Transcript 9 - University of Idaho

Chapter 9
Understanding
Sampling
Distributions:
Statistics as
Random Variables
Copyright ©2011 Brooks/Cole, Cengage Learning
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9.1 Parameters, Statistics,
and Statistical Inference
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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Statistical Inference
Statistical Inference: making conclusions about
population parameters on basis of sample statistics.
Two most common procedures:
Confidence intervals: an interval of values that the
researcher is fairly sure will cover the true,
unknown value of the population parameter.
Hypothesis tests: uses sample data to attempt to
reject a hypothesis about the population.
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9.2 From Curiosity to Questions
about Parameters
The Big Five Parameters
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Paired Differences
Paired data (or paired samples): when pairs of variables
are collected. Only interested in population (and sample)
of differences, and not in the original data.
• Each person measured twice. Two measurements of same
characteristic or trait are made under different conditions.
• Similar individuals are paired prior to an experiment. Each
member of a pair receives a different treatment.
Same response variable is measured for all individuals.
• Two different variables are measured for each individual.
Interested in amount of difference between two variables.
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Independent Samples
Two samples are called independent samples
when the measurements in one sample are not
related to the measurements in the other sample.
• Random samples taken separately from two
populations and same response variable is recorded.
• One random sample taken and a variable recorded,
but units are categorized to form two populations.
• Participants randomly assigned to one of two
treatment conditions, and same response variable
is recorded.
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Familiar Examples Translated into
Questions about Parameters
Situation 1. Estimating the proportion falling into a
category of a categorical variable.
Example research questions:
What proportion of American adults believe there is
extraterrestrial life? In what proportion of British
marriages is the wife taller than her husband?
Population parameter: p = proportion in the population
falling into that category.
Sample estimate: p̂ = proportion in the sample falling
into that category.
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Familiar Examples
Situation 2. Estimating the difference between two
populations with regard to the proportion falling
into a category of a qualitative variable.
Example research questions:
How much difference is there between the proportions that
would quit smoking if taking the antidepressant buproprion
(Zyban) versus if wearing a nicotine patch?
How much difference is there between men who snore
and men who don’t snore with regard to the proportion
who have heart disease?
Population parameter: p1 – p2 = difference between the
two population proportions.
Sample estimate: pˆ1  pˆ 2 = difference between the two
sample proportions.
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Familiar Examples
Situation 3. Estimating the mean of a quantitative variable.
Example research questions:
What is the mean time that college students watch TV
per day? What is the mean pulse rate of women?
Population parameter: m = population mean for the variable
Sample estimate: x = sample mean for the variable
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Familiar Examples
Situation 4. Estimating the mean of paired differences
for quantitative variables.
Example research questions:
What is the mean difference in weights for freshmen at the
beginning and end of the first semester?
What is the mean difference in age between husbands and
wives in Britain?
Population parameter: m d = population mean of differences
Sample estimate: d = mean of differences for paired sample
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Familiar Examples
Situation 5. Estimating the difference between two
populations with regard to the mean
of a quantitative variable.
Example research questions:
How much difference is there in average weight loss for
those who diet compared to those who exercise to lose
weight? How much difference is there between the mean
foot lengths of men and women?
Population parameter: m1 – m2 = difference between the
two population means.
Sample estimate: x1  x2 = difference between the two
sample means.
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9.3 SD Mod 0: An Overview
of Sampling Distribution
Statistics as Random Variables
Each new sample taken 
value of the 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 9.2 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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Standard Deviation and
Standard Error of a Statistic
• Standard deviation of a sampling distribution
measures the variation among possible values of the
sample statistic over all possible random samples.
We include the name of the statistic being studied,
e.g. the standard deviation of the mean.
• Standard error describes the estimated value of the
standard deviation of a statistic. We include the name
of the statistic, e.g. the standard error of the mean.
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9.4 SD Mod 1: Sampling Distribution
for One Sample Proportion
• 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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Sampling Distribution
for a Sample Proportion
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:
1: The Physical Situation
There is an actual population or repeatable situation.
2: Data Collection
A random sample is obtained or situation repeated many times.
3: The Size of the Sample or Number of Trials
The size of the sample or number of repetitions is relatively large,
np and np(1-p) must be at least 5 and preferable 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 9.4 Possible Sample Proportions
Favoring a Candidate
Suppose 40% all voters favor Candidate C. 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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9.5 SD Mod 2: Sampling Distribution
for Diff in Two Sample Proportions
For the populations:
p1 = population proportion for the first population.
p2 = population proportion for the second population.
Parameter: p1 – p2 = difference in popul proportions.
For the samples:
p̂1 = sample proportion for sample from first popul.
p̂2 = sample proportion for sample from second popul.
Statistic: pˆ1  pˆ 2 = difference in sample proportions.
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Conditions
Sampling distribution of difference in two independent sample
proportions is approximately normal when:
Condition 1: Sample proportions are available for two
independent samples, randomly selected from the two
populations of interest.
Condition 2: All of the quantities n1p1, n1(1 – p1), n2p2, and
n1(1 – p2) are at least 10. These quantities represent the
expected numbers of successes and failures in each sample.
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Sampling Distribution for the
Difference in Two Sample Proportions
Mean = p1 – p2
Standard deviation = s.d.( pˆ1  pˆ 2 )
=
p1 (1  p1 )

