1. Chapter 7 (b) – Point Estimation and Sampling Distributions
Point estimation is a form of statistical inference.
In point estimation we use the data from the sample
to compute a value of a sample statistic that serves
as an estimate of a population parameter.
We refer to as the point estimator of the population
mean .
s is the point estimator of the population standard
deviation .
is the point estimator of the population proportion p.

2. Management Training Program
Point Estimation
Example: EAI Employee Data Out of a total number of 2500 employees, a simple random sample of 30 employees and corresponding data on annual salary and management training program participation are shown in the given table. x1, x2 ….. xn is used to denote annual salary of the employees and the participation in the management training program is indicated by a yes / no.
Annual Salary ($)
Management Training Program
X1 =
YES
45,922.60
45,120.90
X2 =
57,268.40 NO
51,753.00
49,643.50
55,688.40
54,391.80
49,894.90
51,564.70
50,164.20
47,621.60
56,188.20
52,973.60
55,924.00
51,766.00
50,241.30
49,092.30
52,541.30
52,793.90
51,404.40
50,979.40
50,957.70
51,932.60
55,860.90
55,109.70
52,973.00
57,309.10

3. Point Estimation

4. Summary of Point Estimates Obtained from a Simple Random Sample
Population Parameter
Parameter value
Point estimator
Point estimate
m = Population mean annual salary
$ 51,800
$ 51,814
s = Population standard deviation for annual salary
$ 4,000
s = Sample standard deviation for annual salary
$ 3,348
p = Population proportion having completed MTP
.60
.63

5. Practical Advice The target population is the population we want to
make inferences about.
The sampled population is the population from
which the sample is actually taken.
Whenever a sample is used to make inferences about
a population, we should make sure that the targeted
population and the sampled population are in close
agreement.

6. Sampling Distribution of
Process of Statistical Inference
Population
with mean
m = ?
A simple random sample
of n elements is selected
from the population.
The value of is used to
make inferences about
the value of m.
The sample data
provide a value for
the sample mean .

7. Sampling Distribution of
Sampling Distribution of
where: µ = the population mean
When the expected value of the point estimator
equals the population parameter, we say the point
estimator is unbiased.

8. Sampling Distribution of
Standard Deviation of
We will use the following notation to define the
standard deviation of the sampling distribution of .
s = the standard deviation of
s = the standard deviation of the population
n = the sample size
N = the population size

9. Sampling Distribution of
Standard Deviation of
Finite Population
Infinite Population
A finite population is treated as being
infinite if n/N < .05.
is the finite population
correction factor.
is referred to as the standard error of the
mean.

10. In cases where the population is highly skewed or outliers are present, samples of size 50 may be needed.

11. Central Limit Theorem

12. Example: EAI Employee Data
Example: EAI Employee Data
Note: n/N = 30/2500 = .012 Because sample is less than
5% of the population size, We ignore the finite population correction factor.

13. Example: EAI Employee Data
Example: EAI Employee Data
Suppose the personnel director believes the sample mean will be an acceptable estimate of population mean if the sample mean is within $500 of the population mean.
What is the probability that the sample mean computed using a simple random sample of 30 employees will be within $500 of population mean?

14. Example: EAI Employee Data
Example: EAI Employee Data
Step 1: Calculate the z-value at the upper endpoint of the interval.
z = (52,300 – 51,800)/ = .68
Step 2: Find the area under the curve to the left of the
upper endpoint.
P(z < .68) = .7517

15. Example: EAI Employee Data Cumulative Probabilities for
Example: EAI Employee Data
Cumulative Probabilities for
the Standard Normal Distribution Z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09 . .5
.6915
.6950
.6985
.7019
.7054
.7088
.7123
.7157
.7190
.7224 .6
.7257
.7291
.7324
.7357
.7389
.7422
.7454
7.486
.7517
.7549 .7
.7580
.7611
.7642
.7673
.7704
.7734
.7764
.7794
.7823
.7852 .8
.7881
.7910
.7939
.7967
.7995
.8023
.8051
.8078
.8106
.8133 .9
.8159
.8186
.8212
.8238
.8264
.8289
.8315
.8340
.8365
.8389

17. Example: EAI Employee Data

18. Function used - NORM.DIST
Function used - NORM.DIST
We do not have to make separate computation of z value.
Evaluating NORM.DIST function at each end point of the interval provides cumulative probability at the specified end point of the interval.
The result obtained using NORM.DIST is more accurate.

21. Example: EAI Employee Data

23. Point Estimation Example: St. Andrew’s College
St. Andrew’s College received 900 applications from prospective students. The application form contains a variety of information including the individual’s Scholastic Aptitude Test (SAT) score and whether or not the individual desires on-campus housing.
At a meeting in a few hours, the Director of Admissions would like to announce the average SAT score and the proportion of applicants that want to live on campus, for the population of 900 applicants.
However, the necessary data on the applicants have not yet been entered in the college’s computerized database. So, the Director decides to estimate the values of the population parameters of interest based on sample statistics. The sample of 30 applicants selected earlier with Excel’s RAND function will be used.

