1. Estimation Basic Concepts & Estimation of Proportions
EMIS 7370 STAT 5340
Department of Engineering Management, Information and Systems
Probability and Statistics for Scientists and Engineers
Estimation Basic Concepts &
Estimation of Proportions
Dr. Jerrell T. Stracener

2. Estimation Basic Concepts & Estimation of Proportions
EMIS 7370 STAT 5340
Department of Engineering Management, Information and Systems
Probability and Statistics for Scientists and Engineers
Estimation Basic Concepts &
Estimation of Proportions
Dr. Jerrell T. Stracener

3. Estimation Types of estimates and methods of estimation
Estimation - Binomial distribution
- Estimation of a Proportion
- Estimation of the difference between two
proportions
Estimation - Normal distribution
- Estimation of the mean
- Estimation of the standard deviation
means

4. Estimation Estimation - Normal distribution (continued)
- Estimation of the ratio of the two standard
deviations
- Tolerance intervals
Estimation - Lognormal distribution
Estimation - Weibull distribution
Estimation - Unknown distribution Types
- Continuous populations
- Finite populations

5. Types of Estimates & Methods of Estimation

6. Definition - Statistic
A statistic is a function of only the values of a
random sample, X1, X2, …, Xn.
For example
is a statistic

7. Properties of Estimates
A statistic is said to be an unbiased estimator
of the parameter  if
If we consider all possible unbiased estimators of
some parameter , the one with the smallest
variance is called the most efficient estimator
of 

8. Types of Estimates Point Estimate
A function of the values of a random sample that
yields a single value, i.e., a point
Interval Estimate
An interval, whose end points are functions of the
values of a random sample, for which one can assert
with a specified confidence that the interval contains
the parameter being estimated

9. Types of Estimates & Methods of Estimation
If we use a sample mean to estimate the mean
of a population, a sample proportion to estimate
the probability of success on an individual trial,
or a sample variance to estimate the variance of
a population, we are in each case using a point
estimate of the parameter in question. These
estimates are called point estimates since they are
single numbers, single points, used, respectively,
to estimate , , and 2. Since we can hardly
expect the point estimates based on samples to
hit the parameters they are supposed to estimate
exactly ‘on the nose’, it is often desirable to give an
interval rather than a single number.

10. Types of Estimates & Methods of Estimation
We can then assert with a certain probability (or
degree of confidence) that such an interval
contains the parameter it is intended to estimate.
For instance, when estimating the average IQ of
all college students in the US, we might arrive at
a point estimate of 117, or we might arrive at an
interval estimate to the effect that the interval
from 113 to 121 contains the ‘true’ average IQ of
all college students in the US.

11. Interval Estimates of  for Different Samples

12. Method of Maximum Likelihood
Given independent observations x1, x2, ..., xn from
a probability density function (continuous case)
f(x; ) or probability mass function (discrete case)
p(x; ) the maximum likelihood estimator is
that which maximizes the likelihood function.
L(X1, X2, ..., Xn; ) = f(X1; )·f(X2; )·...·f(Xn; ),
if x is continuous
= p(X1; )·p(X2; )·...·p(Xn; ),
if x is discrete

13. Method of Maximum Likelihood
Let x1, x2, ..., xn denote observed values in a
sample. In the case of a discrete random variable
the interpretation is very clear. The quantity
L(x1, x2, ..., xn; ), the likelihood of the sample,
is the following joint probability:
P(X1 = x1, X2 = x2, ... , Xn = xn)
This is the probability of obtaining the sample
values x1, x2, ..., xn. For the discrete case the
maximum likelihood estimator is one that results
in a maximum value for this joint probability, or
maximizes the likelihood of the sample.

14. Estimation of Proportions

15. Estimation of Proportions
Estimation of the proportion, p, based on a random sample of the Binomial Distribution B(n,p)
Point Estimation
Interval Estimation
Approximate Method
Sample Size
Exact Method
Estimation of the difference in two proportions P1 - P2 based on random samples from
B(n, P1) and B(n, p2).

