1. 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.

2. Forms of “statistical” relationship Correlation Contingency Cause-and-effect * Causal * Feedback * Multi-directional * Recursive The last two categories are normally dealt with through regression

3. Statistical Data Analysis Methods – A Summary Scale of measurement One-sampleTwo independent Sample K independent Sample Measures of Association Independent Sample Single treatment repeat Measures Multiple treatment repeat Measures NominalBinomial test; one-way contingency Table McNemar test Cochrane Q Test Two-way contingency Table Contingency Table Contingency Coefficients OrdinalRuns testWilcoxon signed rank test Friedman test Mann- Whitney Test Kruskal- Wallis Test Spearman rank Correlation Interval/ratioZ- or t-test of variance Paired t-testRepeat measures ANOVA Unpaired t-test; tests of variance ANOVARegression, Pearson correlation, time series

4. 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 BeforeAfter Project LProject K A = 40B = 60 Project LC = 30D = 50 40 people switch from K to L and 50 people switch from L to K before and after

5. One-Sample Test (contd.) Test statistics: r c Q =  (0 ij – E ij ) 2 /E ij i=1 j=1 where E = (A+D)/2 Therefore, r c Q =  (0 ij – E ij ) 2 /E ij 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 H o not rejected. No influence of advertising on choice of project

6. One-Sample Test (contd.) 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 Man231 Min123 Lee132 Ling312 Dass123 Total811

7. One-Sample Test (contd.) Test statistics: (n-1)k k F r = ----------  R j 2 – 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, F r follows X 2 (k-1);α

8. One-Sample Test (contd.) (5-1)3 F = ------------ [8 2 + 11 2 + 11 2 ] – 3x5(3+1) 5x3(3+1) = 1.2 X 2 (3-1); 0.05 = 5.99 H o not rejected. Buyers do not show different preference for condo type

9. One-Sample Test (contd.) 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)

10. One-Sample Test (contd.) Waiting time at junction (min.)Row meanSum Sq. about row mean(Wi) MorningNoonEvening Junction 14.005.006.005.00 2.00 Junction 25.006.00 5.67 0.67 Junction 36.007.008.007.00 2.00 Junction 45.008.006.006.33 4.67 Junction 55. 004.009.006.00 14.00 Column mean 5.006.007.00M = 6.00W = 23.34

11. One-sample test (contd.) m n T =   (c ij – M) 2 i=1 j=1 = 30 W i =  (c ij – ) 2 = 23.34 B = m  ( - M) 2 = 6.65 t = n  ( - M) 2 = 10 W = t + r r = W – t = 23.34 – 10 = 13.34

12. One-Sample Test (contd.) 10/(3-1) F c = ---------------------- = 2.99 13.34/(5-1)(3-2) F t (3-1),(3-1)(5-1); 0.05 = 4.46 H o not rejected. Congestion is quite the same at all times during the day.

13. 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: GroupTotal (R) Inner suburbs Outer suburbs Terraced5075125 Semi- detached 302555 Total (C)80100180 r c Q =  (0 ij – E ij ) 2 /E ij i=1 j=1

14. Two-Sample Test (contd.) D.o.f. = (r-1)(c-1), where r=number of rows, c=number of columns E ij = R i C j /N Inner suburbsOuter suburbs Terraced125 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 H o not rejected

15. 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? data Formula:

16. K Independent Test - Correlation and regression – matrix approach

17. Correlation and regression – matrix approach

21. 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 ‘000130137128390140241342143 SQ. M OF FLOOR135140100360175270200170 PRICE - RM ‘000 (x)3637383940414243 NO. OF LOCALITIES (f)314103673272017

22. 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, 6.5 B: 3.3, 3.4, 4.2, 5.5, 5.8, 6.8 Decide which investment you would choose.

23. Test yourselves! Q5: Find:  (AGE > “30-34”)  (AGE ≤ 20-24)  ( “35-39”≤ AGE < “50-54”)

24. 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.

25. Perception about Influence of New Neighbourhood Degree of perception Locality Total BblautPatau1Patau2Racha2 Not worried at all 173024980 Not so worried 6021422 Worried 603413 Quite worried 10023 So Worried 00112 Total 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.

26. 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.