24. Point Estimation Using Excel
Excel Value Worksheet (Sorted) A B 1
Applicant
Number 2 3 4 5 6 7 8 9
Random 12
773
408 58
116
185
510
394 C D
SAT
Score
On-Campus
Housing
1207 No
1143
Yes
1091
1108
1227
982
1363
Note: Rows are not shown.

25. Point Estimation s as Point Estimator of 
s as Point Estimator of 
Note: Different random numbers would have
identified a different sample which would have
resulted in different point estimates.

26. Point Estimation Once all the data for the 900 applicants were entered
in the college’s database, the values of the population
parameters of interest were calculated.
Population Mean SAT Score
Population Standard Deviation for SAT Score
Population Proportion Wanting On-Campus Housing

27. Summary of Point Estimates Obtained from a Simple Random Sample
Population
Parameter
Parameter
Value
Point
Estimator
Point
Estimate
m = Population mean
SAT score
1697
1684
s = Population std.
deviation for
SAT score
87.4
s = Sample stan-
dard deviation
for SAT score
85.2
p = Population pro-
portion wanting
campus housing
.72
.67

28. Example: St. Andrew’s College
Example: St. Andrew’s College

29. Sampling Distribution of
Example: St. Andrew’s College
What is the probability that a simple random sample of 30 applicants will provide an estimate of the population mean SAT score that is within +/-10 of the actual population mean  ?
Be between 1687 and 1707?
Step 1: Calculate the z-value at the upper endpoint of
the interval.
z = ( )/15.96 = .63
Step 2: Find the area under the curve to the left of the
upper endpoint.
P(z < .63) = .7357

30. Example: St. Andrew’s College
Cumulative Probabilities for
the Standard Normal Distribution z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09 . .5
.6915
.6950
.6985
.7019
.7054
.7088
.7123
.7157
.7190
.7224 .6
.7257
.7291
.7324
.7357
.7389
.7422
.7454
.7486
.7517
.7549 .7
.7580
.7611
.7642
.7673
.7704
.7734
.7764
.7794
.7823
.7852 .8
.7881
.7910
.7939
.7967
.7995
.8023
.8051
.8078
.8106
.8133 .9
.8159
.8186
.8212
.8238
.8264
.8289
.8315
.8340
.8365
.8389

31. Example: St. Andrew’s College
Example: St. Andrew’s College
Area = .7357
1697
1707

32. Step 3: Calculate the z-value at the lower endpoint of
the interval.
z = ( )/15.96 = - .63
Step 4: Find the area under the curve to the left of the
lower endpoint.
P(z < -.63) = .2643
Area = .2643
1687
1697

33. Example: St. Andrew’s College
Example: St. Andrew’s College
Step 5: Calculate the area under the curve between
the lower and upper endpoints of the interval.
P(-.68 < z < .68) = P(z < .68) - P(z < -.68) = =
The probability that the sample mean SAT score will
be between 1687 and 1707 is:

34. Example: St. Andrew’s College
Area = .4714
1687
1697
1707

35. Increasing the Sampling Size
Example: St. Andrew’s College
Suppose we select a simple random sample of 100
applicants instead of the 30 originally considered.

36. Example: St. Andrew’s College
Example: St. Andrew’s College

37. Example: St. Andrew’s College

38. Example: St. Andrew’s College
Example: St. Andrew’s College
Area = .7776
1687
1697
1707

39. Sampling Distribution of
The sampling distribution of is the probability
distribution of all possible values of the sample
proportion .
Expected Value of
where:
p = the population proportion

40. Sampling Distribution of
Standard Deviation of
Finite Population
Infinite Population
is referred to as the standard error of
the proportion.
is the finite population
correction factor.

41. Form of the Sampling Distribution of
The sampling distribution of can be approximated
by a normal distribution whenever the sample size
is large enough to satisfy the two conditions:
np > and n(1 – p) > 5
. . . because when these conditions are satisfied, the
probability distribution of x in the sample proportion,
= x/n, can be approximated by normal distribution
(and because n is a constant).

42. Example: EAI Employee Data For the EAI study we know that the population proportion of employees who participated in the management training program is p = .6. What is the probability that a simple random sample of 30 employees will provide an estimate of the population proportion of employees attending management program that is within plus or minus .05 of the actual population proportion?

43. Example: EAI Employee Data
Example: EAI Employee Data
For our example, with n = 30 and p = .6, the normal distribution is an acceptable approximation because:
np = 30(.6) = 18 > 5
And
n(1 - p) = 30(.4) = 12 > 5

44. Example: EAI Employee Data
Example: EAI Employee Data
Step 1: Calculate the z-value at the upper endpoint of the interval.
z = ( )/.0894 = .56
Step 2: Find the area under the curve to the left of the upper endpoint.
P(z < .56) = .7123

