1. 1 1 Slide Mátgæði Kafli 11 í Newbold Snjólfur Ólafsson + Slides Prepared by John Loucks © 1999 ITP/South-Western College Publishing

2. 2 2 Slide Tests of Goodness of Fit and Independence Tests of Goodness of Fit and Independence n Goodness of Fit Test: A Multinomial Population n Tests of Independence: Contingency Tables n Goodness of Fit Test: Poisson and Normal Distributions

3. 3 3 Slide Goodness of Fit Test: A Multinomial Population 1. Set up the null and alternative hypotheses. 2. Select a random sample and record the observed frequency, f i, for each of the k categories. 3. Assuming H 0 is true, compute the expected frequency, e i, in each category by multiplying the category probability by the sample size. 4. Compute the value of the test statistic. 5. Reject H 0 if (where  is the significance level and there are k - 1 degrees of freedom).

4. 4 4 Slide Example: Finger Lakes Homes (A) n Multinomial Distribution Goodness of Fit Test Finger Lakes Homes manufactures four models of prefabricated homes, a two-story colonial, a ranch, 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. The number of homes sold of each model for 100 sales over the past two years is shown below. Model Colonial Ranch Split-Level A-Frame # Sold 30 20 35 15

5. 5 5 Slide Example: Finger Lakes Homes (A) n Multinomial Distribution Goodness of Fit Test Let p C = population proportion that purchase a colonial p R = population proportion that purchase a ranch p S = population proportion that purchase a split-level p A = population proportion that purchase an A-frame Hypotheses H 0 : p C = p R = p S = p A =.25 H a : The population proportions are not p C =.25, p R =.25, p S =.25, and p A =.25 p R =.25, p S =.25, and p A =.25

6. 6 6 Slide Example: Finger Lakes Homes (A) n Multinomial Distribution Goodness of Fit Test Expected Frequencies e 1 =.25(100) = 25 e 2 =.25(100) = 25 e 1 =.25(100) = 25 e 2 =.25(100) = 25 e 3 =.25(100) = 25 e 4 =.25(100) = 25 e 3 =.25(100) = 25 e 4 =.25(100) = 25 Test Statistic = 1 + 1 + 4 + 4 = 1 + 1 + 4 + 4 = 10 = 10

7. 7 7 Slide n Multinomial Distribution Goodness of Fit Test Rejection Rule With  =.05 and With  =.05 and k - 1 = 4 - 1 = 3 degrees of k - 1 = 4 - 1 = 3 degrees of freedom freedomConclusion We reject the assumption there is no home style We reject the assumption there is no home style preference, at the.05 level of significance. preference, at the.05 level of significance. 22 22 7.81 Do Not Reject H 0 Reject H 0 Example: Finger Lakes Homes (A)

8. 8 8 Slide Tests of Independence: Contingency Tables 1. Set up the null and alternative hypotheses. 2. Select a random sample and record the observed frequency, f ij, for each cell of the contingency table. 3. Compute the expected frequency, e ij, for each cell. 4. Compute the value of the test statistic. 5. Reject H 0 if (where  is the significance level and with n rows and m columns there are ( n - 1)( m - 1) degrees of freedom).

9. 9 9 Slide Example: Finger Lakes Homes (B) n Contingency Table Test Each home sold can be classified according to price and to style. Finger Lakes Homes’ manager would like to determine if the price of the home and the style of the home are independent variables. Each home sold can be classified according to price and to style. Finger Lakes Homes’ manager would like to determine if the price of the home and the style of the home are independent variables. 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 $65,000 or less or more than $65,000. Price Colonial Ranch Split-Level A-Frame Price Colonial Ranch Split-Level A-Frame < $65,000 18 6 19 12 < $65,000 18 6 19 12 > $65,000 12 14 16 3 > $65,000 12 14 16 3

10. 10 Slide n Contingency Table Test Hypotheses H 0 : Price of the home is independent of the style of H 0 : Price of the home is independent of the style of the home that is purchased the home that is purchased H a : Price of the home is not independent of the H a : Price of the home is not independent of the style of the home that is purchased style of the home that is purchased Expected Frequencies Price Colonial Ranch Split-Level A-Frame Total Price Colonial Ranch Split-Level A-Frame Total < $65K 16,5 11 19,25 8,25 55 > $65K 13,5 9 15,75 6,75 45 Total 30 20 35 15 100 Total 30 20 35 15 100 Example: Finger Lakes Homes (B)

11. 11 Slide n Contingency Table Test Test Statistic =.14 + 2.27 +... + 2.08 = 8.00 =.14 + 2.27 +... + 2.08 = 8.00 Rejection Rule With  =.05 and (2 - 1)(4 - 1) = 3 d.f., With  =.05 and (2 - 1)(4 - 1) = 3 d.f., Reject H 0 if  2 > 7.81 Reject H 0 if  2 > 7.81Conclusion We reject H 0. We reject the assumption that the We reject H 0. We reject the assumption that the price of the home is independent of the style of the price of the home is independent of the style of the home that is purchased. home that is purchased. Example: Finger Lakes Homes (B)

12. 12 Slide Goodness of Fit Test: Poisson Distribution 1. Set up the null and alternative hypotheses. 2. Select a random sample and a. Record the observed frequency, f i, for each of the k values of the Poisson random variable. k values of the Poisson random variable. b. Compute the mean number of occurrences, . 3. Compute the expected frequency of occurrences, e i, for each value of the Poisson random variable. 4. Compute the value of the test statistic. 5. Reject H 0 if (where  is the significance level and there are k - 2 degrees of freedom).

