Chi-Square and Analysis of Variance (ANOVA) Chapter 11 Chi-Square and Analysis of Variance (ANOVA) © McGraw-Hill, Bluman, 5th ed., Chapter 11

Chapter 11 Overview Introduction 11-1 Test for Goodness of Fit 11-2 Tests Using Contingency Tables 11-3 Analysis of Variance (ANOVA) Bluman, Chapter 11

Chapter 11 Objectives Test a distribution for goodness of fit, using chi-square. Test two variables for independence, using chi-square. Test proportions for homogeneity, using chi-square. Use the ANOVA technique to determine if there is a significant difference among three or more means Bluman, Chapter 11

11.1 Test for Goodness of Fit The chi-square statistic can be used to see whether a frequency distribution fits a specific pattern. This is referred to as the chi-square goodness-of-fit test. Bluman, Chapter 11

Test for Goodness of Fit Formula for the test for goodness of fit: where d.f. = number of categories minus 1 O = observed frequency E = expected frequency Bluman, Chapter 11

Assumptions for Goodness of Fit The data are obtained from a random sample. The expected frequency for each category must be 5 or more. Bluman, Chapter 11

Chapter 11 Chi-Square and Analysis of Variance (ANOVA) Section 11-1 Example 11-1 Page #592 Bluman, Chapter 11

Example 11-1: Fruit Soda Flavors A market analyst wished to see whether consumers have any preference among five flavors of a new fruit soda. A sample of 100 people provided the following data. Is there enough evidence to reject the claim that there is no preference in the selection of fruit soda flavors, using the data shown previously? Let α = 0.05. Cherry Strawberry Orange Lime Grape Observed 32 28 16 14 10 Expected 20 Step 1: State the hypotheses and identify the claim. H0: Consumers show no preference (claim). H1: Consumers show a preference. Bluman, Chapter 11

Example 11-1: Fruit Soda Flavors Cherry Strawberry Orange Lime Grape Observed 32 28 16 14 10 Expected 20 Step 2: Find the critical value. D.f. = 5 – 1 = 4, and α = 0.05. CV = 9.488. Step 3: Compute the test value. Bluman, Chapter 11

Example 11-1: Fruit Soda Flavors Step 4: Make the decision. The decision is to reject the null hypothesis, since 18.0 > 9.488. Step 5: Summarize the results. There is enough evidence to reject the claim that consumers show no preference for the flavors. Bluman, Chapter 11

Chapter 11 Chi-Square and Analysis of Variance (ANOVA) Section 11-1 Example 11-2 Page #594 Bluman, Chapter 11

Example 11-2: Retirees The Russel Reynold Association surveyed retired senior executives who had returned to work. They found that after returning to work, 38% were employed by another organization, 32% were self-employed, 23% were either freelancing or consulting, and 7% had formed their own companies. To see if these percentages are consistent with those of Allegheny County residents, a local researcher surveyed 300 retired executives who had returned to work and found that 122 were working for another company, 85 were self-employed, 76 were either freelancing or consulting, and 17 had formed their own companies. At α = 0.10, test the claim that the percentages are the same for those people in Allegheny County. Bluman, Chapter 11

Example 11-2: Retirees New Company Self-Employed Free-lancing Owns Company Observed 122 85 76 17 Expected .38(300)= 114 .32(300)= 96 .23(300)= 69 .07(300)= 21 Step 1: State the hypotheses and identify the claim. H0: The retired executives who returned to work are distributed as follows: 38% are employed by another organization, 32% are self-employed, 23% are either freelancing or consulting, and 7% have formed their own companies (claim). H1: The distribution is not the same as stated in the null hypothesis. Bluman, Chapter 11

Example 11-2: Retirees Step 2: Find the critical value. New Company Self-Employed Free-lancing Owns Company Observed 122 85 76 17 Expected .38(300)= 114 .32(300)= 96 .23(300)= 69 .07(300)= 21 Step 2: Find the critical value. D.f. = 4 – 1 = 3, and α = 0.10. CV = 6.251. Step 3: Compute the test value. Bluman, Chapter 11

Example 11-2: Retirees Step 4: Make the decision. Since 3.2939 < 6.251, the decision is not to reject the null hypothesis. Step 5: Summarize the results. There is not enough evidence to reject the claim. It can be concluded that the percentages are not significantly different from those given in the null hypothesis. Bluman, Chapter 11

Chapter 11 Chi-Square and Analysis of Variance (ANOVA) Section 11-1 Example 11-3 Page #595 Bluman, Chapter 11