45. Example: EAI Employee Data Z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 .
Example: EAI Employee Data Z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09 . .5
.6915
.6950
.6985
.7019
.7054
.7088
.7123
.7157
.7190
.7224 .6
.7257
.7291
.7324
.7357
.7389
.7422
.7454
7.486
.7517
.7549 .7
.7580
.7611
.7642
.7673
.7704
.7734
.7764
.7794
.7823
.7852 .8
.7881
.7910
.7939
.7967
.7995
.8023
.8051
.8078
.8106
.8133 .9
.8159
.8186
.8212
.8238
.8264
.8289
.8315
.8340
.8365
.8389

46. P(-.56 < z < .56) = P(z < .56) - P(z < -.56)
Example: EAI Employee Data
Step 3: Calculate the z-value at the lower endpoint of the interval.
z = ( )/.0894 = - .56
Step 4: Find the area under the curve to the left of the lower endpoint.
P(z < -.56) =
Step 5: Calculate the area under the curve between the lower and upper endpoints of the interval.
P(-.56 < z < .56) = P(z < .56) - P(z < -.56) = =

47. Sampling Distribution of (another example)
Example: St. Andrew’s College
Recall that 72% of the prospective students applying
to St. Andrew’s College desire on-campus housing.
What is the probability that a simple random sample of 30 applicants will provide an estimate of the population proportion of applicant desiring on-campus housing that is within plus or minus .05 of the actual population proportion?
For our example, with n = 30 and p = .72, the
normal distribution is an acceptable approximation
because:
np = 30(.72) = 21.6 > 5
n(1 - p) = 30(.28) = 8.4 > 5

48. Prospective Students Applying to St
Prospective Students Applying to St. Andrew’s College Desiring On-Campus Housing.
Sampling
Distribution of

49. Step 1: Calculate the z-value at the upper endpoint
of the interval.
z = ( )/.082 = .61
Step 2: Find the area under the curve to the left of
the upper endpoint.
P(z < .61) = .7291
Step 3: Calculate the z-value at the lower endpoint of
the interval.
z = ( )/.082 = - .61
Step 4: Find the area under the curve to the left of the
lower endpoint.
P(z < -.61) = .2709

50. Step 5: Calculate the area under the curve between
the lower and upper endpoints of the interval.
P(-.61 < z < .61) = P(z < .61) - P(z < -.61) = =
The probability that the sample proportion of applicants
wanting on-campus housing will be within +/-.05 of the
actual population proportion :
P(.67 < < .77) = .4582

51. Sampling Distribution of
Example: St. Andrew’s College
Sampling
Distribution of
Area = .4582
.67
.72
.77

52. Other Types of Sampling
Stratified Random Sampling
The population is first divided into groups of
elements called strata.
Each element in the population belongs to one and
only one stratum.
Best results are obtained when the elements within
each stratum are as much alike as possible
(i.e. a homogeneous group).

53. Stratified Random Sampling
A simple random sample is taken from each stratum.
Formulas are available for combining the stratum
sample results into one population parameter
estimate.
Advantage: If strata are homogeneous, this method
is as “precise” as simple random sampling but with
a smaller total sample size.
Example: The basis for forming the strata might be
department, location, age, industry type, and so on.

54. Cluster Sampling The population is first divided into separate groups
of elements called clusters.
Ideally, each cluster is a representative small-scale
version of the population (i.e. heterogeneous group).
A simple random sample of the clusters is then taken.
All elements within each sampled (chosen) cluster
form the sample.

55. Cluster Sampling Example: A primary application is area sampling,
where clusters are city blocks or other well-defined
areas.
Advantage: The close proximity of elements can be
cost effective (i.e. many sample observations can be
obtained in a short time).
Disadvantage: This method generally requires a
larger total sample size than simple or stratified
random sampling.

56. Systematic Sampling
If a sample size of n is desired from a population containing N elements,
we might sample one element for every n/N elements in the population.
This method has the properties of a simple random sample,
especially if the list of the population elements is a random ordering.
We randomly select one of the first n/N elements from the population
list.
We then select every n/Nth element that follows in the population list.
Advantage: The sample usually will be easier to identify than it
would be if simple random sampling were used.
Example: Selecting every 100th listing in a telephone book after the
first randomly selected listing

57. Convenience Sampling
It is a nonprobability sampling technique. Items are
included in the sample without known probabilities
of being selected.
The sample is identified primarily by convenience.
Example: A professor conducting research might use
student volunteers to constitute a sample.
Advantage: Sample selection and data collection are
relatively easy.
Disadvantage: It is impossible to determine how
representative of the population the sample is.

58. Judgment Sampling The person most knowledgeable on the subject of the
study selects elements of the population that he or
she feels are most representative of the population.
It is a nonprobability sampling technique.
Example: A reporter might sample three or four senators,
judging them as reflecting the general opinion of the senate.
Advantage: It is a relatively easy way of selecting a sample.
Disadvantage: The quality of the sample results depends
on the judgment of the person selecting the sample.

59. Recommendation It is recommended that probability sampling methods
(simple random, stratified, cluster, or systematic) be
used.
For these methods, formulas are available for
evaluating the “goodness” of the sample results in
terms of the closeness of the results to the population
parameters being estimated.
An evaluation of the goodness cannot be made with
non-probability (convenience or judgment) sampling
methods.