds reasonable
to expect if m = 8 pounds?
t
x m
s
n
.328
 48.74
 0.34
25
The sample mean of 8.32 is only about one-third
of a standard error above 8, which is consistent
with a population mean weight loss of 8 pounds.
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9.8 Statistical Inference
• Confidence Intervals: uses sample data
to provide an interval of values that the
researcher is confident covers the true value
for the population.
• Hypothesis Testing or Significance Testing:
uses sample data to attempt to reject the
hypothesis that nothing interesting is
happening, i.e. to reject the notion that
chance alone can explain the sample results.
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Case Study 9.1 Do Americans Really Vote
When They Say They Do?
Election of 1994:
• Time Magazine Poll: n = 800 adults (two days after election),
56% reported that they had voted.
• Info from Committee for the Study of the American Electorate:
only 39% of American adults had voted.
If p = 0.39 then sample proportions for samples of size
n = 800 should vary approximately normally with …
mean = p = 0.39 and s.d.( p̂ ) =
p(1  p)
n
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
0.39 (1  0.39 )
800
 0.017
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Case Study 9.1 Do Americans Really Vote
When They Say They Do?
If respondents were telling the truth, the sample percent
should be no higher than 39% + 3(1.7%) = 44.1%,
nowhere near the reported percentage of 56%.
If 39% of the population voted, the standardized score
for the reported value of 56% is …
0.56  0.39
z
 10.0
0.017
It is virtually impossible to obtain a standardized score of 10.
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