Download chapter 8—sampling distributions

Chapter 8--Sampling Distributions.Doc
STATISTICS 301—APPLIED STATISTICS, Statistics for Engineers and Scientists, Walpole, Myers, Myers, and Ye, Prentice Hall
Goal:
We next investigate some sample statistics (eg X --the Sample Average and p̂ --the
Sample Proportion or X1 - X2 and pˆ1 - pˆ2 ) as Random Variables and further investigate
the distributions of these random variables. Lastly we use this information to
“ESTIMATE” a population parameter.
Consider the first few paragraphs from an article in the New York Times.
Trends: Halloween, for Skinnier Skeletons
By JOHN O'NEIL
Published: October 21, 2003
Trick-or-treaters can be satisfied with something other than candy, say researchers who conducted an
experiment last Halloween that arose from concern about the nation's obesity epidemic.
In the study, test households in five Connecticut neighborhoods offered children two bowls: one with
lollipops or fruit candy and one containing small, inexpensive Halloween toys, like plastic bugs that glow
in the dark.
Of the 283 children from 3 to 14 whose reactions were tallied, 148 chose candy and 135 toys; the
difference was statistically insignificant, the researchers reported in The Journal of Nutrition
Education and Behavior in the July-August issue.
Consider the following questions:
1. What is the population?
2. What is the parameter?
3. What is the sample?
4. What is the statistic?
5. Can we say anything about the theoretical “distribution” of the statistic?
So our goal is two-fold: Determine the distribution of statistics and then use this
information to provide information about a population parameter.
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Chapter 9—Estimation.Doc
STATISTICS 301—APPLIED STATISTICS, Statistics for Engineers and Scientists, Walpole, Myers, Myers, and Ye, Prentice Hall
Goal:
In this section we will investigate the concept of “Estimation” in which our goal is to use
sample information (assumed to be a random sample from the population of interest) to
arrive at a reasonable guess of a population parameter. Estimation is done in two ways—
point estimation (or single value) and interval estimation (an interval or range of likely
values). We will tackle POINT ESTIMATION at this point and leave INTERVAL
ESTIMATION to the next chapter.
Background
We assume that we are interested in one (or more) population parameter(s), such as the
mean (), the variance (2), the standard deviation (), the median ( μ ), or proportion (p).
In every case we assume the parameter is unknown and the population is so large that it
precludes our measuring every element of the population to obtain the parameter. So we
take a random sample from the population, of size n, and calculate the appropriate sample
statistic.
POINT ESTIMATION
Below we provide a table with each of the population parameters we have discussed along
with the corresponding point estimate or sample statistic of that parameter.
Population
Parameter
μ
Estimator
Estimate
X
x
σ2
S2
σ
S
# of obs with characteristic
ˆ
p=
total
p
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s2 =
1
 (xi -x)2
n-1
s = s2
# of obs with characteristic
ˆ
p=
total
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EXAMPLES
Miami Undergraduates and Graduate School Plans
Miami’s Graduate Dean contacts me about the percentage of students who plan on
pursuing a graduate degree. Using the information that I have collected on STA 301
students, I tell the Dean of Miami’s Graduate School that of the 123 students I
have surveyed 74 indicated they plan to attend graduate school. Hence we would
estimate the proportion of MU undergrads that plan to attend graduate school as
74
pˆ =
= 60.2% .
123
Underweight Milky Way Candy Bars
Recall that we found 20 of 85 candy bars weighed less than Milky Way’s claim that
none should weigh less than 58.1 gms. Hence we would estimate the proportion of
19
= 22.4% , or nearly one quarter!
underweight Milky Ways to be pˆ =
85
62.2
61.6
61.3
58.6
60.5
59.1
57.7
59.6
64.5
57.2
61.9
61.3
57.2
58.1
60.4
57.4
58.4
60.3
60.1
58.4
56.8
59.7
56.0
62.1
60.7
59.4
58.5
58.1
62.4
60.2
59.6
60.0
60.0
60.1
57.0
57.1
60.5
58.3
58.2
60.5
57.5
58.5
61.5
61.3
57.1
57.7
63.4
57.9
58.3
64.6
59.2
58.7
53.2
59.5
55.3
59.4
61.6
59.7
58.6
58.7
59.1
59.7
59.8
59.5
60.7
61.5
61.9
60.6
60.6
59.7
59.7
59.1
59.9
58.8
57.5
57.7
59.7
54.1
57.4
58.4
60.0
60.9
60.8
60.5 56.1
“High School Mathematics Course-Taking by Gender and Ethnicity,” American
Educational Research Journal, Fall 1998, Volume 35, Number 3, pages 497-514.