p2 (1  p2 )
n1
n2
When we don’t know the populations proportions, we use the
sample proportions, resulting in:
Standard error = s.e.( pˆ1  pˆ 2 ) =
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pˆ 1 (1  pˆ 1 )
n1

pˆ 2 (1  pˆ 2 )
n2
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Example 9.6 Men, Women, Death Penalty
Suppose 37% of women and 27% of men oppose death penalty,
p1 = .37 and p2 = .27, for a difference p1 – p2 = .37 – .27 = .10
For independent random samples of 1017 women and 885 men,
the sampling distribution of pˆ1  pˆ 2 is approx normal with
mean .10 and standard deviation:
.37(1  .37) .27(1  .27)

 .021
1017
885
Note: 2008 survey gave observed
difference of .36 – .285 = .075,
which is not unusual.
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9.6 SD Mod 3: Sampling Distribution
for One Sample Mean
• 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 9.8 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.d.( x ) =
s
n

5
 0.5 pounds.
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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9.7 SD Mod 4: Sampling Distribution
for Sample Mean of Paired Differences
Let md = mean for population of differences.
Let sd = standard deviation for population of differences.
Let d = mean for the sample of differences.
Let sd = standard deviation for sample of differences.
If numerous random samples of the same size n are taken, the
distribution of possible values of d is approximately a normal
curve distribution with
• Mean = md
sd
• Standard deviation = s.d.( d ) =
n
This approximate distribution is sampling distribution of d .
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The Normal Curve Approximation Rule
for Sample Mean of Paired Differences
Normal Approximation Rule can be applied in two situations:
Situation 1: The population of differences is
bell-shaped and a random sample of any size is measured.
Situation 2: The population of differences 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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Standard Error of the Mean Difference
Standard deviation of d :
sd
s.d.( d ) =
.
n
Standard error of d :
sd
s.e.( d ) =
.
n
The standard error is used to estimate the standard deviation.
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Example 9.9 No “Freshman 15”
How likely to see a mean weight gain of 4.2 pounds or larger
(for a random sample of 60 freshman) if there is no average
weight gain in the population of all such students? Suppose
the standard deviation for the population of weight gains is
known to be 7 pounds. The sampling distribution of d is …
• Approximately normal
• mean = md = 0 pounds
s
7
• s.d.( d ) = d 
 0.904  0.9 pounds
n
60
Empirical Rule: 95%
of the possible sample
means will be between
-1.8 and 1.8 pounds.
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9.8 SD Mod 5: Sampling Distribution
for Difference in Two Sample Means
Let m1 = population mean for first population.
Let m2 = population mean for second population.
Parameter: m1 – m2 = difference in population means.
Let x1 = sample mean for sample from first population.
Let x2 = sample mean for sample from second population.
Statistic: x1  x2 = difference in sample means.
Let s1 = population standard deviation for first population.
Let s2 = population standard deviation for second population.
Let s1 = sample std deviation for sample from first population.
Let s2 = sample std deviation for sample from second population.
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Conditions for Sampling Distribution
of x1  x2 to be Approx Normal
An important condition in this situation is that the two samples
must be independent. How?
• Take separate random samples from each of two populations
such as men and women.
• Take a random sample from a population and divide the
sample into two groups based on a categorical variable such
as smoker and nonsmoker.
• Randomly assign participants in a randomized experiment to
two treatment groups such as exercise or diet.
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Conditions for Sampling Distribution
of x1  x2 to be Approx Normal
In addition to independent samples,
one of the following two situations must hold:
Situation 1: The populations of measurements are both
bell-shaped and random samples of any size are measured.
Situation 2: Large random samples are measured from each
population. Arbitrary definition of large is both samples are
at least 30, but extreme outliers or extreme skewness in
either sample may require even larger samples.
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Standard Error of the Mean Difference
Standard deviation of x1  x2 :
s 12 s 22
s.d.( x1  x2 ) =
.

n1 n2
Standard error of x1  x2 :
s.e.( x1  x2 ) =
s12 s22

.
n1 n2
The standard error is used to estimate the standard deviation.
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Example 9.10 Who Are the Speed Demons?
What’s the fastest you’ve ever driven a car? ____ mph.
Mean for 87 males = 107 mph, mean for 102 females = 88 mph.
Is this 19 mph difference large enough to convince of real difference
in populations? Suppose standard deviations for each population of
speeds is known to be 15 mph. The sampling distribution of x1  x2 is:
• Approximately normal
• mean = m1 – m2 = 0 mph
2
2
2
2
• s.d.( x1  x2 ) = s 1  s 2  15  15  2.2
n1
n2
87
102
Note: difference of 19 mph almost
impossible in this scenario. Thus,
true difference in population means
almost surely much greater than 0.
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9.9 Preparing for Statistical
Inference: Standardized Statistics
If conditions are met, these standardized
statistics have, approximately, a standard
normal distribution N(0,1).
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Example 9.11 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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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.d .( x )
s/ n
n (x  m)
s
If 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 in Chapters 11 and13.
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Example 9.12 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

8.328
4.74
25
 0.34
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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