13. 13 Slide Example: Troy Parking Garage n Poisson Distribution Goodness of Fit Test In studying the need for an additional entrance to a city parking garage, a consultant has recommended an approach that is applicable only in situations where the number of cars entering during a specified time period follows a Poisson distribution. In studying the need for an additional entrance to a city parking garage, a consultant has recommended an approach that is applicable only in situations where the number of cars entering during a specified time period follows a Poisson distribution. A random sample of 100 one-minute time intervals resulted in the customer arrivals listed below. A statistical test must be conducted to see if the assumption of a Poisson distribution is reasonable. A random sample of 100 one-minute time intervals resulted in the customer arrivals listed below. A statistical test must be conducted to see if the assumption of a Poisson distribution is reasonable. # Arrivals 0 1 2 3 4 5 6 7 8 9 10 11 12 # Arrivals 0 1 2 3 4 5 6 7 8 9 10 11 12 Frequency 0 1 4 10 14 20 12 12 9 8 6 3 1 Frequency 0 1 4 10 14 20 12 12 9 8 6 3 1

14. 14 Slide Example: Troy Parking Garage n Poisson Distribution Goodness of Fit Test Hypotheses H 0 : The number of cars entering the garage during a one-minute interval is Poisson distributed. H a : The number of cars entering the garage during a one-minute interval is not Poisson distributed. Estimate of Poisson Probability Function  otal Arrivals = 0(0) + 1(1) + 2(4) +... + 12(1) = 600 Total Time Periods = 100 Estimate of  = 600/100 = 6

15. 15 Slide Example: Troy Parking Garage n Poisson Distribution Goodness of Fit Test Expected Frequencies x f ( x ) xf ( x ) x f ( x ) xf ( x ) 0.0025.25 7.138913.89 0.0025.25 7.138913.89 1.0149 1.49 8.104110.41 1.0149 1.49 8.104110.41 2.0446 4.46 9.06946.94 2.0446 4.46 9.06946.94 3.0892 8.9210.04174.17 3.0892 8.9210.04174.17 4.133913.3911.02272.27 4.133913.3911.02272.27 5.162016.2012.01551.55 5.162016.2012.01551.55 6.160616.06 Total1.0000100.00 6.160616.06 Total1.0000100.00

16. 16 Slide Example: Troy Parking Garage n Poisson Distribution Goodness of Fit Test Observed and Expected Frequencies i f i e i f i - e i i f i e i f i - e i 0 or 1 or 256.20-1.20 0 or 1 or 256.20-1.20 3108.921.08 41413.39.61 52016.203.80 61216.06-4.06 71213.89-1.89 8910.41-1.41 986.941.06 10 or more107.992.01 10 or more107.992.01

17. 17 Slide n Poisson Distribution Goodness of Fit Test Test Statistic Rejection Rule With  =.05 and k - p - 1 = 9 - 1 - 1 = 7 d.f. With  =.05 and k - p - 1 = 9 - 1 - 1 = 7 d.f. (where k = number of categories and p = number (where k = number of categories and p = number of population parameters estimated), of population parameters estimated), Reject H 0 if  2 > 14.07 Reject H 0 if  2 > 14.07Conclusion We cannot reject H 0. There is no reason to question We cannot reject H 0. There is no reason to question the assumption of a Poisson distribution. the assumption of a Poisson distribution. Example: Troy Parking Garage

18. 18 Slide Mátgæði fyrir normaldreifingu Tvær ólíkar leiðir. A. Nota  2 - próf eins og hér á undan. B. Reikna skekkingu og ferilris og nota Bowman- Shelton próf. Þið þurfið að vita um þessar leiðir og vita hvað skekking og ferilris segja, en þurfið ekki að reikna þetta.

19. 19 Slide A.  2 - próf Meðaltal og staðalfrávik úrtaks notað í tilgátuprófi. H 0 : Þýðið er normaldreift með meðalgildi og staðalfrávik eins og í úrtaki. H 0 : Þýðið er normaldreift með meðalgildi og staðalfrávik eins og í úrtaki. H a : Svo er ekki. H a : Svo er ekki. Rauntöluásnum skipt í hafilega mörg bil, þ.a. væntanlegur fjöldi á bili, samkvæmt H0, sé meiri en 5. Reiknaðo.s.frv.

20. 20 Slide B. Skekking og ferilris Skekking (skewness) mælir hve samhverf tíðnidreifing er (assymmetry of a frequency distribution). Ferilris (kurtosis) mælir hve tíðnidreifing rís hátt (flatness or peakedness of a frequency distribution)

21. 21 Slide Í Bowman-Shelton prófi er skekking og ferilris reiknað fyrir úrtakið og athugað hvort það sé í nægilega góðu samræmi við skekkingu og ferilris fyrir normaldreifingu. Skekking og ferilris - framhald