1. 1. State the null and alternative hypotheses. 2. Select a random sample and record observed frequency f i for the i th category ( k categories). 3. 3.Compute expected frequency e i for the i th category: 4. Compute the value of the test statistic. if e i > 5, this has a chi-square distribution 5. Reject H 0 if df = k – 1 Goodness of Fit Test

2. 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. Split- A- Model Colonial Log Level Frame # Sold 30 20 35 15 The number of homes sold of each model for 100 sales over the past two years is shown below. k = 4

3. 1/4.25 Hypotheses Goodness of Fit Test H 0 : p C = p L = p S = p A = H a : customers prefer a particular style e i = ( n )( p i ) i.e., there is at least one proportion much greater than.25 e 1 = (0.25)(100) = 25 Expected frequencies e 2 = (0.25)(100) = 25 e 3 = (0.25)(100) = 25 e 4 = (0.25)(100) = 25

4. Goodness of Fit Test  =.05 (column) 7.815 Do Not Reject H 0 Reject H 0.05  2 At 5% significance, the assumption that there is no home style preference is rejected. df = 4 – 1 = 3 (row)  = 3

5. 1. State the null and alternative hypotheses. 2. Select a random sample and record observed frequency f i for each cell of the contingency table. 3. 3.Compute expected frequency e ij for each cell 4. Compute the test statistic. 5. Reject H 0 if df = ( m - 1)( k - 1) Independence Test if e i > 5, this has a chi-square distribution

6. 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. 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. Example: Finger Lakes Homes (B) Price Colonial Log Split-Level A-Frame > $99,000 < $99,000 k = 4 Independence Test 18 6 19 12 12 14 16 3 m = 2

7. 45 Price Colonial Log Split-Level A-Frame Total < $99K > $99K Total 30 12 14 16 3 18 6 19 12 55 Price Colonial Log Split-Level A-Frame Total < $99K > $99K Total 16.5 11 19.25 8.25 13.5 9 15.75 6.75 55 30 Expected Frequencies ( e i ) Observed Frequencies ( f i ) Independence Test 20 35 15 100 45 20 35 15 100

8. Compute test statistic Hypotheses H 0 : Price of the home is independent of the style of the home that is purchased H a : Price of the home is not independent of the style of the home that is purchased Independence Test

9.  =.05 (column) 7.815 Do Not Reject H 0 Reject H 0.05  2 At 5% significance, we reject the assumption that the price of the home is independent of the style of home that is purchased. df = (4 – 1)(2 – 1) = 3 (row)  = 3 Independence Test

10. 1. Set up the null and alternative hypotheses. 3. Compute expected frequency of occurrences e i for each value of the Poisson random variable. 2. Select a random sample and a. Record observed frequencies b. Estimate mean number of occurrences  4. Compute the value of the test statistic. 5. Reject H 0 if df = k – p – 1 Goodness of Fit Test: Poisson Distribution

11. Example: Troy Parking Garage In studying the need for an additional entrance to a city parking garage, a consultant has recommended an analysis approach that is applicable only in situations where the number of cars entering during a specified time period follows a Poisson distribution. Goodness of Fit Test: Poisson Distribution A random sample of n = 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. # of 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  otal Arrivals = 0(0) + 1(1) + 2(4) + 3(10) +... + 12(1) = 600 Estimate of  = 600/100 = 6 Total one-minute intervals = n = 0 + 1 + 4 + 10 +... + 1 = 100

12. x f ( x ) n∙ f ( x ) 012345012345 16.06 13.77 10.33 6.88 8.39 0.1606 0.1377 0.1033 0.0688 6 7 8 9 10+ 0.0025 0.25 x f ( x ) n∙ f ( x ) 1.0000 100.00 For x = 0 1 0.0149 0.0446 0.0892 0.1339 0.1606 1.49 4.46 8.92 13.39 16.06 Goodness of Fit Test: Poisson Distribution The hypothesized probability of x cars arriving during the time period is 0.0839

13. i f i e i f i - e i -1.20 1.08 0.61 3.94 -4.06 -1.77 -1.33 1.12 1.61 6.20 8.92 13.39 16.06 13.77 10.33 6.88 8.39 5 10 14 20 12 9 8 10 0 or 1 or 2 3 4 5 6 7 8 9 10 or more Goodness of Fit Test: Poisson Distribution

14. With  =.05 14.067 Do Not Reject H 0 Reject H 0.05  2 At 10% significance, there is no reason to doubt the assumption of a Poisson distribution. (column) and df = 7 (row) Goodness of Fit Test: Poisson Distribution

15. 1. State the null and alternative hypotheses. 3. Compute e i for each interval. 2. Select a random sample and a. Compute the mean and standard deviation (p = 2). b. Define intervals so that e i > 5 is in the i th interval c. For each interval, record observed frequencies f i 4. Compute the value of the test statistic. 5. Reject H 0 if df = k – p – 1 if e i > 5, this has a chi-square distribution Goodness of Fit Test: Normal Distribution

16. Example: IQ Computers IQ Computers (one better than HP?) manufactures and sells a general purpose microcomputer. As part of a study to evaluate sales personnel, management wants to determine, at a 5% significance level, if the annual sales volume (number of units sold by a salesperson) follows a normal probability distribution. A simple random sample of 33 of the salespeople was taken and their numbers of units sold are below. 33 43 44 45 52 52 56 58 63 63 64 64 65 66 68 70 72 73 73 74 74 75 83 84 85 86 91 92 94 98 101 102 105 n = 33, x = 71.76, s = 18.47 Goodness of Fit Test: Normal Distribution

17. z.z. k = 33/5 = 6.6 6 equal intervals. To ensure the test statistic has a chi-square distribution, the normal distribution is divided into k intervals. Expected frequency: e i = 33/6 = 5.5 1/6 =.1667 The probability of being in each interval is equal to Goodness of Fit Test: Normal Distribution

18. = (1)(.1667) =.1667 z.z. –.97.1667 Find the z that corresponds to the red tail probability Goodness of Fit Test: Normal Distribution

19. =.3333 –.43.3333 z.z. Goodness of Fit Test: Normal Distribution Find the z that corresponds to the red tail probability = (2)(.1667)

20. =.5000 0.5000 z.z. Goodness of Fit Test: Normal Distribution = (3)(.1667) Find the z that corresponds to the red tail probability

21. 0.97.43 z.z. –.97 –.43 Goodness of Fit Test: Normal Distribution Find the remaining z values using symmetry

22. Convert the z’s to x’s x z.z. 0.97.43 –.97 –.43 Goodness of Fit Test: Normal Distribution Find the z that corresponds to the red tail probability

23. 33 43 44 45 52 52 56 58 63 63 64 64 65 66 68 70 72 73 73 74 74 75 83 84 85 86 91 92 94 98 101 102 105 Observed and Expected Frequencies 0.5 -1.5 0.5 -1.5 1.5 5.5 33 6 f i e i f i – e i (f i – e i ) 2 /e i Total LL UL 53.84 63.81 71.76 79.70 89.68 ∞ 4 6 6 4 7 33 Data Table   ∞ 53.84 63.81 71.76 79.70 89.68 0.05 0.41 0.05 0.41 1.36 Goodness of Fit Test: Normal Distribution

24.  =.05 (column) 7.815 Do Not Reject H 0 Reject H 0.05  2 At 5% significance, there is no reason to doubt the assumption that the population is normally distributed. df = 3 (row)  = 3 Goodness of Fit Test: Normal Distribution