Chapter Fifteen McGraw-Hill/Irwin © 2005 The McGraw-Hill Companies, Inc., All Rights Reserved.

Chapter Fifteen Nonparametric Methods: Chi-Square Applications GOALS When you have completed this chapter, you will be able to: ONE List the characteristics of the Chi-square distribution. TWO Conduct a test of hypothesis comparing an observed set of frequencies to an expected set of frequencies. THREE Conduct a hypothesis test to determine whether two classification criteria are related. Goals

Characteristics of the Chi-Square Distribution Chi-Square Applications The major characteristics of the chi-square distribution are: It is positively skewed It is non-negative There is a family of chi-square distributions Characteristics of the Chi-Square Distribution

df = 3 df = 5 df = 10 c2 c2 distribution

Goodness-of-Fit Test: Equal Expected Frequencies Let f0 and fe be the observed and expected frequencies respectively. H1: There is a difference between the observed and the expected frequencies. H0: There is no difference between the observed and 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 Goodness-of-Fit Test: Equal Expected Frequencies

The following information shows the number of employees absent by day of the week at a large a manufacturing plant. At the .01 level of significance, is there a difference in the absence rate by day of the week? Day of Week Number Absent Monday 120 Tuesday 45 Wednesday 60 Thursday 90 Friday 130 Total 445 Example 1 continued

Step 1: State the null and alternate hypotheses H0: There is no difference between the observed and expected frequencies. H1: There is a difference between the observed and the expected frequencies Step 2: Select the level of significance. This is given in the problem as .01. Step 3: Select the test statistic. It is the chi-square distribution. Example 1 continued

Step 4: Formulate the decision rule. Assume equal expected frequency as given in the problem fe = (120+45+60+90+130)/5=89 The degrees of freedom: (5-1)=4 The critical value of c2 is 13.28. Reject the null and accept the alternate if Computed c2 > 13.28 or p< .01 EXAMPLE 1 continued

Step Five: Compute the value of chi-square and make a decision. Day Frequency Expected (fo – fe)2/fe Monday 120 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 The p(c2 > 60.9) = .000000000001877 or essentially 0. Example 1 continued

Because the computed value of chi-square, 60 Because the computed value of chi-square, 60.90, is greater than the critical value, 13.28, the p of .000000000001877 < .01, H0 is rejected. We conclude that there is a difference in the number of workers absent by day of the week. Example 1 continued

Goodness-of-fit Test: Unequal Expected Frequencies 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 .02 significance level can we conclude that the Philadelphia area is different from the U.S. as a whole? Example 2

Step 4: H0 is rejected if c2 >9.837, df=3, or if p < a of .02 Step 3: The test statistic is the chi-square. Step 2: The significance level given is .02. Step 1: H0: The distribution has not changed H1: The distribution has changed. Example 2 continued

Calculate the expected frequencies Calculate chi-square values. Married: (.639)500 = 319.5 Widowed: (.077)500 = 38.5 Divorced: (.069)500 = 34.5 Single: (.215)500 = 107.5 Calculate the expected frequencies Calculate chi-square values. Example 2 continued

Step 5: c2 = 2.3814, p(c2 > 2.3814) = .497. The null hypothesis is not rejected. The distribution regarding marital status in Philadelphia is not different from the rest of the United States. Example 2 continued

Contingency Table Analysis Chi-square can be used to test for a relationship between two nominal scaled variables, where one variable is independent of the other. A contingency table is used to investigate whether two traits or characteristics are related. Contingency Table Analysis

Contingency table analysis Each observation is classified according to two criteria. We use the usual hypothesis testing procedure. The degrees of freedom are equal to: (number of rows-1)(number of columns-1). The expected frequency is computed as: Expected Frequency = (row total)(column total) grand total Contingency table analysis

Contingency Table Analysis 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? Example 3

Step 1: H0: Gender and location are not related. H1: Gender and location are related. Step 2: The level of significance is set at .01. Step 3: the test statistic is the chi-square distribution. Step 4: The degrees of freedom equal (r-1)(c-1) or 2. The critical c2 at 2 d.f. is 9.21. If computed c2 >9.21, or if p < .01, reject the null and accept the alternate. Step 5: A data table and the following contingency table are constructed.

Observed frequencies (fo ) Gender Work Home Other Total Male 60 20 10 90 Female 30 80 50 150 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. Example 3 continued

Expected frequencies (fe ) Gender Work Home Other Total Male (80)(90)150 = 48 (50)(90) 150 =30 (20)(90) =12 90 Female (80)(60) =32 (50)(60) =20 (20)(60) =8 60 80 50 20 Example 3 continued

c2: (fo – fe)2/ fe Gender Work Home Other Total c2 Male (60-48)2 48 (20-30)2 30 (10-12)2 12 6.667 Female (20-32)2 32 (30-20)2 20 (12-10)2 10 10.000 Total 16.667 Example 3 continued

The p(c2 > 16.667) = .00024. Since the c2 of 16.667 > 9.21, p of .00024 < .01, reject the null and conclude that there is a relationship between the location of an accident and the gender of the person involved. Example 3 concluded