In this example one of the goals was to determine the mean number of Carnegie
units earned by high-school graduates in seven categories of mathematics courses.
Among all students in the study ( the sample size was 21,784 ), the average number
of math CU’s earned was x = 3.11. Thus we would estimate the mean number of math
CU’s earned by all high-school graduates is 3.11.
While not part of the question or purpose of the study, the authors also report the
sample standard deviation of the number of math CU’s earned, s = 0.93. This is an
estimate the population standard deviation (σ) of the number of math CU’s earned
by HS students.
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METHODS OF ESTIMATION
GOAL: Given a random sample (ie the X1, X2, …, Xn) from a population with mean = popln and
variance = 2popln,
HOW DO WE MANIPULATE THE X’s SO THAT WE ARRIVE AT A “GOOD” GUESS OF THE
POPULATION PARAMETER OF INTEREST? That is:
Sample
X1
X2
.
.
.
Xn
For example, if our goal is to estimate the population mean, popln, what function of the X’s
do we use to arrive at our ESTIMATOR of popln?
Or what function of the X’s do we use to arrive at our ESTIMATOR of 2popln?
DIFFERENT METHODS OF ESTIMATION
There are many different methods of estimation and all yield the function of the X’s that is
used to estimate a population parameter. The three most commonly used methods are:
1. Method of Least Squares Estimation (LSE): you’ll see this used in the regression
and design of experiments courses (STA 463 and 466, respectively).
2. Method of Maximum Likelihood Estimation (MLE): this method will be seen in STA
401 and other more advanced statistical theory classes.
3. Method of Moments: to be seen in STA 401 and infrequently otherwise.
We mention these methods in passing; we will assume for the remainder of this section that
the function and statistic is just given to us.
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PROPERTIES OF ESTIMATORS
We started this section with the goal of obtaining “good” guesses or estimates of population
parameters. There are two main criteria that determine “good” estimators and estimates
and we illustrate these criteria in the following pictures.
In the following “target” picture, think of the bulls-eye as being the “true” or actual
population parameter value and the “hits” on the target as being observations from the
distribution of a statistic used to estimate the parameter. And in the second picture, an
approximation to the distribution of various statistics all estimators of a parameter is given.
In both pictures, the idea of Unbiased and Best Estimators is illustrated and we give the
following definitions:
1. Unbiased Estimators are estimators whose expected value equals the parameter of
interest, so that E [ Estimator ] = parameter.
2. Best Estimators are those that have the smallest variance among a set of
estimators.
Summary of Estimators
1. Since E [ X ] = popln, X is an unbiased estimator of the popln mean, popln.
2. Since E[ p̂ ] = p, p̂ is an unbiased estimator of the popln proportion, p.
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In general, if we let ̂ be a sample statistic, we will show that:
1. ̂ is a Random Variable and hence
2. ̂ has a distribution consisting of i) Center = Mean of the RV =  ̂ = E[ ̂ ],
ii) Spread = Variance of RV = 2̂ = E[ ( ̂ -  ̂ )2 ], and
iii) Shape = Family or Kind of RV .
Recall that Families or Kinds of RV will fall into
one of the distributions that we have
encountered already, namely:
All Random Variables
Discrete
Binomial
Continuous
Normal
Finally recall some properties of functions of
random variables, namely, sums of linear
combinations of random variables.
If X and Y are two INDEPENDENT random
variables with means and variances, X and Y
and 2X and 2Y , respectively, and let a, b, and
c be constants, then we know that:
Chi-Square
T
F
1. The Mean of W (= aX + bY + c) = W = E [ W ] = aX + bY + c ALWAYS!!!
2. The Variance W = V(aX + bY + c) = 2W = a22X + b22Y ONLY IF X AND Y INDEP!!!
3. The Mean of W (= X ± Y) = W = E [ W ] = E [ X ± Y ] = X ± Y ALWAYS!!!
4. The Variance W (= X ± Y) = 2W = V[ W ] = V[ X ± Y ] = 2X + 2Y ONLY IF X & Y INDEP!
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Now consider the following experiment. Take a random sample of size n from a population of
measurements.
