Need to know in order to do the
normal dist problems
• How to calculate Z
• How to read a probability from the table,
knowing Z
**** how to convert table values to area that
you need---you need to DRAW THE ND
AND SHADE WHAT YOU NEED
• How to go from a probability on the table
to Z
• How to convert Z to X
Statistical Inference
Samples & Populations
Differs from the course so far
Up till now, just “descriptive”
statistics.
Just reporting values that were
directly measured or counted.
Estimation
Starting with some theory to
set up a basis.
We plan to use information from sample(s)
to describe a population.
Theory
• Chapter 8, pp 196 – 198 and 204 - 208
Most Significant Parameters
• Mean of the Population
• Standard Deviation of the Population
Sampling Distribution of the
Mean
•
•
•
•
Take many samples, each of size n
Take the mean of each sample, X
Take the mean of these means
The mean of the sampling distribution of
means follows a normal distribution
If we …
Take all possible samples of size n from a
population
The mean of the means of all samples
equals the mean of the population
And the Standard Deviation
Of the population can be calculated from the
distribution of sample means.
Standard Deviation of the sampling
distribution is called the standard error of
the mean
The dispersion of the sampling distribution is
narrower than the values in the original
population
Dispersion of Means narrower than
dispersion of values in the
population
For example
• Serum cholesterol values for all 50-yr old
men – the distribution would follow
something like the low very dispersed ND
shown in the previous slide.
• Samples, 10 men in each – the distribution
of the sample means would follow
something like the high narrow ND.
Standard Error of the Mean
Compare it to the standard deviation of the
population
The standard deviation of the population is
expressed as sigma, б
The standard error = б / n
Standard Error of the Mean
• Its value depends on sigma of the
population
• And on n. The larger the sample, the
smaller the value for S.E.
• S.E. is written as sigma sub X bar. Too
difficult for me to put in power pt.
More about the
Sampling Distribution of the Mean
The mean of the sampling distribution of
means follows a normal distribution.
The basis for this is called
The Central Limit Theorem
Central Limit Theorem
• Even if the values for the original population do
not fit a normal curve, the distribution of the
sample means does fit the normal distribution.
• This is true if the size of the samples is large
enough.
• What is “large enough”?
• N = 30 or larger. Each sample must have 30 or
more observations.
Is n ≥ 30 always necessary?
• No
• If the original population is itself a normal
distribution, then the distribution of sample
means will be normal even if the sample
size is extremely small….even n = 1.
The Original Population
The distribution of sample means is a
normal distribution even if
• The values of the original population follow
a skewed distribution
• The values of the original population are
discrete
The Central Limit Theorem
• How does the sampling distribution compare to the
original population.
• If we take ALL SAMPLES OF SIZE N FROM THE
POPULATION
• Mathematically, the mean of the population equals the
mean of the sampling distribution of means.
• Standard deviation of the distribution of means,
“standard error of the mean” = sigma/ square root of n.
But…
• We are NOT going to be taking all possible
samples
• So….
• We use the mean of a sample as the best
estimate we have of the mean of the
population, 
And…
We may be given the standard deviation of
the population
or
We take the standard deviation of a
sample as the best estimate of the
standard deviation of the population, б.
Estimating the Mean
• The best estimate that we can get is the
mean of the sample
• But that isn’t good enough
• It’s called a point estimate
• And we have no idea of its probability of
being the true mean
Instead
• We look for a range within which we can
expect to find the true mean
• We will also be able to express the
probability that the mean is really within
this range.
An example
• If we find the mean of a sample of insulin levels
is 100 units
• The true mean might be 101, 98, 101.2 or any
number of values
• But if we use a normal distribution, we can
calculate a range for the mean, e.g. 95 – 105
and say that there is a 99% probability that the
true mean falls within this range. (This example
is only concocted numbers. Don’t try to confirm
them.)
The range and the probability are
called
Confidence Intervals
Estimating the Mean
• We are now using Chapter 9
Getting the Range & the Probability
• Use the Normal Distribution
• Use Z and the area under the ND curve
Z  (X μ)
σ/ n
Previously we used
Z  (X  μσ
Compare with Previous Examples
• We did the normal distribution of values,
X, around the mean of a population
• The spread is the standard deviation of the
population.
• Here we are looking at the normal
distribution of sample means, X, around
the mean of the means..
• The spread is the standard error of the
mean
Values of Area and Z
• These are the same as in any other
standard normal distribution
• e.g. 95% of the cases fall within 2
standard deviations of the mean
Note: 2 standard deviations on both sides of the mean
Approximation
We used 2 standard deviations but when
we looked in Table A.3, we found that 95%
of the cases are actually within Z = 1.96,
not 2.00
Let’s check it out. Look at 1.96, what is
the area?
Convert it to the two sides of the mean
Table A.3
• Gives us the area under the “tail”.
• Subtract that area from 0.5000
• Multiply it by two.
• For Z = 1.96, A(under the tail) = 0.025
• 0.5000 – 0.025 = 0.475
• Times two = 0.950 that is, 95%
Example
• We are interested in finding the average
level of enzyme, cut-em-up, in a
population, e.g. patients in Pro-health
Group Practice.
• A sample of 10 patients has an average
level of 22 units.
• It is known from other information that the
level of this enzyme is approximately
normally distributed with a variance of 45.
Find 95% Confidence Interval for 
To find 95% C.I., use Z = 1.96
Z  (X μ)
σ/ n
Rearrange the equation to solve for
Z * σ/ n  X  μ

