# Chapter 11. Chapter 11 Overview. Chapter 11 Objectives 11/24/2015. Other Chi-Square Tests

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1 11/4/015 Chapter 11 Overview Chapter 11 Introduction 11-1 Test for Goodness of Fit 11- Tests Using Contingency Tables Other Chi-Square Tests McGraw-Hill, Bluman, 7th ed., Chapter 11 1 Bluman, Chapter 11 Chapter 11 Objectives 1. Test a distribution for goodness of fit, using chi-square.. Test two variables for independence, using chi-square. 3. Test proportions for homogeneity, using chi-square. Characteristics of the Chi-Square Distribution 1. It is not symmetric. Bluman, Chapter 11 3 Characteristics of the Chi-Square Distribution 1. It is not symmetric.. The shape of the chi-square distribution depends upon the degrees of freedom, just like Student s t-distribution. Characteristics of the Chi-Square Distribution 1. It is not symmetric.. The shape of the chi-square distribution depends upon the degrees of freedom, just like Student s t-distribution. 3. As the number of degrees of freedom increases, the chi-square distribution becomes more symmetric as is illustrated in Figure 1. 1

2 11/4/015 Characteristics of the Chi-Square Distribution 1. It is not symmetric.. The shape of the chi-square distribution depends upon the degrees of freedom, just like Student s t-distribution. 3. As the number of degrees of freedom increases, the chi-square distribution becomes more symmetric as is illustrated in Figure The values are non-negative. That is, the values of are greater than or equal to 0. The Chi-Square Distribution 11.1 Test for Goodness of Fit Reading the Table 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. Find the value of for 7 df and an area of 0.10 in the right tail of the chi-square distribution curve. Bluman, Chapter 11 9 Bluman, Chapter Reading the Table Find the value of for 1 df and an area of 0.05 in the left tail of the chi-square distribution curve. Test for Goodness of Fit Formula for the test for goodness of fit: O where d.f. = k 1 ; number of categories minus 1 O = observed frequency = expected frequency Bluman, Chapter Bluman, Chapter 11 1

3 11/4/015 Assumptions for Goodness of Fit 1. The data are obtained from a random sample.. The expected frequency for each category must be 5 or more. Bluman, Chapter xample 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 = Cherry Strawberry Orange Lime Grape Observed xpected H 0 : Consumers show no preference (claim). (H 0 : p 1 = p = p 3 = p 4 = p 5 ) H 1 : Consumers show a preference. (H 1 : At least of the 5 proportions are 0.0) Bluman, Chapter xample 1: Fruit Soda Flavors Cherry Strawberry Orange Lime Grape Observed xpected xample 1: Fruit Soda Flavors The decision is to reject the null hypothesis, since 18.0 > Step : Find the critical value. D.f. = 5 1 = 4, and α = CV = O Bluman, Chapter There is enough evidence to reject the claim that consumers show no preference for the flavors. Bluman, Chapter xample 1: Using your graphing calc. Cherry Strawberry Orange Lime Grape Observed xpected Step : nter the observed values into L 1 nter the expected values into L calculate first and enter or enter as 100*. Step 3: STATS TST- D: GOF Test Observed: L 1 xpected: L df: 4 Calculate This should be displayed on your screen: =18 p = df = 4 and CNTRB= (7. 3. Bluman, Chapter xample : 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, 3% were self-employed, 3% 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 1 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

4 11/4/015 xample : Retirees New Self- mployed Owns Observed xpected.38(300)= 114.3(300)= 96.3(300)= 69.07(300)= 1 H 0 : The retired executives who returned to work are distributed as follows: 38% are employed by another organization, 3% are self-employed, 3% are either freelancing or consulting, and 7% have formed their own companies (claim). (H 0 : p 1 =.38, p =.3 p 3 =.3 p 4 =.07) H 1 : The distribution is not the same as stated in the null hypothesis. (H 1 : At least one of the percentages is not as stated.) Bluman, Chapter xample : Retirees New Freelancing Self- mployed Owns Observed xpected.38(300)= 114.3(300)= 96.3(300)= 69 Step : Find the critical value. D.f. = 4 1 = 3, and α = CV = (300)= 1 O Bluman, Chapter 11 0 xample : Retirees Since < 6.51, the decision is not to reject the null hypothesis. 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 1 xample : Graphing Calculator New Freelancing Self- mployed Freelancing Owns Observed (300)=.3(300)=.3(300)= xpected Step : nter the observed values into L 1 nter the expected values into L Step 3: STATS TST- D: GOF Test Observed: L 1 xpected: L df: 3 Calculate This should be displayed on your screen: = p = df = 3 and CNTRB= ( (300)= 1 Bluman, Chapter 11 xample 11-3: ATM Usage (p. 495) A bank has an ATM installed inside the bank, and it is available to its customers only from 7 AM to 6 PM Mon Fri. The manager of the bank wanted to investigate if the percentage of transactions made on the ATM is the same for each of the five days (M F) of the week. She randomly selected one week and counted the number of transactions made on this ATM on each of the 5 days. At α = 0.01, can we reject the null hypothesis that the proportion of people who use this ATM each of the 5 days is the same? (Assume this week is typical of all weeks regarding ATM use) DAY # of users Mon Tues Wed Thurs Fri xample 11-3: ATM Usage DAY # of users Mon Tues Wed Thurs Fri H 0 : The proportion of people using the ATM is the same for all 5 days of the week. (H 0 : p 1 = p = p 3 = p 4 = p 5 ) H 1 : The distribution is not the same as stated in the null hypothesis. (H 1 : At least of the 5 proportions are 0.0) Bluman, Chapter 11 3 Bluman, Chapter

