Bài giảng Statistical Techniques in Business and Economics - Chapter 15 Chi-Square Distribution

Chi-Square DistributionChapter 151Chapter GoalsUnderstand the nature and role of chi-square distributionIdentify a wide variety of uses of the chi-square distributionConduct a test of hypothesis comparing an observed frequency distribution to an expected frequency distributionWhen you have completed this chapter, you will be able to:and...2Conduct a hypothesis test to determine whether two attributes are independentChapter Goals10Conduct a test of hypothesis for normality using the chi-square distribution3Characteristics 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.4df = 3df = 5df = 10c2Characteristics of the Chi-Square Distribution5Goodness-of-Fit Test: Equal Expected Frequencies 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 frequencies6 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 ()ồờờởộ-=eeofff22cờờởộ the test statistic is:7 The following information shows the number of employees absent by day of the week at a large a manufacturing plant. Solve Day FrequencyMonday 120Tuesday 45Wednesday 60Thursday 90Friday 130 Total 445Goodness-of-Fit Test: Equal Expected FrequenciesAt the .05 level of significance, is there a difference in the absence rate by day of the week? 8 Hypothesis Test Step 1Step 2Step 3Step 4H0: There is no difference in absence rate by day of the weekH1: Absence rates by day are not all equal  = 0.05Use Chi-Square testReject H0 if 2 > (5-1) = 4Degrees of freedom9.488. (see Appendix I)(120+45+60+90+130)/5 = 89Goodness-of-Fit Test: Equal Expected FrequenciesChi-Square910Right-Tail Area = 0.05Degrees of Freedom 5 – 1 = 4Reject H0 if 2 > 9.488Using the Table11 = 1.98 Day Frequency Expected (fo – fe)2/feMonday 120 89 10.80Tuesday 45 89 21.75Wednesday 60 89 9.45Thursday 90 89 0.01Friday 130 89 18.89 Total 445 445 60.902 Reject the null hypothesis. Absentee rates are not the same for each day of the week. Conclusion:(120-89)2/89Step 5Test StatisticsReject H0 if 2 > 9.48812SolveAt the .05 significance level, can we conclude that the Philadelphia area is different from the U.S. as a whole?MarriedWidowedDivorcedSingle63.9%7.7%6.9%21.5%A U.S. Bureau of the Census indicated thatNot re-marriedNever marriedA sample of 500 adults from the Philadelphia area showed:310403012013Married310 *319.5 ** .2825Widowed4038.5.0584Divorced3034.5.5870Single120107.51.4535Total5002.3814Status continued22.3814* Census figures would predict: i.e. 639*500 = 319.5** Our sample: (310-319.5)2/319.5 = .2825(fo – fe)2/feExpected14Step 1Step 2Step 3Step 4H0: The distribution has not changed continuedH1: The distribution has changed.H0 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. Conclusion:15Goodness-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 fe for each category- Use the chi-square goodness-of-fit test to determine if fo coincides with fe to determine the mean and standard deviation of the frequency distributionProcedure16A sample of 500 donations to the Arthritis Foundation is reported in the following frequency distributionIs 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 levelGoodness-of-Fit Test: Normality17Amount Spent fo Area fe (fo- fe )2/fe $1470Total500 continued18To compute fe for the first class, first determine the z - value continuedNow find the probability of a z - value less than –2.0000.22106-=-=-=smXz.02284772.5000.0)00.2(=-=-$1470.02Total500 continued20 The expected frequency is the probability of a z-value less than –2.00 times the sample size continuedThe other expected frequencies are computed similarly40.11)500)(0228(.==ef21Amount Spent fo Area fe (fo- fe )2/fe $1470.0211.40301.22Total500500336.33 continued22 continuedStep 1Step 2Step 3Step 4H0: The observations follow the normal distributionH0 is rejected if 2 >7.815, df = 6 = 0.052 = 336.33H0: is rejected.The observations do NOT follow the normal distribution Conclusion:H0: The observations do NOT follow the normal distribution23A 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 Contingency TableAnalysis each observation is classified according to two criteriathe usual hypothesis testing procedure is used the degrees of freedom is equal to: (number of rows -1)(number of columns -1)Note:24Is 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?25 continuedSexWorkHomeOtherTotalMale60201090Female20301060Total805020150LocationThe expected frequency for the work-male intersection is computed as (90)(80)/150 =48Similarly, you can compute the expected frequencies for the other cells26Step 1Step 2Step 3Step 4H0: The Gender and Location are NOT relatedH0 is rejected if c 2 >5.991, df = 2 = 0.05H0: is rejected.Gender and Location are related!Conclusion:H0: The Gender and Location are related(there are (3- 1)(2-1) = 2 degrees of freedom)Find the value of c 2()()8810...484860222-++-=c continued667.16=27Test your learning www.mcgrawhill.ca/college/lindClick onOnline Learning Centrefor quizzesextra contentdata setssearchable glossaryaccess to Statistics Canada’s E-Stat dataand much more!28This completes Chapter 1529