1. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 1

2. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 2 Understand the nature and role of chi-square distribution Identify a wide variety of uses of the chi-square distribution Conduct a test of hypothesis comparing an observed frequency distribution to an expected frequency distribution When you have completed this chapter, you will be able to:

3. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 3 Conduct a hypothesis test to determine whether two attributes are independent Conduct a test of hypothesis for normality using the chi-square distribution

4. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 4 Characteristics of the Chi-Square Distribution … 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 …e.g.

5. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 5 df = 3 df = 5 df = 10  Characteristics of the Chi-Square Distribution

6. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 6 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

7. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 7 …t he critical value is a chi-square value with (k-1) degrees of freedom, where k is the number of categories Goodness-of-Fit Test: Equal Expected Frequencies         e eo f ff 2 2      … the test statistic is:

8. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 8 The following information shows the number of employees absent by day of the week at a large a manufacturing plant. Day Frequency Monday120 Tuesday 45 Wednesday 60 Thursday 90 Friday 130 Total445 Day Frequency Monday120 Tuesday 45 Wednesday 60 Thursday 90 Friday 130 Total445 Goodness-of-Fit Test: Equal Expected Frequencies At the.05 level of significance, is there a difference in the absence rate by day of the week?

9. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 9 Hypothesis Test Step 1 Step 2 Step 3 Step 4 H 0 : There is no difference in absence rate by day of the week… H 1 : Absence rates by day are not all equal  = 0.05 Use Chi-Square test Reject H 0 if  2 > (5-1) = 4 Degrees of freedom 9.488. (see Appendix I) (120+45+60+90+130)/5 = 89 Goodness-of-Fit Test: Equal Expected Frequencies Chi-Square

10. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 10

11. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 11 Right-Tail Area  = 0.05 Degrees of Freedom 5 – 1 = 4 Reject H 0 if  2 > 9.488 Using the Table…

12. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 12 = 1.98 Day Frequency Expected (f o – f e ) 2 /f e Monday120 89 10.80 Tuesday 45 89 21.75 Wednesday 60 89 9.45 Thursday 90 89 0.01 Friday 130 89 18.89 Total 445 445 60.90 Day Frequency Expected (f o – f e ) 2 /f e Monday120 89 10.80 Tuesday 45 89 21.75 Wednesday 60 89 9.45 Thursday 90 89 0.01 Friday 130 89 18.89 Total 445 445 60.90 22 Reject the null hypothesis. Absentee rates are not the same for each day of the week. (120-89) 2 /89 Step 5 Test Statistics Reject H 0 if  2 > 9.488

13. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 13 At the.05 significance level, can we conclude that the Philadelphia area is different from the U.S. as a whole? MarriedWidowedDivorcedSingle 63.9%7.7%6.9%21.5% A U.S. Bureau of the Census indicated that… Not re-married Never married A sample of 500 adults from the Philadelphia area showed: 3104030120

14. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 14 Married310 *319.5 **.2825 Widowed4038.5.0584 Divorced3034.5.5870 Single120107.51.4535 Total5002.3814 … continued 22 2.3814 * Census figures would predict: i.e. 639*500 = 319.5 ** Our sample: (310-319.5)2/319.5 =.2825 (fo – fe)2/fe Expected

15. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 15 Step 1 Step 2 Step 3 Step 4 H 0 : The distribution has not changed … continued H 1 : The distribution has changed. H 0 is rejected if  2 >7.815, df = 3  = 0.05  2 = 2.3814 Reject the null hypothesis. The distribution regarding marital status in Philadelphia is different from the rest of the United States.

16. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 16 Goodness-of-Fit Test: Normality … the test investigates if the observed frequencies in a frequency distribution match the theoretical normal 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 …to determine the mean and standard deviation of the frequency distribution

17. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 17  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 Goodness-of-Fit Test: Normality

18. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 18 Amount Spentfofo Area fe fe (f o - f e ) 2 /f e <$620 $6 up to $860 $8 up to $10140 $10 up to $12120 $12 up to $1490 >$1470 Total500 … continued

19. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 19 To compute f e for the first class, first determine the z - value … continued Now… find the probability of a z - value less than –2.00 00.2 2 106        X z.02284772.5000.0)00.2(  zP

20. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 20 Amount Spentfofo Area fe fe (f o - f e ) 2 /f e <$620.02 $6 up to $860.14 $8 up to $10140.34 $10 up to $12120.34 $12 up to $1490.14 >$1470.02 Total500 … continued

21. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 21 The expected frequency is the probability of a z -value less than –2.00 times the sample size … continued The other expected frequencies are computed similarly 40.11)500)(0228(.  e f

22. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 22 Amount Spentfofo Area fe fe (f o - 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 … continued

23. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 23 … continued Step 1 Step 2 Step 3 Step 4 H 0 : The observations follow the normal distribution H 0 is rejected if  2 >7.815, df = 6  = 0.05  2 = 336.33 H0: is rejected. The observations do NOT follow the normal distribution H 0 : The observations do NOT follow the normal distribution

24. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 24 A contingency table is used to investigate whether two traits or characteristics are related … the expected frequency is computed as: Expected Frequency = (row total)(column total)/grand total … each observation is classified according to two criteria …the usual hypothesis testing procedure is used … the degrees of freedom is equal to: (number of rows -1)(number of columns -1)

25. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 25 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?

26. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 26 … continued Sex WorkHomeOther Total Male60201090 Female20301060 Total805020150 Location The expected frequency for the work-male intersection is computed as (90)(80)/150 =48 Similarly, you can compute the expected frequencies for the other cells

27. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 27 Step 1 Step 2 Step 3 Step 4 H 0 : The Gender and Location are NOT related H 0 is rejected if  2 >5.991, df = 2  = 0.05 H 0 : is rejected. Gender and Location are related! H 0 : The Gender and Location are related (…there are (3- 1)(2-1) = 2 degrees of freedom) Find the value of  2  8 810... 48 60 22 2     … continued 667.16 

28. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 28 Test your learning … www.mcgrawhill.ca/college/lind Click on… Online Learning Centre for quizzes extra content data sets searchable glossary access to Statistics Canada’s E-Stat data …and much more!

29. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 29 This completes Chapter 15