1 1 Slide © 2008 Thomson South-Western. All Rights Reserved Slides by JOHN LOUCKS St. Edward’s University

2 2 Slide © 2008 Thomson South-Western. All Rights Reserved Chapter 12 Tests of Goodness of Fit and Independence n Goodness of Fit Test: A Multinomial Population Goodness of Fit Test: Poisson Goodness of Fit Test: Poisson and Normal Distributions and Normal Distributions Test of Independence Test of Independence

3 3 Slide © 2008 Thomson South-Western. All Rights Reserved Hypothesis (Goodness of Fit) Test for Proportions of 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. 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 frequency, e i, in each category by multiplying the category probability by the sample size. category probability by the sample size.

4 4 Slide © 2008 Thomson South-Western. All Rights Reserved Hypothesis (Goodness of Fit) Test for Proportions of a Multinomial Population 4. Compute the value of the test statistic. 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. 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. f i = observed frequency for category i e i = expected frequency for category i k = number of categories where:

5 5 Slide © 2008 Thomson South-Western. All Rights Reserved Hypothesis (Goodness of Fit) Test for Proportions of a Multinomial Population where  is the significance level and there are k - 1 degrees of freedom p -value approach: Critical value approach: Reject H 0 if p -value <  5. Rejection rule: Reject H 0 if

6 6 Slide © 2008 Thomson South-Western. All Rights Reserved Multinomial Distribution Goodness of Fit Test n Example: Finger Lakes Homes (A) Finger Lakes Homes manufactures Finger Lakes Homes manufactures four models of prefabricated homes, four models of prefabricated homes, a two-story colonial, a log cabin, a a two-story colonial, a log cabin, a split-level, and an A-frame. To help split-level, and an A-frame. To help in production planning, management in production planning, management would like to determine if previous would like to determine if previous customer purchases indicate that there customer purchases indicate that there is a preference in the style selected. is a preference in the style selected.

7 7 Slide © 2008 Thomson South-Western. All Rights Reserved Split- A- Split- A- Model Colonial Log Level Frame # Sold The number of homes sold of each The number of homes sold of each model for 100 sales over the past two years is shown below. Multinomial Distribution Goodness of Fit Test n Example: Finger Lakes Homes (A)

8 8 Slide © 2008 Thomson South-Western. All Rights Reserved n Hypotheses Multinomial Distribution Goodness of Fit Test where: p C = population proportion that purchase a colonial p C = population proportion that purchase a colonial p L = population proportion that purchase a log cabin p L = population proportion that purchase a log cabin p S = population proportion that purchase a split-level p S = population proportion that purchase a split-level p A = population proportion that purchase an A-frame p A = population proportion that purchase an A-frame H 0 : p C = p L = p S = p A =.25 H a : The population proportions are not p C =.25, p L =.25, p S =.25, and p A =.25 p C =.25, p L =.25, p S =.25, and p A =.25

9 9 Slide © 2008 Thomson South-Western. All Rights Reserved n Rejection Rule 22 2 Do Not Reject H 0 Reject H 0 Multinomial Distribution Goodness of Fit Test With  =.05 and k - 1 = = 3 k - 1 = = 3 degrees of freedom degrees of freedom if p -value Reject H 0 if p -value

10 Slide © 2008 Thomson South-Western. All Rights Reserved n Expected Frequencies n Test Statistic Multinomial Distribution Goodness of Fit Test 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 = = 10

11 Slide © 2008 Thomson South-Western. All Rights Reserved Multinomial Distribution Goodness of Fit Test n Conclusion Using the p -Value Approach The p -value < . We can reject the null hypothesis. The p -value < . We can reject the null hypothesis. Because  2 = 10 is between and , the Because  2 = 10 is between and , the area in the upper tail of the distribution is between area in the upper tail of the distribution is between.025 and and.01. Area in Upper Tail  2 Value (df = 3) Note: A precise p -value can be found using Note: A precise p -value can be found using Minitab or Excel. Minitab or Excel. Note: A precise p -value can be found using Note: A precise p -value can be found using Minitab or Excel. Minitab or Excel.

