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Unformatted text preview: Statistical Methods I (EXST 7005) Page 37 not, we can write this as two expressions, P(Z ≤ Z1) OR P(Z ≥ Z2).
It is also possible that problems involving sections may not be symmetric, and may occur
entirely in the positive tail, or negative tail. Some Examples
P(Z ≥ Z0), where Z0 = 1.96. Since we are taking the
absolute value of a randomly chosen Z value in either
tail, that value may be either positive or negative and
its absolute value may be greater than or equal to 1.96.
P(Z ≥ Z0), where Z0 = 1.96. 4 3 2 1 3 2 0 1 1 2 3 4 P(Z ≥ Z0) = P(Z ≤ –Z0) + P(Z ≥ Z0), and since it is symmetric
P(Z ≥ Z0) = 2*P(Z ≥ Z0) = 2(0.0250) = 0.050
P(Z ≤ Z0), where Z0 = 2.576. This problem is similar to
the previous, but it describes the area in the middle,
between two limits.
4 P(Z ≤ Z0) = 1 – P(Z ≥ Z0) = 1–2*P(Z ≥ Z0) = 1 –
2(0.0050) = 0.99 0 1 2 3 4 1 2 3 4 1 2 3 4 An asymmetric case
P(–1.96 ≤ Z ≤ 2.576) = ? This is the area in the
middle, the total minus the two tails. We already
know these tails.
P(–1.96 ≤ Z ≤ 2.576) = 1 – P(Z ≥ 1.96) – P(Z ≥
2.576) = 1 – 0.0250 – 0.0050 = 0.97 4 3 2 1 0 Working the Z tables, backward & forward
We have seen how to find a probability from a value of Z0.
Now we need to be able to find a value of Z0 when a
probability is known. Basically, we find the value of
the probability in the body of our Z table, and
determine the corresponding value of Z0.
P(Z ≤ Z0) = 0.1587, find the value of Z0 4 3 2 1 0 This probability is a value less than 0.5, so it is a tail and can be solved directly from our
tables. We only need to find 0.1587 in the table and determine the corresponding value
of Z0. The value in the table occurs in the row corresponding to “1.0” and the column
corresponding to “0.00”. Finally note that the randomly chosen Z was to be less than or
equal to Z0, so we are in the lower tail. Z0 = –1.00
P(Z ≤ Z0) = 0.8413, find the value of Z0
This probability is a value greater than 0.5. To read it
from our tables we must determine the
corresponding tail. The tail would be given by 1–
0.8413 = 0.1587. So this is the same as the value
we just looked up, it occurs in the row
corresponding to “1.0” and the column
4 3 2 1 0 1 2 3 4 James P. Geaghan Copyright 2010 Statistical Methods I (EXST 7005) Page 38 corresponding to “0.00”. Finally, note that the randomly chosen Z was to be less than
or equal to Z0, but since the probability was larger than 0.5 we are in the upper tail.
Z0 = +1.00
P(1 ≤ Z ≤ Z0) = 0.1, find the value of Z0
First note that the lower limit is 1.00, so we are
working with the upper tail of the Z
distribution. We can find the probability that Z
is ≤ 1 as 1 – P(Z ≥ 1) = 1 – 0.1587 = 0.8413.
The area of interest is stated to have an area of
0.1, so the upper tail would be what remains in
the upper tail, 1 – 0.8413 – 0.1 = 0.0587. We
find 0.0587 in our Z tables and note that it is
between 1.56 and 1.57. I would accept either
answer. The actual value, determined by the
EXCEL “NORMDIST” function is Z0 =
1.565781531. 4 3 2 1 0 1 2 3 4 Z 0 s o m e w h e re h e re 0 .1
0 .0 5 8 7 0.8 4 1 3
4 3 2 1 0 1 2 3 4 P(Z ≥ Z0) = 0.05, find the value of Z0
Note that the random value of Z is greater than Z0, so
we are examining an area in the tails, where the
random Z may be either positive or negative and its
absolute value will exceed the value of Z0. Recall
that our tables give only one tail, and this particular case has 0.05 in two tails, so we
want to determine the value of Z0 for only one of those tails to match our tables. Each
tail would have half of the 0.05 since it is a
symmetric problem. So the area in the tail is 0.05/2
= 0.025. We can examine the tables and determine
that for a probability of 0.025 the Z value is 1.96,
so Z0 = 1.96. Some people like to write Z0 = ±1.96,
1.96
1.96
but this isn't really necessary for this case.
