1. Principles of Statistics Assoc. Prof. Dr. Abdul Hamid b. Hj. Mar Iman Former Director, Centre for Real Estate Studies Faculty of Geoinformation Science and Engineering, Universiti Teknologi Malaysia, Skudai, Johor. E-mail: hamid@fksg.utm.my

2. Hypothesis Testing Content: Concepts of hypothesis testing Test of statistical significance Hypothesis testing one variable at a time

3. Hypothesis Unproven proposition Supposition that tentatively explains certain facts or phenomena Assumption about nature of the world E.g. the mean price of a three-bedroom single storey houses in Skudai is RM 155,000.

4. Hypothesis (contd.) An unproven proposition or supposition that tentatively explains certain facts or phenomena: –Null hypothesis –Alternative hypothesis Null hypothesis is that there is no systematic relationship between independent variables (IVs) and dependent variables (DVs). Research hypothesis is that any relationship observed in the data is real.

5. Null Hypothesis Statement about the status quo No difference Statistically expressed as: H o : b=0 where b is any sample parameter used to explain the population.

6. Alternative Hypothesis Statement that indicates the opposite of the null hypothesis There is difference Statistically expressed as: H 1 : b  0 H 1 : b < 0 H 1 : b > 0

7. Significance Level Critical probability in choosing between the H o and H 1. Simply means, the cut-off point (COP) at which a given value is probably true. Tells how likely a result is due to chance Most common level, used to mean “something is good enough to be believed”, is.95. It means, the finding has a 95% chance of being likely true. What is the COP at 95% chance?

8. Significance Level (contd.) Denoted as  Tells how much the probability mass is in the tails of a given distribution Probability or significance level selected is typically.05 or.01 Too low to warrant support for the null hypothesis In other words, high chances to warrant support for alternative hypothesis Main purpose of statistical testing: to reject null hypothesis

9. Significance Level (contd.) P[-1.96  Z  1.96] = 1 -  = 0.95 P[Z  Z c ] = P[Z  -Z c ] =  /2

10. Let say we have the following relationship: Y = β + e i i =1,…, T and e i ~ N(0,σ 2 ) ……………....(1) The least square estimator for β is: T b=  Y i /T ……………………………………………..(2) i=1 with the following properties: 1) E[b] = β ………………………………………….(3a) 2) Var(b)=E[(b-β)] 2 = σ 2 /T ………………………...(3b) 3) b~N(β, σ 2 /T) …………………………………….(3c)

11. The “standardized” normal random variable for β is: b-β Z =-------- ~ N(0,1) ……………………………………..(4)  (σ 2 T) The critical value of Z, i.e. Z c, such that α=0.05 of the probability mass is in the tails of distribution, is given as: P[Z  1.96] = P[Z  -1.96]=0.025 ………………………(5a) and P[-1.96  Z  1.96]=1-0.05=0.95 ………………………(5b)

12. Substituting SND for variable β (Eqn. 4) into Eqn (5a), we get: b-β P[-1.96  ---------  1.96]=0.95 ……………………………..…...(6)  (σ 2 /T) Solving for β, we get: P[b-1.96σ/  T  β  b+1.96σ/  T]=0.95 ………………………… (7) In general: P[b-Z c σ/  T  β  b+Z c σ/  T]= 1-  ……………….. (8a) b-β b -β Also: P[-------  -Z c ] = P[ --------  Z c ] = α/2 (2-tail test)...…(8b) σ/  T σ/  T

13. The null hypothesis that the mean is equal to 3.0: Example You suspect that the mean rental of 225 purpose- built office units in Johor is RM 3.00/sq.ft. If the std. dev. is RM 1.50/sq.ft., what is the 95% confidence interval of the mean? The alternative hypothesis that the mean does not equal to 3.0: H o : μ = 3.0 H 1 : μ  3.0