Example 11-3: Firearm Deaths A researcher read that firearm-related deaths for people aged 1 to 18 were distributed as follows: 74% were accidental, 16% were homicides, and 10% were suicides. In her district, there were 68 accidental deaths, 27 homicides, and 5 suicides during the past year. At α = 0.10, test the claim that the percentages are equal. Accidental Homicides Suicides Observed 68 27 5 Expected 74 16 10 Bluman, Chapter 11

Example 11-3: Firearm Deaths Accidental Homicides Suicides Observed 68 27 5 Expected 74 16 10 Step 1: State the hypotheses and identify the claim. H0: Deaths due to firearms for people aged 1 through 18 are distributed as follows: 74% accidental, 16% homicides, and 10% suicides (claim). H1: The distribution is not the same as stated in the null hypothesis. Bluman, Chapter 11

Example 11-3: Firearm Deaths Accidental Homicides Suicides Observed 68 27 5 Expected 74 16 10 Step 2: Find the critical value. D.f. = 3 – 1 = 2, and α = 0.10. CV = 4.605. Step 3: Compute the test value. Bluman, Chapter 11

Example 11-3: Firearm Deaths Step 4: Make the decision. Reject the null hypothesis, since 10.549 > 4.605. Step 5: Summarize the results. There is enough evidence to reject the claim that the distribution is 74% accidental, 16% homicides, and 10% suicides. Bluman, Chapter 11

Test for Normality (Optional) The chi-square goodness-of-fit test can be used to test a variable to see if it is normally distributed. The hypotheses are: H0: The variable is normally distributed. H1: The variable is not normally distributed. This procedure is somewhat complicated. The calculations are shown in example 11-4 on page 597 in the text. Bluman, Chapter 11

11.2 Tests Using Contingency Tables When data can be tabulated in table form in terms of frequencies, several types of hypotheses can be tested by using the chi-square test. The test of independence of variables is used to determine whether two variables are independent of or related to each other when a single sample is selected. The test of homogeneity of proportions is used to determine whether the proportions for a variable are equal when several samples are selected from different populations. Bluman, Chapter 11

Test for Independence The chi-square goodness-of-fit test can be used to test the independence of two variables. The hypotheses are: H0: There is no relationship between two variables. H1: There is a relationship between two variables. If the null hypothesis is rejected, there is some relationship between the variables. Bluman, Chapter 11

Test for Independence In order to test the null hypothesis, one must compute the expected frequencies, assuming the null hypothesis is true. When data are arranged in table form for the independence test, the table is called a contingency table. Bluman, Chapter 11

Contingency Tables The degrees of freedom for any contingency table are d.f. = (rows – 1) (columns – 1) = (R – 1)(C – 1). Bluman, Chapter 11

Test for Independence The formula for the test for independence: where d.f. = (R – 1)(C – 1) O = observed frequency E = expected frequency = Bluman, Chapter 11

Chapter 11 Chi-Square and Analysis of Variance (ANOVA) Section 11-2 Example 11-5 Page #606 Bluman, Chapter 11

Example 11-5: College Education and Place of Residence A sociologist wishes to see whether the number of years of college a person has completed is related to her or his place of residence. A sample of 88 people is selected and classified as shown. At α = 0.05, can the sociologist conclude that a person’s location is dependent on the number of years of college? Location No College Four-Year Degree Advanced Degree Total Urban 15 12 8 35 Suburban 9 32 Rural 6 7 21 29 24 88 Bluman, Chapter 11

Example 11-5: College Education and Place of Residence Step 1: State the hypotheses and identify the claim. H0: A person’s place of residence is independent of the number of years of college completed. H1: A person’s place of residence is dependent on the number of years of college completed (claim). Step 2: Find the critical value. The critical value is 4.605, since the degrees of freedom are (2 – 1)(3 – 1) = 2. Bluman, Chapter 11

Example 11-5: College Education and Place of Residence Compute the expected values. Location No College Four-Year Degree Advanced Degree Total Urban 15 12 8 35 Suburban 9 32 Rural 6 7 21 29 24 88 (11.53) (13.92) (9.55) (10.55) (12.73) (8.73) (6.92) (8.35) (5.73) Bluman, Chapter 11

Example 11-5: College Education and Place of Residence Step 3: Compute the test value. Bluman, Chapter 11

Example 11-5: College Education and Place of Residence Step 4: Make the decision. Do not reject the null hypothesis, since 3.01<9.488. Step 5: Summarize the results. There is not enough evidence to support the claim that a person’s place of residence is dependent on the number of years of college completed. Bluman, Chapter 11

Chapter 11 Chi-Square and Analysis of Variance (ANOVA) Section 11-2 Example 11-6 Page #608 Bluman, Chapter 11