Population
 Population of
measurements
 Distn of the popln is:
popln
2popln
Shape
Sample
Random
Sample
X1 = first sample value
X2 = first sample value
.
.
.
Xn = first sample value
Some facts!
1. Every sample statistic is a function of X1, X2, …, Xn. Later we’ll briefly discuss this.
2. The values X1, X2, …, Xn are random variables. Why?
3. The random variables X1, X2, …, Xn are independent random variables. Why?
4. Since the X1, X2, …, Xn are random variables, they have a distribution (Center, Spread,
and Shape). It’s easy to “see” that the distribution of X1 is identical to the population
distribution and hence: Center of X1: popln
Spread of X1: 2popln
Shape of X1: Popln Shape
5. Less easy to see, but still true is that each of the X’s has the same distribution as the
population. This results in the following crucial theorem.
Thm: If X1, X2, …, Xn are a random sample from a population with mean = popln,
variance = 2popln, and some shape, then the Xi’s are said to independent and identically
distributed. That is, the Xi’s are i.i.d. (popln, 2popln) with same shape as population.
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NOTATION AND TERMINOLOGY
We will distinguish between ESTIMATORS and ESTIMATES. Recall that the X’s, the
random sample, are random variables. They are random quantities! The values of these X’s,
x’s, are the values that the random variables take on and are constants.
Whereas the X’s are random variables and hence have distributions, the x’s are constants
and DO NOT HAVE A DISTRIBUTION!
Hence we have the following definitions:
Defn:
An ESTIMATOR is a function of the random sample values, X’s, and is a random
variable with a distribution.
Defn:
An ESTIMATE is a value that the estimator can take on and hence is a single
number.
COMMON ESTIMATORS (aka SAMPLE STATISTICS)
Below are the most commonly used ESTIMATORS and ESTIMATES of population
parameters of most interest.
Population
Parameter
μ
σ2
σ
p
Estimator
(aka Sample Statistic)
1
1
X = (X1 + X2 +...+ Xn )=  Xi
n
n
1
S2 =
(Xi -X)2

n-1
Estimate
1
1
x = (x1 + x2 +...+ xn )=  xi
n
n
1
s2 =
(xi -x)2

n-1
S=
S2
s = s2
# of obs with characteristic 1
# of obs with characteristic 1
ˆ
ˆ
p=
=  Xi p=
=  xi
total
n
total
n
Median
X = middle X
x = middle x
Range
R = Max X – Min X
r = Max x – Min x
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Sampling Distributions of Sample Statistics
Population
Parameter of
Interest
popln
Medianpopln
2popln
popln
p
Sample
Random
Sample
Method
Corresponding
Statistic
X
M
S2
S
p̂
Statistical Inference
What does the sample statistic (eg X ) tells us
about the population parameter (eg )?
Foundation (Theory behind) Statistical Inference
Probability, Random Variables, Families of RV’s, and
SAMPLING DISTRIBUTIONS
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Sampling Distributions of Sample Statistics
Defn: The Sampling Distribution of a Statistic is the distribution of statistic (the random
variable) if every possible random sample of the same size were obtained from the
population.
Recall that Distributions have CENTERS, SPREADS, and SHAPES, hence:
SAMPLING DISTRIBUTION will have
CENTER
SPREAD
SHAPE
Typically only two things determine the distribution of a sample statistic:
1. sample size and
2. distribution of the population from which the sample came from.
The sampling distribution for any of the sample statistics that we have seen can be
determined. Thus we can find the sampling distribution of:
CENTER =
 The sample average, X
SPREAD =
SHAPE =
2
 The sample variance, S .

The sample standard deviation, S.

The sample proportion, p̂ .
In each of these cases, the sampling distribution of the sample statistic depends upon
1. The sample size
and
2. The distribution of the population.
In this class we will only investigate the sampling distribution of the sample average and
sample proportion in depth.
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NOTES AND COMMENTS ABOUT SAMPLING DISTRIBUTIONS
1. Since a Sampling Distribution is a distribution, it has a Center, Spread, and Shape.
While we had many ways of describing Center and Spread in general, we will use the
MEAN to define the Center and VARIANCE or STANDARD DEVIATION to define the
Spread for Sampling Distributions.
