1. Ka-fu Wong © 2003 Chap 15- 1 Dr. Ka-fu Wong ECON1003 Analysis of Economic Data

2. Ka-fu Wong © 2003 Chap 15- 2 l GOALS 1.List the characteristics of the Chi-square distribution. 2.Conduct a test of hypothesis comparing an observed set of frequencies to an expected set of frequencies. 3.Conduct a test of hypothesis for normality using the chi-square distribution. 4.Conduct a hypothesis test to determine whether two classification criteria are related. Chapter Fifteen Nonparametric Methods: Chi-Square Applications

3. Ka-fu Wong © 2003 Chap 15- 3 Characteristics of the Chi-Square Distribution The major characteristics of the chi-square distribution are: It is positively skewed. It is non-negative. It is based on degrees of freedom. When the degrees of freedom change a new distribution is created. Chi-square distribution is characterized by only one degree of freedom. F distribution is characterized by two degree of freedom. Similar to F distri- bution

4. Ka-fu Wong © 2003 Chap 15- 4 df = 3 df = 5 df = 10 

5. Ka-fu Wong © 2003 Chap 15- 5 Goodness-of-Fit Test: Equal Expected Frequencies Let f 0 and f e be the observed and expected frequencies respectively. H 0 : There is no difference between the observed and expected frequencies. H 1 : There is a difference between the observed and the expected frequencies.

6. Ka-fu Wong © 2003 Chap 15- 6 Goodness-of-fit Test: Equal Expected Frequencies The test statistic is: The critical value is a chi-square value with (k-1) degrees of freedom, where k is the number of categories

7. Ka-fu Wong © 2003 Chap 15- 7 EXAMPLE 1 The following information shows the number of employees absent by day of the week at a large a manufacturing plant. At the.05 level of significance, is there a difference in the absence rate by day of the week? DayFrequency Monday120 Tuesday45 Wednesday60 Thursday90 Friday130 Total445

8. Ka-fu Wong © 2003 Chap 15- 8 EXAMPLE 1 continued Assume equal expected frequency: (120+45+60+90+130)/5=89. The degrees of freedom is (5-1)=4. The critical value is 9.488. Use Appendix I in the textbook.

9. Ka-fu Wong © 2003 Chap 15- 9 Example 1 continued DayFrequencyExpected(f 0 -f e ) 2 /f e Monday1208910.80 Tuesday458921.75 Wednesday60899.45 Thursday90890.01 Friday1308918.89 Total4458960.90 Because the computed value of chi-square is greater than the critical value (9.488), H 0 is rejected. We conclude that there is a difference in the number of workers absent by day of the week.

10. Ka-fu Wong © 2003 Chap 15- 10 EXAMPLE 2 The U.S. Bureau of the Census indicated that 63.9% of the population is married, 7.7% widowed, 6.9% divorced (and not re-married), and 21.5% single (never been married). A sample of 500 adults from the Philadelphia area showed that 310 were married, 40 widowed, 30 divorced, and 120 single. At the.05 significance level can we conclude that the Philadelphia area is different from the U.S. as a whole?

11. Ka-fu Wong © 2003 Chap 15- 11 EXAMPLE 2 continued

12. Ka-fu Wong © 2003 Chap 15- 12 EXAMPLE 2 continued Step 1: H 0 : The distribution has not changed H 1 : The distribution has changed. Step 2: H 0 is rejected if  2 >7.815, df=3,  =.05 Step 3:  2 = 2.3814 Step 4: The null hypothesis is rejected. The distribution regarding marital status in Philadelphia is different from the rest of the United States.

13. Ka-fu Wong © 2003 Chap 15- 13 Goodness-of-Fit Test for Normality This test investigates if the observed frequencies in a frequency distribution match the theoretical normal distribution. The procedure is to determine the mean and standard deviation of the frequency distribution. Compute the z-value for the lower class limit and the upper class limit for each class. Determine f e for each category Use the chi-square goodness-of-fit test to determine if f o coincides with f e.

14. Ka-fu Wong © 2003 Chap 15- 14 EXAMPLE 3 A sample of 500 donations to the Arthritis Foundation is reported in the following frequency distribution. Is it reasonable to conclude that the distribution is normally distributed with a mean of $10 and a standard deviation of $2? Use the.05 significance level. Amount spentF0F0 <$620 $6 up to $860 $8 up to $10140 $10 up to $12120 $12 up to $1490 > $1470 Total500

15. Ka-fu Wong © 2003 Chap 15- 15 Example 3 continued To compute f e for the first class, first determine the z-value. Find the probability of a z-value less than – 2.00 The expected frequency is the probability of a z-value less that –2.00 times the samples size. f e = (.0228)(500) = 11.4 The other expected frequencies are computed similarly.

16. Ka-fu Wong © 2003 Chap 15- 16 EXAMPLE 3 continued Amount spentF0Areafefe (f 0 -f e ) 2 /f e <$620.0211.406.49 $6 up to $860.1467.95.93 $8 up to $10140.34170.655.50 $10 up to $12120.34170.6515.03 $12 up to $1490.1467.957.16 > $1470.0211.40301.22 Total500 336.33

17. Ka-fu Wong © 2003 Chap 15- 17 EXAMPLE 3 continued Step 1: H 0 : The observations follow the normal distribution. H 1 : The observations do not follow a normal distribution. Step 2: H 0 is rejected if  2 is greater than 7.815. There are 6 degrees of freedom and  is.05. Step 3: The computed value of  2 is 336.33. Step 4: H 0 is rejected. The observations do not follow the normal distribution. Amount spentF0Areafefe (f 0 -f e ) 2 /f e <$620.0211.406.49 $6 up to $860.1467.95.93 $8 up to $10140.34170.655.50 $10 up to $12120.34170.6515.03 $12 up to $1490.1467.957.16 > $1470.0211.40301.22 Total500 336.33

18. Ka-fu Wong © 2003 Chap 15- 18 Contingency Table Analysis A contingency table is used to investigate whether two traits or characteristics are related. Each observation is classified according to two criteria. We use the usual hypothesis testing procedure. The degrees of freedom is equal to: (number of rows- 1)(number of columns-1). The expected frequency is computed as: Expected Frequency = (row total)(column total)/grand total

19. Ka-fu Wong © 2003 Chap 15- 19 EXAMPLE 4 Is there a relationship between the location of an accident and the gender of the person involved in the accident? A sample of 150 accidents reported to the police were classified by type and gender. At the.05 level of significance, can we conclude that gender and the location of the accident are related?

20. Ka-fu Wong © 2003 Chap 15- 20 EXAMPLE 4 continued The expected relative frequency for work is 80/150. The expected relative frequency for male is 90/150. The expected relative frequency for the work-male intersection under the hypothesis that there is no relationship between work and male is (90/150)(80/150). The expected relative frequency for the work-male intersection under the hypothesis that there is no relationship between work and male is (90/150)(80/150)*150 = 48. Similarly, we can compute the expected frequencies for the other cells.

21. Ka-fu Wong © 2003 Chap 15- 21 EXAMPLE 4 continued Step 1: H 0 : Gender and location are not related. H 1 : Gender and location are related. Step 2: H 0 is rejected if the computed value of  2 is greater than 5.991. There are (3- 1)(2-1) = 2 degrees of freedom. Step 3: Find the value of  2.  2 =(60-48) 2 /48 + … + (10-8) 2 /8 = 16.667 Step 4: H 0 is rejected. Gender and location are related. Expected frequency in parentheses

22. Ka-fu Wong © 2003 Chap 15- 22 - END - Chapter Fifteen Nonparametric Methods: Chi-Square Applications