P(Z ≤ Z0) = 0.99, find the value of Z0
Note that the random value of Z is less than Z0, so we are examining an area in the middle
of the distribution. Again, the randomly chosen Z may be either positive or negative.
Since our tables give the tails of the distribution we
need to determine the area in the tails. Since 0.99
occurs in the middle, the tails must have an area of 1
– 0.99 = 0.01. Half of this is in each tail, so the area
in one tail would be 0.01/2 = 0.005. Given that the
area in the tail is 0.005, the Z value is 2.576 (actually
2 .5 7 6
2 .5 7 6
between 2.57 and 2.58; this is a value we have seen
previously). The Z transformation of any normal distribution
Most practical applications of the Z distribution will require that we take a real distribution
{N(μ,σ2)} and convert it to a Z distribution {N(0,1)}, calculate some probability statement,
and then convert the results back to the original distribution.
So we work with two distributions
James P. Geaghan Copyright 2010 Statistical Methods I (EXST 7005) Page 39 Y ~ N(μ,σ2)
Z ~ N(0,1)
In order to convert the observed Y distribution to the more workable Z distribution we need to
transform the distribution. We use the Z transformation.
Zi = (Yi – μ) / σ
For example, suppose we have a distribution where Y ~
N(20,16) and we wish to determine P(Y ≤ 24).
We convert the probability statement for the
original distribution into a Z distribution using our
transformation. 20 Transformation: Zi = (Yi – μ) / σ for N(20, 16) ⎛ (Yi − μ ) P(Y ≤ 24) = P ⎜ ⎝ (
σ ≤ 24 − 20) ⎞
4⎟
⎠ 24 0.8413 = P(Z ≤ 1)
P(Z ≤ 1) = 1 – P(Z ≥ 1)=1 – 0.1587 = 0.8413 20
0 24
1 Original scale
Z distribution scale So, P(Y ≤ 24) = 0.8413
Another example
Using the same distribution Y ~ N(20,16) where we
wish to determine P(Y ≥ 22). ⎛ (Y − μ )
( 22 − 20) ⎞ = P(Z ≥ 0.5)
P(Y ≥ 22) = P ⎜ i
σ ≤
4⎟
⎝
⎠ 0.30 8 5
20
0 22
0.5 From the table we determine that P(Z ≥ 0.5) = 0.3085
Another type of example
Find Y0, where P(Y ≤ Y0) = 0.1515 for the same
distribution, Y~N(20, 16). Again, using our
transformation, ⎛ (Yi − μ ) P(Y ≤ Y0) = 0.1515 = P ⎜ ⎝ σ ≤ (Y0 − μ ) ⎞
σ ⎟=
⎠ Z = 1 .0 3 P(Z ≤ Z0) = 0.1515
Notice that the sign is ≤ and the probability small (less than 0.5), so we are in the lower
half of the distribution. From the table the Z value is 1.03, so Z0 = –1.03. So we
know that Z0 = –1.03, an area in the lower half of the distribution, but we don't know
the value of Y0 yet and our original problem was to find P(Y ≤ Y0) = 0.1515. So we
need to transform back to the Y scale. This backtransformation is a reversal of the Z
transformation.
Transformation: Zi = (Yi – μ) / σ
To transformation back calculate: James P. Geaghan Copyright 2010
Y=15.88 Statistical Methods I (EXST 7005) Page 40 Yi = μ + Zσ = 20 + (–1.03)4 = 20 + (–4.12) =
20 – 4.12 = 15.88.