16. Estimation of Proportions
Point Estimation
Interval Estimation

17. Estimation - Binomial Distribution
Estimation of a Proportion, p
X1, X2, …, Xn is a random sample of size n from
B(n, p), where
Point estimate of p:
where fs = # of successes _ ^

18. Estimation - Binomial Distribution
Approximate (1 - )·100% confidence interval
for p:
where and
where ,
and is the value of the standard normal
random variable Z such that ^ ^ ^ ^

19. Note When n is small and the unknown proportion p is
believed to be close to 0 or to 1, the approximate
confidence interval procedure established here is
unreliable and, therefore, should not be used. To be
on the safe side, one should require or ^ ^

20. Error in Estimating p by p^
Error in Estimating p by p
error ^

21. Error in Estimating p by p^
Error in Estimating p by p
If p is used as an estimate of p, we can be
(1 - )·100% confident that the error will not
exceed
(1 - )·100% confident that the error will be less
than a specified amount e when the sample size is ^ ^ ^ ^ ^

22. Error in Estimating p by p^
Error in Estimating p by p
If p is not used as an estimate of p, we can be
at least (1 - )·100% confident that the error will
not exceed a specified amount e when the sample
size is ^

23. Example In a random sample of n = 500 families owning
television sets in the city of Hamilton, Canada, it
is found that x = 340 subscribed to HBO.
a. How large a sample is required if we want to
be 95% confident that our estimate of p, p, is within
0.02?
b. How large a sample is required if we want to be
95% confident that our estimate of p is within ^

24. Example - Solution a. Let us treat the 500 families as a preliminary
sample providing an estimate p = 0.68. ^
Therefore, if we base our estimate of p on a
random sample of size 2090, we can be 95%
confident that our sample proportion will not
differ from the true proportion by more than
0.02

25. Example - Solution b. We shall now assume that no preliminary
sample has been taken to provide an estimate
of p. Consequently, we can be at least 95%
confident that our sample proportion will not
differ from the true proportion by more than
0.02 if we choose a sample of size

26. Example: Estimation of Binomial parameter p
In a random sample of n = 500 families owning
television sets in the city of Hamilton, Canada, it
was found that fS = 340 owned color sets. Estimate
the population proportion of families with color TV
sets and determine a 95% confidence interval for
the actual proportion of families in this city with
color sets.

27. Example: solution The point estimate of p is = 340/500 = 0.68.
Then, an approximate 95% confidence interval
for p is where
and
so that ^ ^ ^ ^ ^

28. Example: solution and an approximate 95% confidence interval for p is
( , ).
Therefore, our “best” (point estimate) of p is 0.68 and we are about 95% confident that p is between 0.64 and 0.72.

29. ( ) Estimation - Binomial Population
Exact (1 - )·100% Confidence Interval for p:
, where F1 =
and
, where F2 = ,
and is the value of x for which
P(X> )= 
NOTE: Use the ‘FINV’ function in Excel to get the values of F1 and F2 ( ) U L P ,

30. Example A random sample of 25 vehicle records are
selected for audit from a large number of county
records. It is found that 5 have errors. Estimate
the population proportion of vehicle records
having errors in terms of a point estimate and
95% confidence interval.

31. Example - solution An approximate 95% confidence interval for p is
where
 = and ^

32. Example - solution
Then ^ ^ ^

33. Example - solution ^

34. Example - solution
An exact 95% confidence interval for p is
where

35. Example - solution
and

36. Estimation - Binomial Populations
Estimation of the difference between two
proportions
Let X11, X12, …, , and X21, X22, …, ,
be random samples from B(n1, p1) and
B(n2, p2) respectively
Point estimation of p1 – p2 = p ^ ^ ^

37. Estimation - Binomial Populations
Approximate (1 - )·100% confidence interval
for
where
and ^ ^ ^ ^ ^ ^ ^ ^ ^ ^

38. Example: Estimation of P1 - P2
A certain change in a manufacturing process for
component parts is being considered. Samples
are taken using both the existing and the new
procedure in order to determine if the new
procedure results in an improvement. If 75 of
1500 items from the existing procedure were
found to be defective and 80 of 2000 items from
the new procedure were found to be defective,
find a confidence interval for the true difference
in the fraction of defectives between the existing
and the new process.

39. Example: solution Let p1 and p2 be the true proportions of defectives
for the existing and new procedures, respectively.
Hence
and
and the point estimate of p = p1 - p2 is =
= 0.01 ^ ^ ^ ^ ^

40. Example: solution An approximate 90% confidence interval for
p = p1 - p2 is where

41. Example: solution ^ ^ ^ ^ ^ ^

42. Example: solution Then
Therefore an approximate 90% confidence interval
for  = p1 - p2 is ( , ).
or about 93% of the length of the confidence
Interval favors the new procedure