IS 310 Business Statistics CSU Long Beach

1. IS 310 Business Statistics CSU Long Beach

2. Additional Uses of Chi-Square Distribution In Chapter 11, we discussed the use of Chi-Square distribution in testing population variances. Here in chapter 12, we will cover two additional testing procedures by using the Chi-Square distribution. o Goodness of Fit Test o Test of Independence

3. Additional Uses of Chi-Square Distribution Goodness of Fit Test How close are sample results to the expected results? Example: In tossing a coin, you expect half heads and half tails. You tossed a coin 100 times. You expected 50 heads and 50 tails. However, you obtained 48 heads and 52 tails. Are 48 heads and 52 tails close enough to call the coin fair?

4. Additional Uses of Chi-Square Distribution Test of Independence Are two variables of interest independent of each other? Examples: Is starting salary of fresh graduates independent of graduates’ field of study? Is beer preference independent of the gender of the beer drinker?

5. Hypothesis (Goodness of Fit) Testfor Proportions of a Multinomial Population 1. Set up the null and alternative hypotheses. 2. Select a random sample and record the observed frequency, fi , for each of the k categories. 3. Assuming H0 is true, compute the expected frequency, ei , in each category by multiplying the category probability by the sample size.

6. Hypothesis (Goodness of Fit) Testfor Proportions of a Multinomial Population 4. Compute the value of the test statistic. where: fi = observed frequency for category i ei = expected frequency for category i k = number of categories Note: The test statistic has a chi-square distribution with k – 1 df provided that the expected frequencies are 5 or more for all categories.

7. Reject H0 if Hypothesis (Goodness of Fit) Testfor Proportions of a Multinomial Population 5. Rejection rule: Reject H0 if p-value <a p-value approach: Critical value approach: where  is the significance level and there are k - 1 degrees of freedom

8. Multinomial Distribution Goodness of Fit Test • Example: Finger Lakes Homes (A) Finger Lakes Homes manufactures four models of prefabricated homes, a two-story colonial, a log cabin, a split-level, and an A-frame. To help in production planning, management would like to determine if previous customer purchases indicate that there is a preference in the style selected.

9. Multinomial Distribution Goodness of Fit Test • Example: Finger Lakes Homes (A) The number of homes sold of each model for 100 sales over the past two years is shown below. Split- A- Model Colonial Log Level Frame # Sold 30 20 35 15

10. Multinomial Distribution Goodness of Fit Test • Hypotheses H0: pC = pL = pS = pA = .25 Ha: The population proportions are not pC = .25, pL = .25, pS = .25, and pA = .25 where: pC = population proportion that purchase a colonial pL = population proportion that purchase a log cabin pS = population proportion that purchase a split-level pA = population proportion that purchase an A-frame

11. Hypotheses H : There is no preference in the home styles or all o home styles have equal preferences H : All home styles do not have equal preferences a

12. Multinomial Distribution Goodness of Fit Test • Rejection Rule Reject H0 if p-value < .05 or c2 > 7.815. With  = .05 and k - 1 = 4 - 1 = 3 degrees of freedom Do Not Reject H0 Reject H0 2 7.815

13. Multinomial Distribution Goodness of Fit Test • Expected Frequencies • Test Statistic • e1 = .25(100) = 25 e2 = .25(100) = 25 e3 = .25(100) = 25 e4 = .25(100) = 25 = 1 + 1 + 4 + 4 = 10

14. Multinomial Distribution Goodness of Fit Test • Conclusion Using the p-Value Approach Area in Upper Tail .10 .05 .025 .01 .005 c2 Value (df = 3) 6.251 7.815 9.348 11.345 12.838 Because c2= 10 is between 9.348 and 11.345, the area in the upper tail of the distribution is between .025 and .01. The p-value <a . We can reject the null hypothesis. Note: A precise p-value can be found using Minitab or Excel.

15. Multinomial Distribution Goodness of Fit Test • Conclusion Using the Critical Value Approach c2 = 10 > 7.815 We reject, at the .05 level of significance, the assumption that there is no home style preference.