2. The Standard Deviation (SD for short) of the Sampling Distribution of a statistic is
also known as the Standard Error (SE for short) of the Statistic.
Interpretation: The Standard Error of a Statistic is a measure of how variable the
statistic is.
Example
Suppose a research paper reports that the sample
of 56 parents of 7th graders in a particular school
system was taken. It also reports that the
incomes of parents had a sample average income of
$45,000 with a standard deviation of $12,000. It
also reports that the Standard Error of the
Average was $1,600. Pictorially:
Here’s what this means.
1. The $12,000 standard deviation is a measure of how variable the collective 56
incomes were. This is a pretty large spread.
2. The $1,600 standard error of the average is interpreted as if we were to take
repeated random samples of 56 parents, measure their incomes, calculate the
sample average in each case, the standard deviation for all of the average
incomes that we calculated would be approximately $1,600.
The Standard Error of the (sample) Average is also more commonly referred to as the
Standard Error of the Mean. Again, poor choice of words, but you’ll see it routinely
used. Below is an example of Minitab descriptive statistics output that uses these
terms.
Descriptive Statistics: Score
Variable
Score
N
121
Mean
90.339
Median
92.000
TrMean
91.138
Variable
Score
Minimum
36.000
Maximum
99.000
Q1
88.000
Q3
94.000
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StDev
6.947
SE Mean
0.632
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HOW ARE SAMPLING DISTRIBUTIONS DETERMINED?
Sampling distributions are a statement about the distribution of every possible value of a
statistic. That is, suppose we wanted to determine the sampling distribution of the sample
average. It might seem that we would have to take every sample of size n from the
population, observe, then summarize these values. That is:
RS1
S1
RS2
RS3
S2
S3
RS
S
The most surprising of all is the fact that to determine the sampling distribution of any
statistic requires ONLY ONE RANDOM SAMPLE, and contrary to first thoughts, NOT
MANY RANDOM SAMPLES!!!!
In fact, using statistical theory, we can determine what the distribution of any statistic
will be base on two things:
1. The size of the random sample only, and
2. Facts and characteristics OR assumptions about the population.
Lastly, we note that we will say or assume NOTHING ABOUT THE POPLN SIZE!!!
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SAMPLING DISTRIBUTION OF THE SAMPLE AVERAGE— X
Background/Assumptions—We will assume the following:
1. We have a population of measurements with a mean, popln, and standard deviation, popln.
2. We have a random sample of size n ( X1, X2, …, Xn ) from this population.
Facts—Here’s what we know:
1
1
1. Defn: The Sample Average is X = (X1 + X2 +...+ Xn )=  Xi .
n
n
2.  Xi = E [ Xi ] = popln
3. 2Xi = Var [ Xi ] = 2popln
4. The Xi’s are independent random variables.
Facts concerning Expectation Function.
5. E [ aX + bY + c ] = aX + bY + c
6. Variance of [ aX + bY + c ] = a22X + b22Y if X and Y are independent.
Using these facts:
Thm: If X1, X2, …, Xn are a RS from a population with mean, popln, and variance, 2popln, then
1.  X = E [ X ] = popln and
2.  = Var [ X ] =
2
X
Pf:
2
σpopln
n
.
1. E [ X ] =
2. Var [ X ] =
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Example #1
Suppose we take a random sample of 10 from a population that has a mean of 4 and a
variance of 5. We now know that:
n=
popln =
2popln =
X has a mean of 4 and a variance of
5
= ½ = 0.5.
10
Implications
1. The rather small variance of X , only 0.5, ( the standard deviation of X or standard
error of X is likewise small at 0.71) implies that the random variable X has a very
small spread and hence its values are very tightly grouped together.
The important implication is that no matter what value we obtain in our sample, it’s
not very far away from the center or mean of X or popln. As a result we could
conclude that our actual sample value of X will probably be very close to the
population mean, popln value of 4.
2. The other important point to take from the theorem is that it says nothing about
the shape of the population. The result is true FOR ANY POPULATION OF ANY
SHAPE!
3. What is the SE of X ?
Could you draw a picture of the distribution of X for this example? What’s missing?
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Shape
One missing detail about the distribution of X that is not addressed in the theorem is the
shape of the distribution. Recall that the shape of the distribution is equivalent to stating
the “kind” of random variable or distribution. To this end we have the following three
results that we combine into one theorem on the Shape of X .
