some application of statistical methods in data analysis assoc. prof. dr. abdul hamid b. hj. mar...

Some Application of Statistical Methods in

Data Analysis

Assoc. Prof. Dr. Abdul Hamid b. Hj. Mar Iman,

Former Director,

Centre for Real Estate Studies,

Universiti Teknologi Malaysia.

Forms of “statistical” relationship

Correlation Contingency Cause-and-effect * Causal * Feedback * Multi-directional * Recursive The last two categories are normally dealt with

through regression

Statistical Data Analysis Methods – A Summary

Scale of measurement

One-sample Two independent Sample

K independent Sample

Measures of Association

Independent Sample

Single treatment repeat Measures

Multiple treatment repeat Measures

Nominal Binomial test; one-way contingency Table

McNemar test

Cochrane Q Test

Two-way contingency Table

Contingency Table

Contingency Coefficients

Ordinal Runs test Wilcoxon signed rank

test

Friedman test

Mann-Whitney Test

Kruskal-Wallis Test

Spearman rank Correlation

Interval/ratio Z- or t-test of variance

Paired t-test Repeat measures ANOVA

Unpairedt-test; tests of variance

ANOVA Regression, Pearson correlation, time series

One-Sample Test

McNemar Test: tests for change in a sample upon a “treatment”.

Example. Two condominium projects K&L. Respondents decide their preferences for K or L before and after “advertising”.

Hypothesis: Advertising does not influence buyers to change their mind on product choice

Before After

Project L Project K

Project K A = 40 B = 60

Project L C = 30 D = 50

40 people switch from K to L and 50 people switch from L to K before and after

One-Sample Test (contd.)

Test statistics:r c

Q = (0ij – Eij)2/Eij

i=1 j=1

where E = (A+D)/2

Therefore,

r c


i=1 j=1

[A-(A+D)/2]2 [D-(A+D)/2]2 (A-D)2

----------------- + ---------------- = -------

(A+D)/2 (A+D)/2 A+D

Thus, Q = (40-45)2/(40+45)

= 25/85

= 0.29

(2-1)(2-1); 0.05 = 3.84

Ho not rejected. No influence of advertising on choice of project


Friedman Test: tests for equal preferences for something of various characteristics.

Example. Buyers’ rank of preference for three condominium types A, B, C.

Hypothesis: Buyers’ preferences for all condo type do not differ

Resp. Type A

Type B

Type C

Man 2 3 1

Min 1 2 3

Lee 1 3 2

Ling 3 1 2

Dass 1 2 3

Total 8 11 11


Test statistics:

(n-1)k k

Fr = ---------- Rj2 – 3n(k+1)

nk(k+1) j=1

where n = sample size, k = number of categories; R = is column’s total

For large n and k, Fr follows X2(k-1);α


(5-1)3 F = ------------ [82 + 112 + 112] – 3x5(3+1) 5x3(3+1) = 1.2

X2(3-1); 0.05 = 5.99

Ho not rejected. Buyers do not show different preference for condo type


Repeated measures ANOVA: tests outcome of a phenomenon under different conditions.

Example. Waiting time at junctions in the city area to determine level of congestion at different times of the day.

Test statistics: t/(m-1) F = ---------------- r/[(n-1)(m-1)

where t = sum of squares due to treatment, r =sum of squares of residual, m = number of treatment, n = number of observations.

Critical region based on: F v1. v2; α

where v1 = (m-1), v2 = (n-1)(m-1)

One-Sample Test (contd.)Waiting time at junction (min.) Row mean Sum Sq.

about row mean(Wi)

Morning Noon Evening

Junction 1 4.00 5.00 6.00 5.00 2.00

Junction 2 5.00 6.00 6.00 5.67 0.67

Junction 3 6.00 7.00 8.00 7.00 2.00

Junction 4 5.00 8.00 6.00 6.33 4.67

Junction 5 5. 00 4.00 9.00 6.00 14.00

Column mean

5.00 6.00 7.00 M = 6.00 W = 23.34

One-sample test (contd.)

m n

T = (cij – M)2

i=1 j=1 = 30Wi = (cij – )2

= 23.34B = m( - M)2

= 6.65

t = n ( - M)2

= 10

W = t + r

r = W – t

= 23.34 – 10

= 13.34


10/(3-1) Fc = ---------------------- = 2.99 13.34/(5-1)(3-2)

Ft (3-1),(3-1)(5-1); 0.05 = 4.46

Ho not rejected. Congestion is quite the same at all times during the day.