The interpretation of this problem is that there is a probability of 0.1515 of randomly
drawing a value from this distribution and getting a value that is less than 15.88.
One last example
Determine within what limits 50% of the distribution lies. We will use the same
hypothetical distribution Y~N(20,16). We will assume that the objective here is to
form symmetric limits as shown below. Otherwise, the problem has no answer. After
all, half of the distribution (50%) lies above Z0 = 0, so the limits could be (0, ∞). Half
the distribution also lies below Z0 = 0, so the limits could also be (–∞, 0). And there
are an infinite number of other such limits in between. So, assuming symmetry, the
problem becomes to determine the values of Y0 such that half of the distribution lies
between –Z1 and +Z2, where Z1 = Y2 due to
symmetry and we will refer to this value as Y0.
Once we determine the values of Z that meet the
probability limits we will transform the upper
and lower value back to the Y distribution limits.
0.50
Determine: 0.25 0.25 P(Z ≤ Z0) = 0.5 Z0 Z 0 Since our tables give area in the tails, we need to change this from P(Z ≤ Z0) = 0.5 to
12P(Z ≥ Z0) = 0.5 and find P(Z ≥ Z0) = (1–0.5)/2 or P(Z ≥ Z0) = 0.25.
From our table: Z0 = 0.675
Transforming Z0 = 0.675 back to the Y scale gives,
–Z0, Y1 = 20 – 0.675(4) = 20 – 2.7 = 17.3
Z0, Y2 = 20 + 0.675(4) = 20 + 2.7 = 22.7
The final probability statement is best given in
the form, P(17.3 ≤ Y ≤ 22.7) = 0.50 Summary on use of Z tables
Values of the relative cumulative frequency are given in
the table. 0.50
0.25 0.25
Z 0 =0.675
Y 1 =17.3 Z 0=0.675
Y 2 =22.7 the table is one – sided
the value given in the table is for the upper tail of the distribution
The total area under the curve is 1.0
The distribution is symmetrical.
e.g. for Z = –1 r.c.f. = 0.1587 (in the left tail)
for Z = 1 r.c.f. = 0.1587 (in the right tail)
We can transform from any normal distribution to the Z distribution by using the
transformation Zi = (Yi – μ) / σ
We can transform back from the Z distribution to any normal distribution by using the
transformation Yi = μ + Zσ
James P. Geaghan Copyright 2010 Statistical Methods I (EXST 7005) Page 41 Memorable values of Z (for 2 sided evaluations)
μ ± 1σ = 0.68
μ ± 1.645σ = 0.90
μ ± 1.96σ = 0.95
μ ± 2.576σ = 0.99
A tip. Some tables give the one tailed probabilities at the top, some the two tailed probabilities.
Some are cumulative from the lower end, some cumulative from the zero value (middle)
up, and some give the cumulative area in the tails (like mine).
If you know that 1.96 leaves 2.5% in one tail and 5% in two tails you can look for this
value and figure out what kind of tables you have. Distribution of sample means
Sample means are the basis for testing hypotheses about μ, the most common types of hypothesis
tests. Before discussing hypothesis tests we will need some additional information about the
nature of sample means.
Imagine we are drawing samples from a population with the following characteristics.