Thm: If X1, X2, …, Xn are a RS from a population with mean = popln, and variance = 2popln, then
1. If the population is Normally distributed then X is also Normally distributed.
2. If the population is Normal and 2popln, is unknown then
X  popln
S
= T(n-1) .
n
3. If n is large then X is approximately Normally distributed!
Implications
1. The first result is important since we note that if we sample from a Normal
population, then the sample average is exactly normally distributed, regardless of
the sample size.
As a follow-up, we also know that: If the population is approx Normal, then X ≈
Normal regardless of sample size.
2. In the second result we see our first encounter with another of the continuous
distributions, the T distribution.
3. The last result of the Theorem is the most important theorem in statistics and is
known as the CENTRAL LIMIT THEOREM. We restate here to emphasize it’s
importance!
CENTRAL LIMIT THEOREM: Let X be the sample average based on a large
random sample, then X ≈ Normal!
What makes this result so important is that it says nothing about the shape of the
population—NOTHING! The only requirement is that we have a “large” sample size.
But then we need to ask what’s large? It turns out that n > 30 will yield acceptable
results in most instances. However, if the population is know to be terribly skewed
(in either direction) then a larger sample size will be required; how large depends on
how skewed the population is.
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Example #2
Suppose X is the sample average lifetime ( in hours ) based on a random sample of n = 16
light bulbs from a population of light bulbs that is approximately Normally distributed with
a mean of 800 hours and a standard deviation of 40 hours.
What would a picture of the population of lifetimes look like?
What is the probability that a random chosen light bulb from the population lasts between
760 and 840 hours?
What is the distribution of X for this example? What does the picture look like? What
is the SE of X ?
What is the probability that our sample average is between 760 and 840 hours?
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Example #3
Suppose we take a random sample of 40 college students and obtain their cell phone usage
records for the last month and determine the # of minutes each student used that month.
What are the chances that our sample average is within ¼ a population standard deviation
of the population mean time spent on a cell phone by college students?
What do we know about X ?
What does it mean to say “our sample average is within ¼ a population standard deviation
of the population mean?”
What are the chances that our sample average is within ¼ a population standard deviation
of the population mean time spent on a cell phone by college students?
As a final question, if an event has an 88% chance of happening, would you guess that it
WOULD or WOULD NOT happen? Why?
Now suppose we actually take our sample of 40 college students and find that
x  749.5min s . What could you conclude about the mean time that college students spend
on their cell phones?
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NOTES AND COMMENTS
1. The sample size need only be greater than 30 for this result to hold. IT DOES NOT
MATTER HOW LARGE OR SMALL THE POPULATION IS.
2. For the Sample Average to be approximately Normal, we only need the sample size to
be bigger than 30. We do not need to make any assumption about the shape of the
population!
Example #4
Suppose we take a random sample of 150 students and measure the IQ of each student.
The population from which we selected the 150 students have a mean IQ of 115 and a
standard deviation of 18 or popln = 115 and popln = 18. Then, since the sample size of 150 is
> 30, we know the distribution of the average IQ in our sample is Normally Distributed
σpopln
18
18
with a mean of 115 and a standard deviation of
=
=
=1.4697.
n
150 12.1474
What does the picture of the distribution of X look like? What is the SE of X ?
If you were to chose one value, at random, from this population of X values, where would
it be likely to lie? Why?
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SAMPLING DISTRIBUTION OF THE SAMPLE PROPORTION— p̂
Background/Assumptions We will assume the following:
1. We have a population and in the population the proportion of items with a given
characteristic of interest is p.
2. We have a random sample of size n ( X1, X2, …, Xn ) from this population, where each of
the Xi is 1 if the ith observation has the characteristic and 0 if not.
Facts Here’s what we know:
1. Recall that the Binomial Experiment was an experiment with n trials, success or failure
at each trial, a constant probability of success at each trial, and independent outcomes
at each trial.
2. Recall that a Binomial random variable was the number of successes in a Binomial
Experiment.
3. Recall that if X was Binomial with parameter n and p, then X = np and 2X = np(1-p).
4. Note that if we let X be the sum of the Xi’s, then we have a Binomial Experiment and
we have a Binomial random variable. Hence
X = Sum of the Xi’s = Bin (n, p).