Two-Sample Test

Two-way Contingency Table: test whether two independent groups differ on a given characteristic.

Hypothesis: choice for type of house does not relate to location.

Test:

Group Total(R)

Inner suburbs

Outer suburbs

Terraced 50 75 125

Semi-detached

30 25 55

Total (C) 80 100 180

r c


i=1 j=1

Two-Sample Test (contd.)

D.o.f. = (r-1)(c-1), where r=number of rows, c=number of columns

Eij = RiCj/N

Inner suburbs Outer suburbs

Terraced 125 x 80/180 = 55.6

125 x 100/180= 69.4

Semi-detached

55 x 80/180 = 24.4

55 x 100/180 = 30.6

Q = (50-55.6)2/55.6 + (30-24.4)2/24.4 + (75-69.4)2/69.4 + (25-30.6)2/30.6

= 3.33

(2-1)(2-1); 0.05 = 3.84

Ho not rejected

K Independent Test - Correlation “Co-exist”.E.g.

* left shoe & right shoe, sleep & lying down, food & drink Indicate “some” co-existence relationship. E.g.

* Linearly associated (-ve or +ve)

* Co-dependent, independent But, nothing to do with C-A-E r/ship!

Example: After a field survey, you have the following data on the distance to work and distance to the city of residents in J.B. area. Interpret the results?

Formula:

K Independent Test - Correlation and regression – matrix approach

Correlation and regression – matrix approach

Test yourselves!

Q1: Calculate the min and std. variance of the following data:

Q2: Calculate the mean price of the following low-cost houses, in various

localities across the country:

PRICE - RM ‘000 130 137 128 390 140 241 342 143

SQ. M OF FLOOR 135 140 100 360 175 270 200 170

PRICE - RM ‘000 (x) 36 37 38 39 40 41 42 43

NO. OF LOCALITIES (f) 3 14 10 36 73 27 20 17

Test yourselves!

Q3: From a sample information, a population of housing estate is believed have a “normal” distribution of X ~ (155, 45). What is the general adjustment to obtain a Standard Normal Distribution of this population?

Q4: Consider the following ROI for two types of investment:

A: 3.6, 4.6, 4.6, 5.2, 4.2, 6.5B: 3.3, 3.4, 4.2, 5.5, 5.8, 6.8

Decide which investment you would choose.

Test yourselves!

Q5: Find:

(AGE > “30-34”)

(AGE ≤ 20-24)

( “35-39”≤ AGE < “50-54”)

Test yourselves!

Q6: You are asked by a property marketing manager to ascertain whether

or not distance to work and distance to the city are “equally” important factors influencing people’s choice of house location.

You are given the following data for the purpose of testing:

Explore the data as follows:• Create histograms for both distances. Comment on the shape of the

histograms. What is you conclusion?• Construct scatter diagram of both distances. Comment on the output.• Explore the data and give some analysis.• Set a hypothesis that means of both distances are the same. Make

your conclusion.

Perception about Influence of New Neighbourhood

Degree of perception

Locality

TotalBblaut Patau1 Patau2 Racha2

Not worried at all 17 30 24 9 80

Not so worried 6 0 2 14 22

Worried 6 0 3 4 13

Quite worried 1 0 0 2 3

So Worried 0 0 1 1 2

Total 30 30 30 30 120

Q 7. You have surveyed a group of local people and asked them to express their feeling about a new project that will attract a new population and thus a new neighbourhood. You believe that the local people are concerned about the negative influence the new neighbourhood will have on them as a result of the proposed project. Using the collected data, test your hypohesis.

Test yourselves! (contd.)

Q7: From your initial investigation, you belief that tenants of “low-quality” housing choose to rent particular flat units just to find shelters. In this context ,these groups of people do not pay much attention to pertinent aspects of “quality life” such as accessibility, good surrounding, security, and physical facilities in the living areas.

(a) Set your research design and data analysis procedure to address the research issue(b) Test your hypothesis that low-income tenants do not perceive “quality life” to be important in paying their house rentals.

some application of statistical methods in data analysis assoc. prof. dr. abdul hamid b. hj. mar...

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