Download Sampling Distribution of a Sample Mean

Statistical inference uses
impersonal chance to draw
conclusions about a
population or process based
on data drawn from a random
sample or randomized
experiment.
 When data are produced by random
sampling or randomized experiment, a
statistic is a random variable that
obeys the laws of probability.
A sampling distribution shows
how a statistic would vary with
repeated random sampling of the
same size and from the same
population.

 A sampling distribution, therefore, is
a probability distribution of the results
of an infinitely large number of such
samples.
 A population distribution of a
random variable is the distribution of its
values for all members of the
population.
 Thus a population distribution is also
the probability distribution of the
random variable when we choose one
individual (i.e. observation or subject)
from the population at random.
 Recall that a sampling distribution is a
conceptual ideal: it helps us to understand the
logic of drawing random samples of size-n
from the same population in order to obtain
statistics by which we make inferences about
a parameter.
 Population distribution is likewise a
conceptual ideal: it tells us that sample
statistics are based on probabilities attached
to the population from which random samples
are drawn.
Counts & Sample Proportions
 Count: random variable X is a count of
the occurrences of some outcome—of some
‘success’ versus a corresponding ‘failure’—
in a fixed number of observations.
 A count is a discrete random variable that
describes categorical data (concerning
success vs. failure).
 Sample proportion: if the number of
observations is n, then the sample
proportion of observations is X/n.
 A sample proportion is also a discrete
random variable that describes categorical
data (concerning success vs. failure).
 Inferential statistics for counts &
proportions are premised on a
binomial setting.
The Binomial Setting
1. There are a fixed number n of observations.
2. The n observations are all independent.
3. Each observation falls into one of just two
categories, which for convenience we call ‘success’
or ‘failure.’
4. The probability of a success, p, is the same for each
observation.
5. Strictly speaking, the population must be at least 20
times greater than the sample for counts, 10 times
greater for proportions.
Counts
 The distribution of the count X of
successes in the binomial setting is called
the binomial distribution with
parameters n & p (i.e. number of
observations & probability of success on
any one observation).
X is B(n, p)
 Finding binomial probabilities: use
factorial, binomial table, or software.
 Binomial mean & standard deviation:
  np
  np( 1  p )
 Example: An experimental study
finds that, in a placebo group of 2000
men, 84 got heart attacks, but in a
treatment group of another 2000, just
56 got heart attacks.
 That is, 2000 independent observations of
men have found count X of heart attacks is
B(2000, 0.04), so that:
. mean=np=(2000)(.04)=80
. sd=sqrt(2000)(.04)(.96)=8.76
 Treatment group
. bitesti 2000 56 .04
N Observed k Expected k Assumed p Observed p
-------------------------------------------------------------------2000
56
80
0.04000
0.02800
Pr(k >= 56)
= 0.998333 (one-sided test)
Pr(k <= 56)
= 0.002497 (one-sided test)
Pr(k <= 56 or k >= 106) = 0.005090 (two-sided test)
 So, it’s quite unlikely (p=.002) that there
would be <=56 heart attacks by chance:
the treatment looks promising. What about
the placebo group?
 Placebo group
. bitesti 2000 84 .04
N
Observed k Expected k Assumed p Observed p
----------------------------------------------------------------2000
84
80
0.04000
0.04200
Pr(k >= 84)
= 0.339428 (one-sided test)
Pr(k <= 84)
= 0.700670 (one-sided test)
Pr(k <= 75 or k >= 84) = 0.647786 (two-sided test)
 By contrast, it’s quite likely (p=.70) that
the heart attack count in the placebo group
would occur by chance. By comparison,
then, the treatment looks promising.
Required Sample Size,
Unbiased Estimator
 Strictly speaking, he population must
be at least 20 times greater than the
sample for counts (10 times greater for
proportions).
 The formula for the binomial mean
signifies that np is an unbiased estimator
of the population mean.

Binomial test example (pages 370-71): Corinne is
a basketball player who makes 75% of her free
throws. In a key game, she shoots 12 free throws
but makes just 7 of them. What are the chances
that she would make 7 or fewer free throws in any
sample of 12?
. bitesti 12
N
Observed k
7
.75
Expected k
Assumed p
Observed p
------------------------------------------------------------------------------12
7
9
0.75000
0.58333
Pr(k >= 7)
= 0.945598 (one-sided test)
Pr(k <= 7)
= 0.157644 (one-sided test)
Pr(k <= 7 or k >= 12) = 0.189320 (two-sided test)
Note: ‘bitesti…, detail’ gives k==.103
 See Stata ‘help bitest’.
 We’ve just considered sample
counts.
 Next let’s considered sample
proportions.
Sample Proportion
 Count of successes in a
sample divided by sample
size-n.
Whereas a count has wholenumber values, a sample
proportion is always between
0 & 1.