5. Finally note that since the Sample Proportion becomes
1
1
# with the characteristic
pˆ = (X1 +X2 +...+Xn )=  Xi =
n
n
n
pˆ =
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X
.
n
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Thm: If X1, X2, …, Xn are a random sample from a population with proportion p, then
The Mean of p̂ = E[ p̂ ] = p & The Variance of p̂ = Var[ p̂ ] =
Pf:
p(1-p)
n
1. The Mean of p̂ = E[ p̂ ] =
2. The Variance of p̂ = Var[ p̂ ] =
Unfortunately this theorem does not say anything about the “shape” of the distribution of p̂ .
However, the following theorem does!
Thm: If X1, X2, …, Xn are a random sample from a population with proportion p, then
p̂ ≈ Normal ( p,
if np > 5 AND n(1-p) > 5
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p(1-p)
)
n
or # successes in sample > 5 AND # failures in sample > 5
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Implication
If the sample size is “large,” then the Sampling Distribution of the Sample Proportion
p̂ will have an approximately Normal Distribution with a mean equal to the proportion in
the population, p, and standard deviation equal to the population standard deviation divided
by the square root of the sample size, or
p(1-p)
.
n
What does the distribution of p̂ look like?
Example #1
A random sample of 1,108 adults yielded 78% that said they believe there is a heaven.
Letting pH be the proportion of adults who believe in heaven, p̂H is the 78%.
What can we say about the distribution of p̂H in this example? What do we need to check?
What is the probability that p̂H is within 3% of pH?
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Example #2
Consider the survey from class and whether a student drinks or not. Let pND be the
proportion of MU students who do not drink alcohol.
The results from our class of n = __ who
filled out the survey indicated that only
__ does not drink. Letting p̂ND 1 be the
sample proportion of students who do not
drink out of __, this distribution of p̂ND 1
is:
Can you draw a picture of it?
Now suppose we use all the other STA
301 students that I have survey
information on. I have information on 118
students (the six from this class are NOT
part of the 124). Of this 124, 19
indicated they do not drink. What is the
distribution of this sample proportion,
p̂ND 2 ?
Can you draw a picture of it?
How do these two distributions compare? What are the differences between them?
Why do they differ?
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Interesting fact about the SE of p̂
Recall that the Standard Error of a statistic is nothing more than the standard deviation
of the statistic. Also recall that the standard deviation of the sample proportion is given
by
p(1-p)
.
n
Consider the numerator of this expression and plot this function versus p.
p(1-p)
p
Hence we can conclude that the maximum that p(1-p) can be is 0.25. Hence without
knowing what the value of the population proportion actually is, we can argue that the
standard deviation of p̂ is less than or equal to
0.25
0.25 0.5
1
.



n
n
n 2 n
Hence a conservative estimate of the SD or SE of any p̂ is
1
2 n
.
So if n is large, we can conclude the p̂ ≈
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ASSESSING THE NORMALITY OF A POPULATION
Goal:
We consider as a quick aside of whether a population is Normally distributed or not.
The importance of this question should be obvious, since many of the techniques we
have seen are based on the premise that the population is Normal.
Normality was required in most of the inferences about a single population mean or the
difference of two means. It is also required for inferences about a population variance
and in the Regression and Analysis of Variance methods that we will see.
The basic premise of checking Normality of a population is that a random sample should
exhibit the same characteristics of the population. In particular, if the population is Normal,
then the sample should have “Normal” characteristics. In fact, the sample should have a
“bell-shape.”
Illustration: Consider the following histograms of data. The first two data sets are the soda
pop two-liter contents data and the Lithium Ion cell phone battery data. Based on the
histograms would you be willing to conclude the population is Normal or approximately Normal?
How about the last one?
Histogram of Pop Contents
Histogram of Lithium Ion Batteries:
12
2.0
10
1.5
Frequency
Frequency
8
1.0
6
4
0.5
2
0.0
1.6
1.8
2.0
Pop Contents
2.2
2.4
0
10
12
14
Lithium Ion Batteries:
16
18
Histogram of ShawnGreggResids
40
Frequency
30
20
10
0
-15
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-5
0
5
10
ShawnGreggResids
15
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METHODS OF ASSESSING THE NORMALITY OF A POPULATION
We will consider three methods of determining whether a population can be considered to
be Normally distributed. Two of the methods are graphical in nature and the last is a
hypothesis test.