Example 11-6: Alcohol and Gender A researcher wishes to determine whether there is a relationship between the gender of an individual and the amount of alcohol consumed. A sample of 68 people is selected, and the following data are obtained. At α = 0.10, can the researcher conclude that alcohol consumption is related to gender? Gender Alcohol Consumption Total Low Moderate High Male 10 9 8 27 Female 13 16 12 41 23 25 20 68 Bluman, Chapter 11

Example 11-6: Alcohol and Gender Step 1: State the hypotheses and identify the claim. H0: The amount of alcohol that a person consumes is independent of the individual’s gender. H1: The amount of alcohol that a person consumes is dependent on the individual’s gender (claim). Step 2: Find the critical value. The critical value is 9.488, since the degrees of freedom are (3 – 1 )(3 – 1) = (2)(2) = 4. Bluman, Chapter 11

Example 11-6: Alcohol and Gender Compute the expected values. Gender Alcohol Consumption Total Low Moderate High Male 10 9 8 27 Female 13 16 12 41 23 25 20 68 (9.13) (9.93) (7.94) (13.87) (15.07) (12.06) Bluman, Chapter 11

Example 11-6: Alcohol and Gender Step 3: Compute the test value. Bluman, Chapter 11

Example 11-6: Alcohol and Gender Step 4: Make the decision. Do not reject the null hypothesis, since 0.283 < 4.605. . Step 5: Summarize the results. There is not enough evidence to support the claim that the amount of alcohol a person consumes is dependent on the individual’s gender. Bluman, Chapter 11

Test for Homogeneity of Proportions Homogeneity of proportions test is used when samples are selected from several different populations and the researcher is interested in determining whether the proportions of elements that have a common characteristic are the same for each population. Bluman, Chapter 11

Test for Homogeneity of Proportions The hypotheses are: H0: p1 = p2 = p3 = … = pn H1: At least one proportion is different from the others. When the null hypothesis is rejected, it can be assumed that the proportions are not all equal. Bluman, Chapter 11

Assumptions for Homogeneity of Proportions The data are obtained from a random sample. The expected frequency for each category must be 5 or more. Bluman, Chapter 11

Chapter 11 Chi-Square and Analysis of Variance (ANOVA) Section 11-2 Example 11-7 Page #610 Bluman, Chapter 11

Example 11-7: Lost Luggage A researcher selected 100 passengers from each of 3 airlines and asked them if the airline had lost their luggage on their last flight. The data are shown in the table. At α = 0.05, test the claim that the proportion of passengers from each airline who lost luggage on the flight is the same for each airline. Airline 1 Airline 2 Airline 3 Total Yes 10 7 4 21 No 90 93 96 279 100 300 Bluman, Chapter 11

Example 11-7: Lost Luggage Step 1: State the hypotheses. H0: p1 = p2 = p3 = … = pn H1: At least one mean differs from the other. Step 2: Find the critical value. The critical value is 5.991, since the degrees of freedom are (2 – 1 )(3 – 1) = (1)(2) = 2. Bluman, Chapter 11

Example 11-7: Lost Luggage Compute the expected values. Airline 1 Airline 2 Airline 3 Total Yes 10 7 4 21 No 90 93 96 279 100 300 (7) (7) (7) (93) (93) (93) Bluman, Chapter 11

Example 11-7: Luggage Step 3: Compute the test value. Bluman, Chapter 11

Example 11-7: Lost Luggage Step 4: Make the decision. Do not reject the null hypothesis, since 2.765 < 5.991. . Step 5: Summarize the results. There is not enough evidence to reject the claim that the proportions are equal. Hence it seems that there is no difference in the proportions of the luggage lost by each airline. Bluman, Chapter 11

11-3 Analysis of Variance (ANOVA) The F test, used to compare two variances, can also be used to compare three of more means. This technique is called analysis of variance or ANOVA. For three groups, the F test can only show whether or not a difference exists among the three means, not where the difference lies. Bluman, Chapter 11

Analysis of Variance (ANOVA) When an F test is used to test a hypothesis concerning the means of three or more populations, the technique is called analysis of variance (commonly abbreviated as ANOVA). Although the t test is commonly used to compare two means, it should not be used to compare three or more. Bluman, Chapter 11

Assumptions for the F Test The following assumptions apply when using the F test to compare three or more means. The populations from which the samples were obtained must be normally or approximately normally distributed. The samples must be independent of each other. The variances of the populations must be equal. Bluman, Chapter 11

The F Test In the F test, two different estimates of the population variance are made. The first estimate is called the between-group variance, and it involves finding the variance of the means. The second estimate, the within-group variance, is made by computing the variance using all the data and is not affected by differences in the means. Bluman, Chapter 11