Form of Equation to use for
Confidence Intervals
Z * σ/ n  X  μ
μ  X  Z * σ/ n
But this would give us a point estimate of mu, so have
to change this a little more.
Just look at the part after the equal sign
Mu is between the two values calculated by:
X  Z * σ/ n
X  Z * σ/ n
Always draw the N.D.
The shaded area can help us to see what
we mean by X  Z * σ/ n
It is the border of the shaded area to the right of the mean .
We are saying that the mean lies between that border and the
corresponding left-side border
Write the equation


X  Z * σ/ n 
μ  X  Z * σ/
n
Mu lies between the two values within the
parentheses

Practical Statement of Result
• With 95% confidence, we can say that u
will be ≤ 1.96 *S.E. and
• ≥ -1.96 * S.E.
• We call 1.96 the reliability coefficient
The Math
•
•
•
•
•
•
•
X = 22,
n = 10,
б2 = 45
Review what the symbols mean?
Find the quantity Z * б / Γn
б = Γ45 = 6.7
б / Γn = 6.7/ Γ10 = 6.7 / 3.16 = 2.12
1.96 * 2.12 = 4.16
Mu is between 22 – 4.16 and 22 + 4.16
• 17.84 ≤ µ ≤ 26.16 with 95% confidence
Another Example
• Page 230, #13
•
•
•
•
•
Take out diskette from back of book
Insert into computer
Click on Install
Check ASCII, excel, SPSS
Install to hard drive
Excel
•
•
•
•
Go to “exercise”
Find “lowbwt”
Save to hard drive to work on
File will probably be gone next time you
return. Save the data set we want onto
your own floppy
Do Problem 13, just the males
Separate male & female,
how?
Important Statements in the
Problem
1. Large sample
Applications
• If we know a population mean & st. dev.,
we can calculate the probability that any
sample will have a stated mean.
• A certain large human pop’n has a cranial
length that is approx’ly normally distributed
with mean 185.6 mm and б of 12.7 mm.
µ = 185.6 mm
б = 12.7 mm
• What is the probability that a random
sample of size 10 from this population will
have a mean greater than 190?
• We can calculate this probability but why
would we?
Usefulness??
• Let’s say that it is accepted knowledge
that the population has a certain mean.
• I am working with a group of people.
• I want to know if they fit into this
population with regard to the particular
parameter. If the probability of the mean
of the sample is very low, perhaps it is not
really from the same population
Education Example
• Third-graders in the U.S. have an average
reading score of 124.
• Third-graders in a particular school have a
mean reading score of 120. What’s the
probability that they are from the same
population?
Back to Cranial Length
• µ = 185.6 mm б = 12.7 mm
• random sample of size 10 from this
population will have a mean greater
than 190?
• Have to find how far 190 is from 185.6
in units of standard error of the mean
Z  (190 185.6)
12.7/ 10
Probability of Mean of 190
Did you draw a normal
dist???
Z = 4.4
/ 12.7 / 3.16
Z = 1.09
0.138
Area = 0.138
185.6 190
The probability is 13.8%
0
1.09
The mean & st. dev. of serum iron values are
120 & 15 micrograms per 100 ml. What is the
probability that a random sample of 50 normal
men will yield a mean between 115 & 125
µg/100ml?
µ = 120
б = 15
Z1 = (115-120) / 15 / sqrt of 50
Z2 = (125-120) / 15 / sqrt of 50
Z1 = (115-120) / 15 / sqrt of 50
Z2 = (125-120) / 15 / sqrt of 50
Draw the Normal Distribution