Population size = N
Mean = μ
Variance = σ2
The individual observations from this parent population are:
Yi = Y1, Y2, Y3, ... , YN
The set of samples of size n from the parent population form a derived population
There are Nn possible samples of size n that can be drawn from a population of size N
(sampling with replacement, which simulates a very large population).
n For each sample we calculate a mean Yk = ∑Y
i =1 ik n , where k = 1, 2, 3, ... , Nn The Derived Population of means of samples of size n
The sample size = n
Population size = Nn
Mean = μ Y
Variance = σ Y2
Derived population values Yk = Y1 , Y2 , Y3 , Y4 ..., YN n ,
Nn Mean of the derived population μY = ∑Y
k =1 k Nn where k = 1, 2, 3, ... , Nn James P. Geaghan Copyright 2010 Statistical Methods I (EXST 7005) Page 42
Nn 2
Variance of the derived population σ Y = ∑ (Y
k =1 k − μ )2 where k = 1, 2, 3, ... , Nn Nn Original Population Example of a Derived Population
Parent Population: Yi = 0, 1, 2, 3 0.25 r.f. n = 2 and Nn = 42 = 16 where 0.00 Draw all possible samples of size 2 from the Parent
0 1 2 3
Population (sampling with replacement, so that
values will occur more than once), and calculate a mean for each of the Nn samples.
For this discrete uniform distribution the Mean = (Max + Min)/2 = (30)/2 = 1.5 and the
variance = (Max–Min+1)2 /12 = (3–0+1)2/12 = 1.25 and the standard deviation = 1.1180.
Sampling results for all possible means with replacement.
Sample
0, 0
0, 1
0, 2
0, 3
1, 0
1, 1
1, 2
1, 3
2, 0
2, 1
2, 2
2, 3
3, 0
3, 1
3, 2
3, 3 Mean
0.0
0.5
1.0
1.5
0.5
1.0
1.5
2.0
1.0
1.5
2.0
2.5
1.5
2.0
2.5
3.0 Deviation from 1.5 Squared Deviation 1.5
1
0.5
0
1
0.5
0
0.5
0.5
0
0.5
1
0
0.5
1
1.5 2.25
1
0.25
0
1
0.25
0
0.25
0.25
0
0.25
1
0
0.25
1
2.25 Distribution of the derived population of sample means
Means
0.0
0.5
1.0
1.5
2.0
2.5
3.0
Sum = Frequency
1
2
3
4
3
2
1
16 Relative Freq
0.0625
0.1250
0.1875
0.2500
0.1875
0.1250
0.0625
1 0.30
0.25
0.20
0.15
0.10
0.05
0.00 Derived Population 0.0 0.5 1.0 1.5 2.0 2.5 3.0 Nn μY = ∑ Yk N n = 2416 = 1.5
k =1
Nn σ = ∑ (Yk − μ )
2
Y k =1 2 N n ( 4(0.0)
= 2 + 6(±0.5)2 + 4(±1.0)2 + 2(±1.5) 2 )
16 = James P. Geaghan Copyright 2010 Statistical Methods I (EXST 7005) Page 43 ( 4(0.0) + 6(0.25) + 4(1.0) + 2(2.25) ) = 10 = 0.625
16 16 σ = 0.7906
Note that the histogram of the derived population shows that the population is shaped more like
the normal distribution than the original population.
Probability statement from the two distributions: Find P(1 ≤ Y ≤ 2)
For the original, uniform population, P(1 ≤ Y ≤ 2) = 0.5000
For the derived population, P(1 ≤ Y ≤ 2) = 0.6250 THEOREM on the distribution of sample means
Given a population with mean μ and variance σ2, if we draw all possible samples of size n
(with replacement) from the population and calculate the mean, then the derived
population of all possible sample means will have
Mean: μY = μ Variance: 2
σY = σ n Standard error: 2
σY = σ n = σ
n 2 Notice that the variance and standard deviation of the mean have “n” in the denominator. As
a result, the variance of the derived population becomes smaller as the sample size
increases, regardless of the value of the population variance. Central Limit Theorem
As the sample size (n) increases, the distribution of sample means of all possible samples, of a
given size from a given population, approaches a normal distribution if the variance is
finite. If the base distribution is normal, then the means are normal regardless of n.
Why is this important? (It is very important!)
If we are more interested in the MEANS (and therefore the distribution of the means)
than the original distribution, then normality is a more reasonable assumption.