12 Slide © 2008 Thomson South-Western. All Rights Reserved n Conclusion Using the Critical Value Approach Multinomial Distribution Goodness of Fit Test We reject, at the.05 level of significance, We reject, at the.05 level of significance, the assumption that there is no home style preference.  2 = 10 > 7.815

13 Slide © 2008 Thomson South-Western. All Rights Reserved Test 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. frequency, f ij, for each cell of the contingency table. 3. Compute the expected frequency, e ij, for each cell.

14 Slide © 2008 Thomson South-Western. All Rights Reserved Test of Independence: Contingency Tables 5. Determine the rejection rule. Reject H 0 if p -value <  or. 4. Compute the test statistic. where  is the significance level and, with n rows and m columns, there are ( n - 1)( m - 1) degrees of freedom.

15 Slide © 2008 Thomson South-Western. All Rights Reserved Each home sold by Finger Lakes 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. Contingency Table (Independence) Test n Example: Finger Lakes Homes (B)

16 Slide © 2008 Thomson South-Western. All Rights Reserved Price Colonial Log Split-Level A-Frame Price Colonial Log Split-Level A-Frame The number of homes sold for 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. > $99, < $99, Contingency Table (Independence) Test n Example: Finger Lakes Homes (B)

17 Slide © 2008 Thomson South-Western. All Rights Reserved n Hypotheses Contingency Table (Independence) Test H 0 : Price of the home is independent of the style of the home that is purchased 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 style of the home that is purchased

18 Slide © 2008 Thomson South-Western. All Rights Reserved n Expected Frequencies Contingency Table (Independence) Test Price Colonial Log Split-Level A-Frame Total Price Colonial Log Split-Level A-Frame Total < $99K > $99K Total Total

19 Slide © 2008 Thomson South-Western. All Rights Reserved n Rejection Rule Contingency Table (Independence) Test With  =.05 and (2 - 1)(4 - 1) = 3 d.f., Reject H 0 if p -value = = n Test Statistic

20 Slide © 2008 Thomson South-Western. All Rights Reserved n Conclusion Using the p -Value Approach The p -value < . We can reject the null hypothesis. The p -value < . We can reject the null hypothesis. Because  2 = is between and 9.348, the Because  2 = is between and 9.348, the area in the upper tail of the distribution is between area in the upper tail of the distribution is between.05 and and.025. Area in Upper Tail  2 Value (df = 3) Contingency Table (Independence) Test Note: A precise p -value can be found using Note: A precise p -value can be found using Minitab or Excel. Minitab or Excel. Note: A precise p -value can be found using Note: A precise p -value can be found using Minitab or Excel. Minitab or Excel.

21 Slide © 2008 Thomson South-Western. All Rights Reserved n Conclusion Using the Critical Value Approach Contingency Table (Independence) Test We reject, at the.05 level of significance, 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.  2 = > 7.815

22 Slide © 2008 Thomson South-Western. All Rights Reserved Goodness of Fit Test: Poisson Distribution 1. Set up the null and alternative hypotheses. H 0 : Population has a Poisson probability distribution H 0 : Population has a Poisson probability distribution H a : Population does not have a Poisson distribution H a : Population does not have a Poisson distribution 3. Compute the expected frequency of occurrences e i for each value of the Poisson random variable. for each value of the Poisson random variable. 2. Select a random sample and a. Record the observed frequency f i for each value of a. Record the observed frequency f i for each value of the Poisson random variable. the Poisson random variable. b. Compute the mean number of occurrences . b. Compute the mean number of occurrences .

