1. CLASS 6 CLASS 7 Tutorial 2 (EXCEL version)
URBP 204 A Class 6
CLASS 6
Tutorial 2 (EXCEL version)
Term Project Introduction – survey area; survey data file; variable definition file; instruction sheets (for the term project, and term project analysis report)
Give Back Exercise Set 1
CLASS 7
Remaining Tutorial 2; SPSS file for FWBT 2008 Survey
Factorial ANOVA
Chi Square test
One sample/way
Two factor/way
Correlation Coefficient
Introduce Ex Set 2
Tutorial 3
Note: the class notes summarize Salkind (2004) Chapters 12, 13 and 15

2. Factorial ANOVA Source: Salkind, pg. 215
Note: the class notes summarize Salkind (2004) Chapters 12, 13 and 15

3. Education level – low education and high education
Factorial ANOVA
More than 2 factors (independent variables)
Education level – low education and high education
Gender – male and female
4 groups
Low educated male
Low educated female
High educated female
High educated male
Dependent variable - minutes biked per day (ratio level variable / continuous variable)
Test effect of main independent variables + the interaction term
Note: the class notes summarize Salkind (2004) Chapters 12, 13 and 15

4. Steps for testing 1. Statement of null and research hypothesis
H0: Цloweduc = Цhigheduc
H0: Цmale = Цfemale
H0: Цloweduc X male = Цhigheduc x male = Цhigheduc x female = Цloweduc x female
H1: Xloweduc = Xhigheduc
H1: Xmale = Xfemale
H1: X loweduc X male = X higheduc x male = X higheduc x female = X loweduc x female
2. Set level of risk
Level of risk of type I error = 5%, or level of significance = 0.05
3. Selection of appropriate test statistic
Choose Factorial ANOVA
Note: the class notes summarize Salkind (2004) Chapters 12, 13 and 15

5. 4. Compute the F statistic (Obtained Value)
Note: the class notes summarize Salkind (2004) Chapters 12, 13 and 15

6. 5. Determination of the value needed for rejection of null hypothesis
(critical value)
See table B3, pg
Critical value = 3.99 (see Salkind, p.363)
6. Comparison of obtained and critical value
Obtained value more extreme than the critical value
(F= , , and 11.24, are all greater than 3.99)
7. Decision
Reject all the three null hypotheses (null hypotheses - there is NO
difference between the mean of groups)
Probability is less than 5% on any one test of the null hypothesis that the
difference between the mean of groups is due to chance alone.
Df for numerator:
Number of factors-1 = 2-1 =1
Df for denominator:
No of observations- no if groups
= 68-4 = 64
Note: the class notes summarize Salkind (2004) Chapters 12, 13 and 15

7. Correlation How the value of one variable
changes if the value of the other variable changes.
For example, correlation between:
Distance from city center and
housing price
Both variables need to be ratio or interval level.
Note: the class notes summarize Salkind (2004) Chapters 12, 13 and 15

8. Correlation Coefficient
Q: How do we know of the correlation is statistically significant?
A: Test for significance of the correlation coefficient
Source: Salkind, p 230
Note: the class notes summarize Salkind (2004) Chapters 12, 13 and 15

9. Steps for testing 1. Statement of null and research hypothesis
H0: ρxy = 0
No relationship between variables x and y
H1: rxy = 0
There is relationship between variables.
2. Set level of risk
Level of risk of type I error = 5%, or
level of significance = 0.05
3. Selection of appropriate test statistic
Choose t-test for the significance of the
correlation coefficient
Note: the class notes summarize Salkind (2004) Chapters 12, 13 and 15

10. Coefficient of correlation
4. Obtained value
For relationship between
Density and distance , r =
Density and price, r =
Distance and price, r =
Note: the class notes summarize Salkind (2004) Chapters 12, 13 and 15

11. 5. Determination of the value needed for rejection of null hypothesis
(critical value)
See table B4, pg. 365
Critical value = (see Salkind, p.363)
6. Comparison of obtained and critical value
Obtained value more extreme than the critical value for following
relationships:
Density and distance , r =
Density and price, r = 0.81
Distance and price, r =
7. Decision
Reject the null hypothesis (null hypothesis - there is NO relationship
between the variables). The relationship is not due to chance alone.
Degree of freedom
= n-2
= 30-2
= 28
Note: the class notes summarize Salkind (2004) Chapters 12, 13 and 15

12. Non parametric tests
When assumption of normality does not hold (small sample size – less than 30 observations)
Need ordinal or nominal level data.
Note: the class notes summarize Salkind (2004) Chapters 12, 13 and 15

13. One factor/sample chi square
Is the distribution of frequencies what you would expect by chance alone?
1. Statement of null and research hypothesis
Proportion of occurrence under each category is equal.
H1:
Proportion of occurrence under each category is not equal.
2. Set level of risk
Level of risk of type I error = 5%, or
level of significance = 0.05
3. Selection of appropriate test statistic
Choose chi square test
H0: P1 = P2 = P3 = P4 = P5
P1 = P2 = P3 = P4 = P5

14. 4. Obtained value Note: Should have minimum
of 5 observations under each
category
Note: the class notes summarize Salkind (2004) Chapters 12, 13 and 15

15. 5. Determination of the value needed for rejection of null hypothesis
(critical value)
See table B5, pg. 367
Critical value = 9.49 (see Salkind, p.363)
6. Comparison of obtained and critical value
Obtained value more extreme than the critical value
7. Decision
Reject the null hypothesis (null hypothesis - that the distribution of
frequencies is equal).
Degrees of freedom
= number of categories of data – 1
= 5-1
= 4

16. Two factor/way chi square
Explore relationships when both the dependent and the independent variables are nominal or ordinal level, that is “Categorical data”
Does the respondents’ age affect their perception about the condition of street lighting?
1. Statement of null and research hypothesis
Proportion of occurrence under each category is equal.
H1:
Proportion of occurrence under each
category is not equal.
2. Set level of risk
Level of risk of type I error = 5%, or
level of significance = 0.05
3. Selection of appropriate test statistic
Choose chi square test
H0: P1 = P2 = P3 = P4
P1 = P2 = P3 = P4

17. 4. Obtained value Expected Frequency Table Observed Frequency Table
Note: Should have minimum of 5 observations under each category
Expected Frequency Table
Observed Frequency Table
9.75 = (26 x 15) / 40
16.25 = (26 x 25) / 40
5.25 = (14 x 15) / 40
8.75 = (14 x 25) / 40
λ2 = (6-9.75)2 / ( )2 /16.25
+ (5.25-9)2 / (8.75-5)2 /8.75 =

18. URBP 204 A Class 6
5. Determination of the value needed for rejection of null hypothesis
(critical value)
See table B5, pg. 367
Critical value = 3.84 (see Salkind, p.363)
6. Comparison of obtained and critical value
Obtained value more extreme than the critical value
7. Decision
Reject the null hypothesis (null hypothesis - that the distribution of
frequencies is equal).
Degrees of freedom
= (row-1) x (column -1)
= (2-1) x (2-1)
= 1 x 1
= 1