1. Reading Property Data Analysis – A Primer, Ch.8
OMGT LECTURE 8: Deductive Statistics & Selected Sampling Distributions
Reading
Property Data Analysis – A Primer, Ch.8

2. Objectives
Revisit the relationship between deductive and inferential statistics.
Discuss the main results of the Central Limit Theorem in respect of the sampling distribution for the sample mean as well as the sample proportion.
Discuss the meaning and significance of the sampling distribution of the sample mean and sample proportion.
Discuss the meaning and significance of the standard errors of the sample mean and sample proportion
Discuss the student t-distribution
Use the attributes of sampling distributions to engage in deductive statistics

3. Relationship between Deductive and Inferential Statistics
A major focus of this lecture is Deductive Statistics
This branch of statistics is concerned with predicting the nature of randomly selected samples from a known population
As such Deductive Statistics helps one to understand the logic underpinning Inductive Statistics
This 2nd branch of statistics is concerned with predicting the nature of an unknown population from information obtained from a randomly selected sample

4. POPULATION
The large egg shaped set represents a the complete Population of gross turnover figures for retail properties in a city

5. The Central Limit Theorem
The Central Limit Theorem reveals something about the nature of the Sampling Distribution of the Mean The main result of this important theorem is that as the sample size n increases this distribution approaches a normal distribution as depicted below where:

6. The Central Limit Theorem Continued

7. The Central Limit Theorem Continued

8. The Central Limit Theorem Continued

9. The Central Limit Theorem Continued

10. The Central Limit Theorem Continued

16. The Student t Distribution
Assume X is ND and consider all samples of size n that may be drawn randomly from a population. Then the statistic.
which measures sampling error in estimated standard error units varies from sample to sample. In other words there is a distribution for the t-statistic in much the same way as there was a distribution for the sample mean. The following slide will help crystallize the argument.

17. The Student t Distribution Continued
Imagine that for each possible sample of size n, a t statistic is calculated. Since t varies from sample to sample there will be a distribution of t-values known as the Student t -Distribution.

18. The Student t Distribution Continued
From the previous slides we have learned that there is a distribution for the statistic:
If the population of X-values is normally distributed then the distribution of t-values will take the appearance of what is known as the Student t - Distribution with n - 1 degrees of freedom.

19. The Student t Distribution Continued
Properties of the Student t-Distribution:
The student t-distribution:
Is bell-shaped and symmetrical like the z-distribution
shares the same mean as the z-distribution because mt = 0 = mz
Is flatter and has a wider spread than the z-distribution because: st ≥ sz = 1

20. The Student t Distribution Continued
Properties of the Student t-Distribution Continued:
The student t-distribution:
is really a family of distributions each member of which is distinguished by its degrees of freedom df. For the present application, df = n – 1, that is the number of sample observations minus 1.
Is reasonably approximated by the z-distribution when df > 30
NOTE: The Greek letter n (or nu) for the letter n is used to denote degrees of freedom: df = n -1

21. The Student t Distribution Continued
General Structure of Student t Distribution Area Tables: For a Student t distribution with degrees of freedom df = n = n - 1, the Area Tables will indicate t* such that : Pr(-t* < t < t*) = 1 – a where :a denotes the combined area in the tails.
a/2 a t*
df = n
Degrees of Freedom
Combined Area in Tails
Area in One Tail
Stylised Representation of Student t Area Tables
a/2
a/2

22. The Student t Distribution Continued
Example of Use of Student t Distribution Area Tables: Suppose for a Student t Distribution with 12 (=n) degrees of freedom, one wishes to find (-t* t*) such that: Pr(-t* < t < t*) = .95 where : a = .05 denotes the combined area in the tails.
.025
.05
2.18 12
Degrees of Freedom
Combined Area in Tails
Area in One Tail
Stylised Representation of the Student t Area Tables t*
a/2
df = n a
At the intersection of the row labelled 12 and the column labelled a = 05, find that t* = Hence it follows that:
Pr(-2.18 < t n=12 < 2.18) = 1 – a = =.95