14.   A Sampling Distribution -X L = ?X U = ?

15. Critical values of  Critical value - upper limit

16. Critical values of 

17. Critical value - lower limit Critical values of 

19. Region of Rejection LOWER LIMIT UPPER LIMIT 

20. Hypothesis Test  2.804 3.196  3.78

21. Accept nullReject null Null is true Null is false Correct- no error Type I error Type II errorCorrect- no error Type I and Type II Errors

22. in Hypothesis Testing State of Null Hypothesis Decision in the PopulationAccept HoReject Ho Ho is trueCorrect--no errorType I error Ho is falseType II errorCorrect--no error

23. Example You estimate that the average price, μ, of single- and double-storey houses in Malaysia’s major industrialised towns to be RM 1,600/sq.m. Based on a sample of 101 houses, you found that the mean price,, is 1,579.44/sq.m. with a std dev. of RM 350.13/sq.m. (a)Would you reject your initial estimate at 0.05 significance level? (b)What is the confidence interval of rental at 5% s.l.?

24. Answer (a) H o = 1,600 H 1  1,600 1,579.44 – 1,600 Test statistic: Z = -------------------- 350.13/  101 ≈ -0.59 P[Z  Z c ] = P[Z  -Z c ] = 0.05 P[0.59  Z c ] = 0.05 From Z-table, Z c = 1.645 Since Z < Z c,do not reject H o. ∴ Rental = RM 1,600/sq.m.

25. Answer (b) 1,579.13-1.645(34.84)=RM 1,521.82 (lower limit) 1,579.13+1.645(34.84)=RM 1,636.44 (upper limit)

26. PARAMETRIC STATISTICS NONPARAMETRIC STATISTICS

27. t-Distribution Symmetrical, bell-shaped distribution Mean of zero and a unit standard deviation Shape influenced by degrees of freedom

28. Degrees of Freedom Abbreviated d.f. Number of observations Number of constraints

29. or Confidence Interval Estimate Using the t-distribution

30. = population mean = sample mean = critical value of t at a specified confidence level = standard error of the mean = sample standard deviation = sample size Confidence Interval Estimate Using the t-distribution

34. Hypothesis Test Using the t-Distribution

35. Suppose that a production manager believes the average number of defective assemblies each day to be 20. The factory records the number of defective assemblies for each of the 25 days it was opened in a given month. The mean was calculated to be 22, and the standard deviation,,to be 5. Univariate Hypothesis Test Utilizing the t-Distribution

38. The researcher desired a 95 percent confidence, and the significance level becomes.05.The researcher must then find the upper and lower limits of the confidence interval to determine the region of rejection. Thus, the value of t is needed. For 24 degrees of freedom (n-1, 25-1), the t-value is 2.064. Univariate Hypothesis Test Utilizing the t-Distribution

41. Univariate Hypothesis Test t-Test

42. Testing a Hypothesis about a Distribution Chi-Square test Test for significance in the analysis of frequency distributions Compare observed frequencies with expected frequencies “Goodness of Fit”

43. Chi-Square Test

44. x² = chi-square statistics O i = observed frequency in the i th cell E i = expected frequency on the i th cell Chi-Square Test

45. Chi-Square Test Estimation for Expected Number for Each Cell

46. R i = total observed frequency in the i th row C j = total observed frequency in the j th column n = sample size

47. Univariate Hypothesis Test Chi-square Example

49. Hypothesis Test of a Proportion  is the population proportion p is the sample proportion  is estimated with p

50. 5. :H 5. :H 1 0   Hypothesis Test of a Proportion

53. 0115.S p  000133.S p  1200 16. S p  1200 )8)(.2(. S p  n pq S p  20.p  200,1n  Hypothesis Test of a Proportion: Another Example

54. 0115.S p  000133.S p  1200 16. S p  1200 )8)(.2(. S p  n pq S p  20.p  200,1n  Hypothesis Test of a Proportion: Another Example

55. Indeed.001 the beyond t significant is it level..05 the at rejected be should hypothesis null the so 1.96, exceeds value Z The 348.4Z 0115. 05. Z 0115. 15.20. Z S p Z p       Hypothesis Test of a Proportion: Another Example