5 11/4/015 xample 11-3: ATM Usage DAY Mon Tues Wed Thurs Fri # of users xample 11-3: ATM Usage Reject the null hypothesis, since > Step : Find the critical value. Df = k - 1 D.f. = 5 1 = 4, and α = CV = O = There is enough evidence to reject the claim that the proportion of people using the ATM is the same for all 5 days of the week. Bluman, Chapter 11 5 Bluman, Chapter 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. Test for Independence The chi-square goodness-of-fit test can be used to test the independence of two variables. The hypotheses are: H 0 : There is no relationship between two variables. H 1 : There is a relationship between two variables. If the null hypothesis is rejected, there is some relationship between the variables. Bluman, Chapter 11 7 Bluman, Chapter 11 8 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. Contingency Tables The degrees of freedom for any contingency table are d.f. = (rows 1) (columns 1) = (R 1)(C 1). Bluman, Chapter 11 9 Bluman, Chapter

6 11/4/015 Test for Independence The formula for the test for independence: O where d.f. = (R 1)(C 1) O = observed frequency row sumcolumn sum = expected frequency = grand total xample 5: College ducation and 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 Suburban Rural Total Bluman, Chapter Bluman, Chapter 11 3 xample 5: College ducation and H 0 : A person s place of residence is independent of the number of years of college completed. H 1 : A person s place of residence is dependent on the number of years of college completed (claim). Step : Find the critical value. The critical value is 9.488, since the degrees of freedom are (3 1)(3 1) = 4. xample 5: College ducation and Compute the expected values. row sumcolumn sum grand total Location Urban Suburba n Rural No College Four-Year Degree Advanced Degree (11.53) (13.9) (9.55) (10.55) (1.73) (8.73) (6.9) (8.35) (5.73) Total Total , Bluman, Chapter Bluman, Chapter xample 5: College ducation and O Bluman, Chapter xample 5: College ducation and Do not reject the null hypothesis, since 3.01< A person s place of residence is independent of the number of years of college completed. Bluman, Chapter

7 11/4/015 Graphing Calculator Press nd x -1 for MATRIX and move the cursor to edit, then press enter nter the number of rows & columns then enter nter the values in the matrix as they appear in the contingency table Press STAT TST C: Test Make sure the observed matrix is [A] and the expected matrix is [B] Calculate, enter Bluman, Chapter You should see Test = P = df = 4 Bluman, Chapter xample6: 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 Low Moderate High Total Male Female Total Bluman, Chapter Graphing Calculator Press nd x -1 for MATRIX and move the cursor to edit, then press enter nter the number of rows & columns then enter nter the values in the matrix as they appear in the contingency table Press STAT TST C: Test Make sure the observed matrix is [A] and the expected matrix is [B] Calculate, enter Bluman, Chapter xample 6: Alcohol and Gender H 0 : The amount of alcohol that a person consumes is independent of the individual s gender. H 1 : The amount of alcohol that a person consumes is dependent on the individual s gender (claim). Step : Find the critical value. The critical value is 9.488, since the degrees of freedom are (3 1 )(3 1) = ()() = 4. xample 6: Alcohol and Gender Compute the expected values. row sumcolumn sum grand total Gender Male Female Alcohol Consumption Low Moderate High (9.13) (9.93) (7.94) (13.87) (15.07) (1.06) Total Total , Bluman, Chapter Bluman, Chapter

8 11/4/015 xample 6: Alcohol and Gender O Bluman, Chapter xample 6: Alcohol and Gender Do not reject the null hypothesis, since 0.83 < 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 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. Test for Homogeneity of Proportions The hypotheses are: H 0 : p 1 = p = p 3 = = p n H 1 : 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 Bluman, Chapter Assumptions for Homogeneity of Proportions 1. The data are obtained from a random sample.. The expected frequency for each category must be 5 or more. xample 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 Airline 3 Total Yes No Total Bluman, Chapter Bluman, Chapter

9 11/4/015 Graphing Calculator Press nd x -1 for MATRIX and move the cursor to edit, then press enter nter the number of rows & columns then enter nter the values in the matrix as they appear in the contingency table Press STAT TST C: Test Make sure the observed matrix is [A] and the expected matrix is [B] Calculate, enter Bluman, Chapter xample 7: Lost Luggage Step 1: State the hypotheses. H 0 : p 1 = p = p 3 = = p n H 1 : At least one mean differs from the other. Step : Find the critical value. The critical value is 5.991, since the degrees of freedom are ( 1 )(3 1) = (1)() =. Bluman, Chapter xample 7: Lost Luggage Compute the expected values. row sumcolumn sum grand total Yes 1, Airline 1 Airline Airline 3 Total (7) (7) (7) No 79 (93) (93) (93) Total xample 7: Luggage O Bluman, Chapter Bluman, Chapter 11 5 xample 7: Lost Luggage Do not reject the null hypothesis, since.765 < There is no difference in the proportions of the luggage lost by each airline. Bluman, Chapter

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