 This is another example of
categorical data (success vs. failure).
 Mean & standard deviation of a
sample proportion:
 p
  p( 1  p ) / n
 The population must be at least 10 times
greater than the sample.
 Formula for a proportion’s mean: unbiased
estimator of population mean.
 Sample proportion example (pages 373-
74): A survey asked a nationwide sample
of 2500 adults if they agreed or disagreed
that “I like buying new clothes, but
shopping is often frustrating & timeconsuming.”
 Suppose that 60% of all adults would
agree with the question. What is the
probability that the sample proportion
who agree is at least 58%?
 Step 1: compute the mean &
standard deviation.
p  p  0.6
p  p( 1  p ) / n
 0.6 * 0.4/2500
 0.0098

Step 2: solve the problem.
p̂ - 0.6
0.58 - 0.6
P( p̂  0.58)  P(

)
0.00098 0.00098
 P(Z  - 2.04)  0.979
 How to do it in Stata:
. prtesti 2500
.58
.60
One-sample test of proportion
x: Number of obs =
Variable
x
Mean
Std. Err.
[95% Conf. Interval]
.58
.0098712
.5606529
2500
.5993471
P(Z>z) = 0.9794
 That is, there is a 98% probability that the percent
of respondents who agree is at least 58%: this is
quite consistent with the broader evidence.
 See Stata ‘help prtest’.
 We’ve just considered sample
proportions.
 Next let’s consider sample
means.
Sampling Distribution of a Sample
Mean
 This is an example of quantitative data.
 A sample mean is just an average of
observations (based on a variable’s
expected value).
 There are two reasons why sample
means are so commonly used:
(1) Averages are less variable
than individual observations.
(2) Averages are more normally
distributed than individual
observations.
Sampling distribution of a sample mean
 Sampling distribution of a sample mean: if
a population has a normal distribution, then
the sampling distribution of a sample mean
of x for n independent observations will also
have a normal distribution.
 General fact: any linear combination of
independent normal random variables is
normally distributed.
Standard deviation of a sample
mean: ‘Standard error’
 Divide the standard deviation of the
sample mean by the square root of sample
size-n. This is the standard error.
 Doing so anchors the standard deviation
to the sample’s size-n: the sampling
distribution of the sample mean across
relatively small samples has larger spread
& across relatively large samples has
smaller spread.
Sampling distribution of a sample
mean:

If population’s distribution =
N (  , )
then the sampling distribution of a
sample mean =
N (  ,
n)
Why does the the sampling
distribution of the sample mean in
relatively small samples have
larger spread & in relatively large
samples have smaller spread?

Because the standard deviation of
the mean is divided by the square
root of sample size-n.

 So, if you want the sampling
distribution of sample means (i.e. the
estimate of the population mean)to
be less variable, what’s the most
basic thing to do?
 Make the sample size-n larger.
 But there are major costs involved,
not only in obtaining a larger sample
size per se, but also in the amount of
increase needed.
 This is because the standard
deviation of the sample mean is
divided by the square root of n.
What does dividing the mean’s
standard deviation by the square root
of n imply?

 It implies that we’re estimating the
variability of the sampling
distribution of sample means from
the expected value of the
population, for an average sample
of size n.
 In short, we’re using a sample to
estimate the population’s standard
deviation of the sampling
distribution of sample means.
Here’s another principle—one
that’s even more important to
the sampling distribution of
sample means than the Law of
Large Numbers.

Central Limit Theorem
 As the size of a random sample
increases, the sampling
distribution of the sample mean
gets closer to a normal
distribution.
 This is true no matter what shape
the population distribution has.
 The following graphs illustrate
the Central Limit Theorem.
 The first sample sizes are very
small small; the sample sizes
become progressively larger.
Note: the Central Limit Theorem
applies to the sampling
distribution of not only sample
means but also sample sums.

 Other statistics (e.g., standard
deviations) have their own
sampling distributions.
 The Central Limit Theorem allows us to
use normal probability calculations to
answer questions about sample means
from many observations, even when the
population distribution is not normal.
 Thus, it justifies reliance of inferential statistics on
the normal distribution.
 N=30 (but perhaps up to 100 or more, depending on
the population’s standard deviation) is a common
benchmark threshold for the Central Limit Theorem—
although a far larger sample is usually necessary for
other statistical reasons. The larger the population’s
standard deviation, the larger N must be.
Why not estimate a parameter on
the basis of just one observation?

 First, because the sample mean is an
unbiased estimator of the population
mean & is less variable than a single
observation.
 Recall that averages are less variable
than individual observations.
 And recall that averages are more
normally distributed than individual
observations.
 Second, because a sample
size of just one observations
yields no measure of
variability.
 That is, we can’t estimate where
the one observed value falls in a
sampling distribution of values.
In summary, the sampling distribution
of sample means is:
 Normal if the population distribution is
normal (i.e. a sample mean is a linear
combination of independent normal random
variables).
 Approximately normal for large samples in
any case (according to the Central Limit
Theorem).
How can we confirm these
pronouncements?

 By drawing simulated samples from
the sampling distribution applet, or by
simulating samples of varying sizes via
a statistics software program (see
Moore/McCabe, chapter 3, for review).
Let’s briefly review several principles of
probability that are strategic to doing
inferential statistics:
(1) In random samples, the sample mean, the
binomial count, & the sample proportion
are unbiased estimators of the population mean;
& they can be made less variable by
substantially increasing sample size-n.
(2) The Law of Large Numbers (which is
based on the sample size-n, not on the
proportion of the population that is sampled).
(3) Averages are less variable than
individual observations & are more
normally distributed than individual
observations.
(4) The sampling distribution of
sample means is normal if the
population distribution is normal. Put
differently, the sample mean is a linear
combination of independent normal
random variables.
(5) The Central Limit Theorem:
the sampling distribution of sample
means is approximately normal for
large samples, even if the
underlying population distribution
is not normal.
These principles become
additionally important because—by
justifying the treatment of means
drawn from relatively large samples
as more or less normal
distributions—they underpin two
more fundamental elements of
inferential statistics: confidence
intervals & significance tests.

 What problems could bias your
predictions, even if your sample
is well designed?
Answer
Non-sampling problems such
as undercoverage, nonresponse, response bias, &
poorly worded questions.