In reality I typically use the graphical methods when the sample size is large (over 50 and
usually in the hundreds) and the testing method when the sample is smallish (less than 30).
The three methods are:
1. A histogram or stem and leaf of the data and if the population is Normal, then the
histogram should also be Normal or bell-shaped.
2. A Normal Probability Plot is a plot that will yield a straight line if the data came from
a Normal population.
3. A test in which Ho: The data came from a Normal population and HA: The data came
from a Non-Normal population.
The following examples will illustrate all three methods.
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EXAMPLE #1: Consider the Heart Rates (based on the 30sec measurement) from the
class survey for STA 261 sections in Winter ’98, Fall ’99, Fall ’00, & Fall ’02. Below are
the graphical descriptive statistics and the Normal Probability Plot for these data.
Descriptive Statistics
Variable: HR2
Anderson-Darling Normality Test
A-Squared:
P-Value:
30
50
70
90
110
130
150
95% Confidence Interval for Mu
1.285
0.002
Mean
StDev
Variance
Skewness
Kurtosis
N
71.8182
13.2895
176.610
1.29951
7.83947
231
Minimum
1st Quartile
Median
3rd Quartile
Maximum
30.000
64.000
72.000
80.000
160.000
95% Confidence Interval for Mu
70.095
69.8
70.8
71.8
72.8
73.8
73.541
95% Confidence Interval for Sigma
12.178
14.626
95% Confidence Interval for Median
95% Confidence Interval for Median
70.000
72.000
Normal Probability Plot for HR2
ML Estimates - 95% CI
ML Estimates
Mean 71.8182
99
StDev 13.2607
Percent
95
90
Goodness of Fit
80
70
60
50
40
30
20
AD*
1.37
10
5
1
50
100
150
Data
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EXAMPLE #2: This data is the parking ticket from the STA 261 Fall 2002 class.
Descriptive Statistics
Variable: PTickets
Anderson-Darling Normality Test
0
200
400
600
800
95% Confidence Interval for Mu
A-Squared:
P-Value:
13.060
0.000
Mean
StDev
Variance
Skewness
Kurtosis
N
60.517
150.128
22538.3
3.13743
10.9226
58
Minimum
1st Quartile
Median
3rd Quartile
Maximum
0.000
0.000
0.000
0.000
800.000
95% Confidence Interval for Mu
21.043
0
50
100
99.991
95% Confidence Interval for Sigma
126.919
183.803
95% Confidence Interval for Median
95% Confidence Interval for Median
0.000
0.000
Normal Probability Plot for PTickets
ML Estimates - 95% CI
99
ML Estimates
Mean 60.5172
95
StDev 148.828
90
Goodness of Fit
Percent
80
AD*
70
60
50
40
30
13.03
20
10
5
1
0
400
800
Data
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EXAMPLE #3: This data is the My Age from the STA 261 Fall 2002 class.
Descriptive Statistics
Variable: My Age
Anderson-Darling Normality Test
A-Squared:
P-Value:
35
39
43
47
51
55
95% Confidence Interval for Mu
0.660
0.081
Mean
StDev
Variance
Skewness
Kurtosis
N
47.0690
3.8428
14.7671
-1.7E-01
1.32698
58
Minimum
1st Quartile
Median
3rd Quartile
Maximum
35.0000
45.0000
47.5000
49.0000
57.0000
95% Confidence Interval for Mu
46.0586
46
47
48
48.0794
95% Confidence Interval for Sigma
3.2487
4.7048
95% Confidence Interval for Median
95% Confidence Interval for Median
46.0000
48.0000
Normal Probability Plot for My Age
ML Estimates - 95% CI
99
ML Estimates
Mean 47.0690
95
StDev 3.80953
90
Goodness of Fit
Percent
80
AD*
70
60
50
40
30
0.827
20
10
5
1
35
45
55
Data
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EXAMPLE #4: This data is pop data.