The F Test If there is no difference in the means, the between-group variance will be approximately equal to the within-group variance, and the F test value will be close to 1—do not reject null hypothesis. However, when the means differ significantly, the between-group variance will be much larger than the within-group variance; the F test will be significantly greater than 1—reject null hypothesis. Bluman, Chapter 11

Chapter 11 Chi-Square and Analysis of Variance (ANOVA) Section 11-3 Example 11-8 Page #620 Bluman, Chapter 11

Example 11-8: Lowering Blood Pressure A researcher wishes to try three different techniques to lower the blood pressure of individuals diagnosed with high blood pressure. The subjects are randomly assigned to three groups; the first group takes medication, the second group exercises, and the third group follows a special diet. After four weeks, the reduction in each person’s blood pressure is recorded. At α = 0.05, test the claim that there is no difference among the means. Bluman, Chapter 11

Example 11-8: Lowering Blood Pressure Step 1: State the hypotheses and identify the claim. H0: μ1 = μ2 = μ3 (claim) H1: At least one mean is different from the others. Bluman, Chapter 11

Example 11-8: Lowering Blood Pressure Step 2: Find the critical value. Since k = 3, N = 15, and α = 0.05, d.f.N. = k – 1 = 3 – 1 = 2 d.f.D. = N – k = 15 – 3 = 12 The critical value is 3.89, obtained from Table H. Bluman, Chapter 11

Example 11-8: Lowering Blood Pressure Step 3: Compute the test value. Find the mean and variance of each sample (these were provided with the data). Find the grand mean, the mean of all values in the samples. c. Find the between-group variance, . Bluman, Chapter 11

Example 11-8: Lowering Blood Pressure Step 3: Compute the test value. (continued) c. Find the between-group variance, . Find the within-group variance, . Bluman, Chapter 11

Example 11-8: Lowering Blood Pressure Step 3: Compute the test value. (continued) e. Compute the F value. Step 4: Make the decision. Reject the null hypothesis, since 9.17 > 3.89. Step 5: Summarize the results. There is enough evidence to reject the claim and conclude that at least one mean is different from the others. Bluman, Chapter 11

ANOVA The between-group variance is sometimes called the mean square, MSB. The numerator of the formula to compute MSB is called the sum of squares between groups, SSB. The within-group variance is sometimes called the mean square, MSW. The numerator of the formula to compute MSW is called the sum of squares within groups, SSW. Bluman, Chapter 11

ANOVA Summary Table Source Sum of Squares d.f. Mean Squares F Between Within (error) SSB SSW k – 1 N – k MSB MSW Total Bluman, Chapter 11

ANOVA Summary Table for Example 11-8 Source Sum of Squares d.f. Mean Squares F Between Within (error) 160.13 104.80 2 12 80.07 8.73 9.17 Total 264.93 14 Bluman, Chapter 11

Chapter 11 Chi-Square and Analysis of Variance (ANOVA) Section 11-3 Example 11-9 Page #622 Bluman, Chapter 11

Example 11-9: Toll Road Employees A state employee wishes to see if there is a significant difference in the number of employees at the interchanges of three state toll roads. The data are shown. At α = 0.05, can it be concluded that there is a significant difference in the average number of employees at each interchange? Bluman, Chapter 11

Example 11-9: Toll Road Employees Step 1: State the hypotheses and identify the claim. H0: μ1 = μ2 = μ3 H1: At least one mean is different from the others (claim). Bluman, Chapter 11

Example 11-9: Toll Road Employees Step 2: Find the critical value. Since k = 3, N = 18, and α = 0.05, d.f.N. = 2, d.f.D. = 15 The critical value is 3.68, obtained from Table H. Bluman, Chapter 11

Example 11-9: Toll Road Employees Step 3: Compute the test value. Find the mean and variance of each sample (these were provided with the data). Find the grand mean, the mean of all values in the samples. c. Find the between-group variance, . Bluman, Chapter 11

Example 11-9: Toll Road Employees Step 3: Compute the test value. (continued) c. Find the between-group variance, . Find the within-group variance, . Bluman, Chapter 11

Example 11-9: Toll Road Employees Step 3: Compute the test value. (continued) e. Compute the F value. Step 4: Make the decision. Reject the null hypothesis, since 5.05 > 3.68. Step 5: Summarize the results. There is enough evidence to support the claim that there is a difference among the means. Bluman, Chapter 11

ANOVA Summary Table for Example 11-9 Source Sum of Squares d.f. Mean Squares F Between Within (error) 459.18 682.5 2 15 229.59 45.5 5.05 Total 1141.68 17 Bluman, Chapter 11