《商务统计学概论》（英文版） CHAPTER 13 Chi-Square Applications

CHAPTER 13 Chi-Square applications to accompany Introduction to business statistics fourth edition by ronald M. Weiers Presentation by Priscilla Chaffe-Stengel Donald n tenge o 2002 The Wadsworth Group

CHAPTER 13 Chi-Square Applications to accompany Introduction to Business Statistics fourth edition, by Ronald M. Weiers Presentation by Priscilla Chaffe-Stengel Donald N. Stengel © 2002 The Wadsworth Group

l Chapter 13-Learning objectives Explain the nature of the chi-square distribution Apply the chi-square distribution to Goodness-of-fit tests Tests of independence between 2 variables Tests comparing proportions from multiple populations Tests of a single population variance o 2002 The Wadsworth Group

Chapter 13 - Learning Objectives • Explain the nature of the chi-square distribution. • Apply the chi-square distribution to: – Goodness-of-fit tests – Tests of independence between 2 variables – Tests comparing proportions from multiple populations – Tests of a single population variance. © 2002 The Wadsworth Group

l Chapter 13-Key Terms Observed versus expected frequencies Number of parameters estimated, m Number of categories used, k Contingency table Independent variables o 2002 The Wadsworth Group

Chapter 13 - Key Terms • Observed versus expected frequencies • Number of parameters estimated, m • Number of categories used, k • Contingency table • Independent variables © 2002 The Wadsworth Group

l Goodness-of-Fit Tests The Question: Does the distribution of sample data resemble a specifie probability distribution, such as > the binomial h ypergeometric, or Poisson discrete distributions >> the uniform, normal, or exponential continuous distributions >>a predefined probability distribution ypotheses Ho: T-values expected H: f* values expected where丌 o 2002 The Wadsworth Group

Goodness-of-Fit Tests • The Question: – Does the distribution of sample data resemble a specified probability distribution, such as: »the binomial, hypergeometric, or Poisson discrete distributions. »the uniform, normal, or exponential continuous distributions. »a predefined probability distribution. • Hypotheses: – H0 : pi = values expected H1 : pi  values expected where p j  = 1 . © 2002 The Wadsworth Group

l Goodness-of-Fit Tests Rejection regions Degrees of Freedom=k-1-m >>where k=# of categories m=# of parameters >Uniform Discrete: m=0 so df =k-1 Binomial: m=0 when T is known, so df =k-1 m=1 when T is unknown, so df=k-2 Poisson: m=1 since u usually estimated df =k-2 Normal: m=2 when u and o estimated df=k-3 Exponential: m= 1 since u usually estimated df=k-2 o 2002 The Wadsworth Group

Goodness-of-Fit Tests • Rejection Region: – Degrees of Freedom = k – 1 – m »where k = # of categories, m = # of parameters »Uniform Discrete: m = 0 so df = k – 1 »Binomial: m = 0 when p is known, so df = k – 1 m = 1 when p is unknown, so df = k – 2 »Poisson: m = 1 since µ usually estimated, df = k – 2 »Normal: m = 2 when µ and s estimated, df = k – 3 »Exponential: m = 1 since µ usually estimated, df = k – 2 © 2002 The Wadsworth Group

l Goodness-of-Fit Tests · Test statistic: O-E x 2 E where 0: actual number observed in each class E;= Expected number, T;.n o 2002 The Wadsworth Group

Goodness-of-Fit Tests • Test Statistic: where Oj = Actual number observed in each class Ej = Expected number, pj • n =  j E j E j (O – )2 c 2 © 2002 The Wadsworth Group

llGoodness-of-Fit: An Example Problem 13.20: In a study of vehicle ownership it has been found that 13.5% of u. s households do not own a vehicle with 33.7% owning 1 vehicle, 33.5% owning 2 vehicles and 19.3% owning 3 or more vehicles. The data for a random sample of 100 households in a resort community are summarized below. At the 0.05 level of significance, can we reject the possibility that the vehicle-ownership distribution in this community differs from that of the nation as a whole? #Ⅴ ehicles owned井 Households 20 1 35 23 3 or more 22 o 2002 The Wadsworth Group

Goodness-of-Fit: An Example • Problem 13.20: In a study of vehicle ownership, it has been found that 13.5% of U.S. households do not own a vehicle, with 33.7% owning 1 vehicle, 33.5% owning 2 vehicles, and 19.3% owning 3 or more vehicles. The data for a random sample of 100 households in a resort community are summarized below. At the 0.05 level of significance, can we reject the possibility that the vehicle-ownership distribution in this community differs from that of the nation as a whole? # Vehicles Owned # Households 0 20 1 35 2 23 3 or more 22 © 2002 The Wadsworth Group

l Goodness-of-Fit: An Example Vehicles O O -E;2/E 20 13.5 3.1296 3533.7 0.0501 2333.5 3.2910 3 19.3 0.3777 sum=6.8484 I.H0:7=0.135,丌=0.37,n2=0.335,7+=0.193 Vehicle-ownership distribution in this community is the same as it is in the nation as a whole H,: At least one of the proportions does not equal the stated value. Vehicle-ownership distribution in this community is not the same as it is in the nation as a whole 2002 The Wadsworth Group

Goodness-of-Fit: An Example # Vehicles Oj Ej [Oj– Ej ] 2/ Ej 0 20 13.5 3.1296 1 35 33.7 0.0501 2 23 33.5 3.2910 3+ 22 19.3 0.3777 Sum = 6.8484 I. H0 : p0 = 0.135, p1 = 0.337, p2 = 0.335, p3+ = 0.193 Vehicle-ownership distribution in this community is the same as it is in the nation as a whole. H1 : At least one of the proportions does not equal the stated value. Vehicle-ownership distribution in this community is not the same as it is in the nation as a whole. © 2002 The Wadsworth Group

Goodness-of-Fit: An example II. Rejection Region c=0.05 f=k-1-m=4-1-0=3 Do Not Reject H Reject h II. Test Statistic. 0.95 0.05 x2=68484 2=7815 IV. Conclusion: Since the test statistic of x2=6.8484 falls below the critical value of x2=7.815, we do not reject Ho with at least 95 confidence V Implications: There is not enough evidence to show that vehicle ownership in this community differs from that in the nation as a whole o 2002 The Wadsworth Group

Goodness-of-Fit: An Example II. Rejection Region: a = 0.05 df = k – 1 – m = 4 – 1 – 0 = 3 III. Test Statistic: c 2 = 6.8484 IV. Conclusion: Since the test statistic of c 2 = 6.8484 falls below the critical value of c 2 = 7.815, we do not reject H0 with at least 95% confidence. V. Implications: There is not enough evidence to show that vehicle ownership in this community differs from that in the nation as a whole.  0.95 Do Not Reject H0 Reject H 0 c =7.815 2 © 2002 The Wadsworth Group

l Chi-Square Tests of independence Between two variables The Question Are the two variables independent? If the two variables of interest are independent then >> the way elements are distributed across the various levels of one variable does not affect how they are distributed across the levels of the other o the probability of an element falling in any level of the second variable is unaffected by knowing its level on the first dimension o 2002 The Wadsworth Group

Chi-Square Tests of Independence Between Two Variables • The Question: – Are the two variables independent? If the two variables of interest are independent, then »the way elements are distributed across the various levels of one variable does not affect how they are distributed across the levels of the other. »the probability of an element falling in any level of the second variable is unaffected by knowing its level on the first dimension. © 2002 The Wadsworth Group