Often, perhaps even usually, we will be more interested in characteristics of the
distribution, especially the mean, than in the distributions of the individuals. Since
the mean is often the statistic of interest it is useful to know that it is possibly
normally distributed regardless of the parent distribution.
NOTES on the distribution of sample means
Another property of sample means
as n increases, σ Y2 and σ Y decrease.
σ Y2 ≤ σ 2 σ Y2 < σ 2 for any n
for any n > 1 James P. Geaghan Copyright 2010 Statistical Methods I (EXST 7005) Page 44 as n increases and σ Y becomes smaller, the distribution of the means gets closer to μ Y .
(i.e. we get a better estimate). Some new terms
Reliability (as a statistical concept) – the less scatter that occurs in a set of data, the more
“reliable” the estimate. This term is also associated with the word “precision” in
statistics where high reliability comes from low variance and high precision. This
concept is important in estimation because it suggests that data is reproducible or
repeatable.
Accuracy (as a statistical concept) – this term refers to the lack of bias in the estimate, and not
the smallness of the variance (e.g. reliability or precision). An accurate, or unbiased,
estimated may have considerable scatter among the points, but on average the center of
the distribution is neither overestimated or underestimated. This aspect of data is related
to the validity of data.
2
Estimation of σY and σY In practice we cannot draw all possible samples.
Recall that E(S2) = σ 2
2
so, SY = S 2 2
n is an estimate of σ Y where;
S Y2 is an estimate of the variance of sample means of size n S2 is the estimate of the variance of observations
2
SY = S =S
is called the standard error to distinguish it from the standard
n
n
deviation it is also called the standard deviation of the means
Notice that this is a division by “n “ for both populations and samples, not by “ n – 1” as
with the calculation of variance for samples.
S Y2 is a measure of the reliability of the sample means as an estimate of the true population mean.
i.e. the smaller S Y2 , the more reliable Y as an estimate of μ
Ways of increasing reliability
Basically, anything that decreases our estimate of S Y2 makes our estimate more reliable.
How do we decrease our estimate of S Y2 ?
Increase the sample size; if n increases then S Y2 decreases.
Decrease the variance; if our estimate of S 2 decreases then S Y2 decreases.
This can sometimes be done:
James P. Geaghan Copyright 2010 Statistical Methods I (EXST 7005) Page 45 • by refining our measurement techniques • by finding a more homogeneous population to measure (stratification) • by removing exogenous sources variance (blocking) Notation: The variance of a variable Y can be denoted S 2 or S Y2 . The subscript is only
needed to clarify which variable is indicated, so it is often not needed and omitted.
However, the variance of the means should be subscripted, S Y2 , to distinguish it from the
variance of the observations. The Z transformation for a derived population
We will use the Z transformation for two applications, individuals and means.
Applications to individual observations – to make statements about individual members of
the population, such as “What percentage of the individuals would be expected to have
values greater than 17?”
Zi = (Yi − μ )
σ Applications to means – to make statements about population means such as “What is the
probability that the true population mean is greater than 17?” Zi = (Y − μ )
i Y σY Summary
Most testing of hypotheses will concern tests of a derived population of means.
The mean of the derived population of sample means is μ Y
The Variance of the derived population of sample means is σ Y2
The CENTRAL LIMIT THEOREM is an important aspect of hypothesis testing because it
states that sample means tend to be more nearly normally distributed than the parent
population. We will often work with distributions that are not normally distributed,
but the fact that we are often interested in the more normally distributed means instead
of the original observations allows the use of parametric statistics.
Reliability and accuracy are statistical concepts relating to variability and lack of bias,
respectively.
Mean: μY = μ Variance: 2
σY = σ n Standard error: 2
σY = σ n = σ
n 2 James P. Geaghan Copyright 2010 ...
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This note was uploaded on 12/29/2011 for the course EXST 7005 taught by Professor Geaghan,j during the Fall '08 term at LSU.
 Fall '08
 Geaghan,J

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