Summary for Pop Contents
A nderson-Darling N ormality Test
1.6
1.8
2.0
2.2
2.4
A -S quared
P -V alue
0.25
0.626
M ean
S tDev
V ariance
S kew ness
Kurtosis
N
2.0000
0.2828
0.0800
0.0
-1.2
7
M inimum
1st Q uartile
M edian
3rd Q uartile
M aximum
1.6000
1.8000
2.0000
2.2000
2.4000
95% C onfidence Interv al for M ean
1.7384
2.2616
95% C onfidence Interv al for M edian
1.7467
2.2533
95% C onfidence Interv al for S tDev
95% Confidence Intervals
0.1823
0.6228
Mean
Median
1.7
1.8
1.9
2.0
2.1
2.2
2.3
Probability Plot of Pop Contents
Normal - 95% CI
99
Mean
StDev
N
AD
P-Value
95
90
2
0.2828
7
0.248
0.626
Percent
80
70
60
50
40
30
20
10
5
1
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1.0
1.5
2.0
Pop Contents
2.5
3.0
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EXAMPLE #5: This data is the Lithium Ion data.
Summary for Lithium Ion Batteries:
A nderson-Darling N ormality Test
10
12
14
16
A -S quared
P -V alue
0.35
0.463
M ean
S tDev
V ariance
S kew ness
Kurtosis
N
14.348
2.069
4.282
-0.077778
-0.658949
45
M inimum
1st Q uartile
M edian
3rd Q uartile
M aximum
18
9.750
12.705
14.200
16.155
17.900
95% C onfidence Interv al for M ean
13.727
14.970
95% C onfidence Interv al for M edian
13.630
14.939
95% C onfidence Interv al for S tDev
95% Confidence Intervals
1.713
2.614
Mean
Median
13.50
13.75
14.00
14.25
14.50
14.75
15.00
Probability Plot of Lithium Ion Batteries:
Normal - 95% CI
99
Mean
StDev
N
AD
P-Value
95
90
14.35
2.069
45
0.348
0.463
Percent
80
70
60
50
40
30
20
10
5
1
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10.0
12.5
15.0
Lithium Ion Batteries:
17.5
20.0
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EXAMPLE #6: This data is the last data set.
Summary for ShawnGreggResids
A nderson-Darling N ormality Test
-15.0
-7.5
0.0
7.5
15.0
A -S quared
P -V alue <
1.46
0.005
M ean
S tDev
V ariance
S kew ness
Kurtosis
N
0.0000
6.0080
36.0955
0.80960
1.98405
144
M inimum
1st Q uartile
M edian
3rd Q uartile
M aximum
22.5
-15.5694
-3.4826
-0.6528
3.2708
22.6528
95% C onfidence Interv al for M ean
-0.9897
0.9897
95% C onfidence Interv al for M edian
-1.1023
0.0865
95% C onfidence Interv al for S tDev
95% Confidence Intervals
5.3850
6.7951
Mean
Median
-1.0
-0.5
0.0
0.5
1.0
Probability Plot of ShawnGreggResids
Normal - 95% CI
99.9
Mean
StDev
N
AD
P-Value
99
Percent
95
90
8.333331E-11
6.008
144
1.465
<0.005
80
70
60
50
40
30
20
10
5
1
0.1
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0
10
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SUMMARY OF SAMPLING DISTRIBUTIONS AND ESTIMATION
Defn: A Sampling Distribution is the distribution of an estimator or sample statistic (ie RV),
such as X , S2, S, p̂ , M
SAMPLE AVERAGE
Defn: X is the sample average of a RS of size n from a popln with mean, popln and
variance, popln
2
Distribution of X
Mean of X =  X = E[ X ] = popln
Variance of X = σ =
2
X
Shape:
2
σpopln
n
X will be exactly or approximately Normally distributed if the popln is
exactly or approximately distributed, regardless of the sample size, n
X will be approximately Normally distributed when n  30
X-μpopln
is Z if  is known and popln is Normal
σpopln
n
T(n-1) if popln is Normal
X-μpopln

.
is 
S
 Z if n large
n

SAMPLE PROPORTION
Defn: P̂ =
X
, where X = number in random sample with characteristic = Bin ( n, p ) where
n
p is the proportion with characteristic in population.
Distribution of P̂
Mean of P̂ = μP̂ = E[ P̂ ] = p = population proportion with characteristic
p(1 - p)
n
x
n!
px (1 - p)n-x or
Shape of P̂ : Pr{ P̂ =
} = Pr{ X = x } =
n
x!(n- x)!
Variance of P̂ = σP̂ =
2
Normal if np 5 & n(1-p) 5
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