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15 - 1 Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 2 When you have completed this chapter, you will be able to: 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 Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 3 Conduct a test of hypothesis for normality using the chi-square distribution Conduct a hypothesis test to determine whether two attributes are independent Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. Characteristics of the Chi-Square Distribution 15 - 4 … 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. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. Characteristics of the Chi-Square Distribution 15 - 5 df = 3 df = 5 df = 10 c2 Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. Goodness-of-Fit Test: Equal Expected Frequencies 15 - 6 Let f0 and fe be the observed and expected frequencies respectively H0: There is no difference between the observed and expected frequencies H1: There is a difference between the observed and the expected frequencies Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. Goodness-of-Fit Test: Equal Expected Frequencies 2 = (fo - fe fe )2 …the critical value is a chi-square value with (k-1) degrees of freedom, where k is the number of categories Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. … the test statistic is: c 15 - 7 Goodness-of-Fit Test: Equal Expected Frequencies 15 - 8 The following information shows the number of employees absent by day of the week at a large a manufacturing plant. Day Frequency Monday 120 Tuesday 45 Wednesday 60 Thursday 90 Friday 130 Total 445 At the .05 level of significance, is there a difference in the absence rate by day of the week? Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. Goodness-of-Fit Test: Equal Expected Frequencies 15 - 9 Hypothesis Test Step 1 H0: There is no difference in absence rate by day of the week… (120+45+60+90+130)/5 = 89 H1: Absence rates by day are not all equal Step 2 = 0.05 Step 3 Step 4 Use Chi-Square test Degrees of freedom (5-1) = 4 Reject H0 if c2 > 9.488. (see Appendix I) Chi-Square Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 10 Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. Reject H0 if c2 > 9.488 Using the Table… Degrees of Freedom 5–1=4 Right-Tail Area = 0.05 Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 11 Test Statistics c Day Step 5 2 = ( f o - f e )2 fe 15 - 12 Frequency Expected (fo – fe)2/fe Monday Tuesday Wednesday Thursday Friday Total 120 45 60 90 130 445 89 89 89 89 89 445 10.80 21.75 9.45 = 1.98 0.01 18.89 60.90 (120-89)2/89 c2 Reject H0 if c2 > 9.488 Reject the null hypothesis. Absentee rates are not the same for each day of the week. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 13 A U.S. Bureau of the Census indicated that… Married Widowed Divorced Single 63.9% 7.7% 6.9% 21.5% Not re-married Never married A sample of 500 adults from the Philadelphia area showed: 310 40 30 120 At the .05 significance level, can we conclude that the Philadelphia area is different from the U.S. as a whole? Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 14 … continued Expected (fo – fe)2/fe Married 310 *319.5 ** .2825 Widowed 40 38.5 .0584 Divorced 30 34.5 .5870 Single 120 107.5 1.4535 Total 500 2.3814 * Census figures would predict: i.e. 639*500 = 319.5 ** Our sample: (310-319.5)2/319.5 = .2825 Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. c2 15 - 15 … continued Step 1 H0: The distribution has not changed H1: The distribution has changed. Step 2 = 0.05 H0 is rejected if c2 >7.815, df = 3 Step 3 Step 4 c2 = 2.3814 Reject the null hypothesis. The distribution regarding marital status in Philadelphia is different from the rest of the United States. Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. Goodness-of-Fit Test: Normality 15 - 16 … the test investigates if the observed frequencies in a frequency distribution match the theoretical normal distribution …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 fe for each category - Use the chi-square goodness-of-fit test to determine if fo coincides with fe Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. Goodness-of-Fit Test: Normality 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 Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 18 … continued Amount Spent fo <$6 20 $6 up to $8 60 $8 up to $10 140 $10 up to $12 120 $12 up to $14 90 >$14 70 Total 500 Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. Area fe (fo- fe )2/fe 15 - 19 … continued z To compute fe for the first class, first determine the z - value X - m 6 - 10 = = = - 2 . 00 s 2 Now… find the probability of a z - value less than –2.00 P( z < -2.00) = 0.5000 - .4772 = .0228 Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 20 … continued Amount Spent fo Area <$6 20 .02 $6 up to $8 60 .14 $8 up to $10 140 .34 $10 up to $12 120 .34 $12 up to $14 90 .14 >$14 70 .02 Total 500 Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. fe (fo- fe )2/fe 15 - 21 … continued The expected frequency is the probability of a z-value less than –2.00 times the sample size f e = (. 0228 )( 500 ) = 11 . 40 The other expected frequencies are computed similarly Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 22 … continued Amount Spent fo Area fe (fo- fe )2/fe <$6 20 .02 11.40 6.49 $6 up to $8 60 .14 67.95 .93 $8 up to $10 140 .34 170.65 5.50 $10 up to $12 120 .34 170.65 15.03 $12 up to $14 90 .14 67.95 7.16 >$14 70 .02 11.40 301.22 Total 500 500 336.33 Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 23 … continued Step 1 Step 2 H0: The observations follow the normal distribution H0: The observations do NOT follow the normal distribution = 0.05 Step 3 H0 is rejected if c2 >7.815, df = 6 Step 4 c2 = 336.33 H0: is rejected. The observations do NOT follow the normal distribution 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 … 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) … the expected frequency is computed as: Expected Frequency = (row total)(column total)/grand total 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? Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 26 … continued Sex Location Total Work Home Other Male 60 20 10 90 Female 20 30 10 60 Total 80 50 20 150 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 Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 27 … continued c Step 1 H0: The Gender and Location are NOT related H0: The Gender and Location are related Step 2 = 0.05 Step 3 H0 is rejected if c 2 >5.991, df = 2 (…there are (3- 1)(2-1) = 2 degrees of freedom) Step 4 Find the value of c 2 2 (60 2 ( - 48 )2 10 - 8 ) = + ... + 48 8 = 16 . 667 H0: is rejected. Gender and Location are related! Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. Test your learning… www.mcgrawhill.ca/college/lind Online Learning Centre for quizzes extra content data sets searchable glossary access to Statistics Canada’s E-Stat data …and much more! Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved. 15 - 28 15 - 29 This completes Chapter 15 Copyright © 2004 McGraw-Hill Ryerson Limited. All rights reserved.