16. Sample Problem Problem # 3 (10-Page 462; 11-Page 477) H : Viewing audience proportions are the same 0 H : Viewing audience proportions are not the same a Category Frequencies Frequencies Differences Squared Observed Expected /Expected Frequencies ABC 95 87 0.74 CBA 70 84 2.33 NBC 89 75 2.61 Independents 46 54 1.19

17. Sample Problem Continued Decision Rule: Reject Null Hypothesis if 2 2 Χ -statistic > Χ 0.05, 3 2 2 Χ -statistic = 6.87 Χ = 7.815 0.05,3 Decision: Do not reject the null hypothesis Interpretation: Viewing audience proportions have not changed.

18. Test of Independence: Contingency Tables 1. Set up the null and alternative hypotheses. 2. Select a random sample and record the observed frequency, fij , for each cell of the contingency table. 3. Compute the expected frequency, eij , for each cell.

19. Reject H0 if p -value <a or . Test of Independence: Contingency Tables 4. Compute the test statistic. 5. Determine the rejection rule. where  is the significance level and, with n rows and m columns, there are (n - 1)(m - 1) degrees of freedom.

20. Contingency Table (Independence) Test • Example: Finger Lakes Homes (B) Each home sold by Finger Lakes Homes can be classified according to price and to style. Finger Lakes’ manager would like to determine if the price of the home and the style of the home are independent variables.

21. Contingency Table (Independence) Test • Example: Finger Lakes Homes (B) The number of homes sold for each model and price for the past two years is shown below. For convenience, the price of the home is listed as either $99,000 or less or more than $99,000. Price Colonial Log Split-Level A-Frame < $99,000 18 6 19 12 > $99,000 12 14 16 3

22. Contingency Table (Independence) Test • Hypotheses H0: Price of the home is independent of the style of the home that is purchased Ha: Price of the home is not independent of the style of the home that is purchased

23. Contingency Table (Independence) Test • Expected Frequencies Price Colonial Log Split-Level A-Frame Total < $99K > $99K Total 18 6 19 12 55 12 14 16 3 45 30 20 35 15 100

24. With  = .05 and (2 - 1)(4 - 1) = 3 d.f., Contingency Table (Independence) Test • Rejection Rule Reject H0 if p-value < .05 or 2> 7.815 • Test Statistic = .1364 + 2.2727 + . . . + 2.0833 = 9.149

25. Contingency Table (Independence) Test • Conclusion Using the p-Value Approach Area in Upper Tail .10 .05 .025 .01 .005 c2 Value (df = 3) 6.251 7.815 9.348 11.345 12.838 Because c2= 9.145 is between 7.815 and 9.348, the area in the upper tail of the distribution is between .05 and .025. The p-value <a . We can reject the null hypothesis. Note: A precise p-value can be found using Minitab or Excel.

26. Contingency Table (Independence) Test • Conclusion Using the Critical Value Approach c2 = 9.145 > 7.815 We reject, at the .05 level of significance, the assumption that the price of the home is independent of the style of home that is purchased.

27. Sample Problem Problem # 12 (10-Page 469; 11-Page 484) H : Method of payment is independent of age group 0 H : Method of payment is NOT independent of age a Payment Age Observed Expected Squared Squared Difference/ Exp. Method Group Frequency Frequency Difference Frequency Plastic 18-24 21 15.5 30.25 1.95 Plastic 25-34 27 23.3 13.69 0.59 Plastic 35-44 27 25.5 2.25 0.09 Plastic 45-up 36 46.6 112.36 2.41 Cash/Chk 18-24 21 26.4 29.2 1.1 Cash/Chk 25-34 36 39.7 13.69 0.34 Cash/Chk 35-44 42 43.5 2.25 0.05 Cash/Chk 45-up 90 79.4 112.36 1.42

28. Sample Problem Contd 2 Χ -statistic = 7.95 2 Critical Χ = 7.815 .05,3 2 2 Since Χ –statistic > Critical Χ , we reject the Null Hypothesis Interpretation: Method of payment is NOT independent of age group. p-value is between 0.025 and 0.05 The age group 18-24 uses plastic more than any other group

29. End of Chapter 12