23. The Student t Distribution Continued
Worked Problem 6: Find the probability that average gross turnover for a random sample of n = 26 enterprises does not differ from the true population mean m by more than 2.06 estimated standard error units.
Stylised Representation of the Student t Area Tables
Pr(-2.06 < t n=25 < 2.06) = 1 – a = = .95
a : Combined Area in Tails
.05
df = n : Degrees of Freedom 25
2.06
.025
.95
.025
.025
a/2 : Area in One Tail
Note: n or df = n – 1 = = 25

24. The Sampling Distribution for the Sample Proportion p
Formulae for Proportions:
Sampling Distribution for the Sample Proportion p:
Consider all simple random samples of size n that could be drawn from a population
and that for each such sample, the sample proportion p is calculated. This procedure is illustrated on the next slide.

25. The Sampling Distribution for the Sample Proportion p Continued
The large egg shaped set represents the Population
of retail enterprises in a city, a proportion p of which have sustained losses
POPULATION
Consider all samples of size n that may be randomly selected from the population and for each of these samples, one records the proportion p of retail enterprises that have sustained losses. Clearly p will vary from sample to sample. When these p-values are organised into a distribution, one obtains what is called the Sampling Distribution for the Sample Proportion p

26. The Central Limit Theorem Revisited
As with the Sampling Distribution of the Mean , the Central Limit Theorem indicates something about the attributes of the Sampling Distribution for the Sample Proportion p.

27. The Central Limit Theorem Revisited Continued
If the above described sampling is undertaken with replacement or from an infinite population, then:

28. The Central Limit Theorem Revisited Continued
On the other hand, if sampling were undertaken without replacement or from a finite population for which n > .05N, Then:

29. The Central Limit Theorem Revisited Continued
Worked Problem 7:
Suppose that p = .36, n = 36 and sampling is undertaken with replacement. What does the sampling distribution look like? ANS: Since np > 5 and n(1 – p) > 5, the sampling distribution of p is normally distributed with mean and standard deviation given by:
Worked Problem 8:
Suppose that p = .36, n = 900 and N = 2500 and sampling is undertaken without replacement. What does the sampling distribution look like? ANS: Since n > .05N,
np > 5 and n(1 – p) > 5, the sampling distribution of p is normally distributed with mean and standard deviation now given by:

30. The Central Limit Theorem Revisited Continued
Worked Problem 9:
Suppose that p = .36, n = 64 and N = 2500 and sampling is undertaken without replacement. What does the sampling distribution now look like? ANS: Since np > 5 and n(1 – p) > 5 and n < .05N, the sampling distribution of p is normally distributed with mean and standard deviation given by:

32. Significance of the Sampling Distribution of the Sample Proportion p
Knowledge of the ND sampling distribution for p enables one to make meaningful probabilistic statements regarding the magnitude of sampling error p - p.
For instance, 90% of the sampling errors will have a magnitude:
Moreover, if the sampling distribution of p is ND with mean mp = p =.36 and sp =.08. then, 90% of the sample proportions fall in the range:
mp ± 1.65sp or
p ± 1.65sp or
.36 ± 1.65(.08) or
.228  .492
Pr(p sp < p < p sp) = .90
.228
.36
.492

33. Significance of the Sampling Distribution of the Sample Proportion p
Worked Problem 10:
Find Pr(.28 < p < .32) if p is ND with:
mp = p =.36 and sp = .08
Solution:
For this problem : z = (p – p)/sp is standard normally distributed. Hence:
p1 = .28  z1 = ( )/.08 = -1.0
p2 = .32  z2 = ( )/.08 = -.50
This means that:
Pr(p1 < p < p2 ) = Pr(.28 < p < .32)
is equivalent to:
Pr(z1 < z < z2) = Pr( -1.0 < z < -.50)
Because of symmetry:
Pr(-1.0 < z < -.50) = Pr(.50 < z < 1.0)
In turn, Pr (.50 < z < 1.0) equals:
Pr(0 < z < 1.0) – Pr(0 < z < .50)
Which (from area tables) equals:
= =  .15