23 Slide © 2008 Thomson South-Western. All Rights Reserved Goodness of Fit Test: Poisson Distribution 4. Compute the value of the test statistic. f i = observed frequency for category i e i = expected frequency for category i k = number of categories where:

24 Slide © 2008 Thomson South-Western. All Rights Reserved where  is the significance level and there are k - 2 degrees of freedom there are k - 2 degrees of freedom p -value approach: Critical value approach: Reject H 0 if p -value <  5. Rejection rule: Reject H 0 if Goodness of Fit Test: Poisson Distribution

25 Slide © 2008 Thomson South-Western. All Rights Reserved n Example: Troy Parking Garage In studying the need for an 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

26 Slide © 2008 Thomson South-Western. All Rights Reserved A random sample of 100 one- 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. Goodness of Fit Test: Poisson Distribution n Example: Troy Parking Garage # Arrivals Frequency

27 Slide © 2008 Thomson South-Western. All Rights Reserved n Hypotheses Goodness of Fit Test: Poisson Distribution H a : Number of cars entering the garage during a one-minute interval is not Poisson distributed one-minute interval is not Poisson distributed H 0 : Number of cars entering the garage during a one-minute interval is Poisson distributed a one-minute interval is Poisson distributed

28 Slide © 2008 Thomson South-Western. All Rights Reserved n Estimate of Poisson Probability Function Goodness of Fit Test: Poisson Distribution  otal Arrivals = 0(0) + 1(1) + 2(4) (1) = 600 Hence, Estimate of  = 600/100 = 6 Total Time Periods = 100

29 Slide © 2008 Thomson South-Western. All Rights Reserved n Expected Frequencies Goodness of Fit Test: Poisson Distribution x f ( x ) nf ( x ) Total x f ( x ) nf ( x )

30 Slide © 2008 Thomson South-Western. All Rights Reserved n Observed and Expected Frequencies Goodness of Fit Test: Poisson Distribution i f i e i f i - e i i f i e i f i - e i or 1 or or more

31 Slide © 2008 Thomson South-Western. All Rights Reserved n Test Statistic Goodness of Fit Test: Poisson Distribution With  =.05 and k - p - 1 = = 7 d.f. With  =.05 and k - p - 1 = = 7 d.f. (where k = number of categories and p = number of population parameters estimated), Reject H 0 if p -value n Rejection Rule

32 Slide © 2008 Thomson South-Western. All Rights Reserved n Conclusion Using the p -Value Approach The p -value > . We cannot reject the null hypothesis. There is no reason to doubt the assumption of a Poisson distribution. The p -value > . We cannot reject the null hypothesis. There is no reason to doubt the assumption of a Poisson distribution. Because  2 = is between and in the Chi-Square Distribution Table, the area in the upper tail Because  2 = is between and in the Chi-Square Distribution Table, the area in the upper tail of the distribution is between.90 and.10. Area in Upper Tail  2 Value (df = 7) Goodness of Fit Test: Poisson Distribution Note: A precise p -value can be found Note: A precise p -value can be found using Minitab or Excel. using Minitab or Excel. Note: A precise p -value can be found Note: A precise p -value can be found using Minitab or Excel. using Minitab or Excel.

33 Slide © 2008 Thomson South-Western. All Rights Reserved Goodness of Fit Test: Normal Distribution 1. Set up the null and alternative hypotheses. 3. Compute the expected frequency, e i, for each interval. 2. Select a random sample and a. Compute the mean and standard deviation. a. Compute the mean and standard deviation. b. Define intervals of values so that the expected b. Define intervals of values so that the expected frequency is at least 5 for each interval. frequency is at least 5 for each interval. c. For each interval record the observed frequencies c. For each interval record the observed frequencies

34 Slide © 2008 Thomson South-Western. All Rights Reserved 4. Compute the value of the test statistic. Goodness of Fit Test: Normal Distribution 5. Reject H 0 if (where  is the significance level and there are k - 3 degrees of freedom). and there are k - 3 degrees of freedom).

35 Slide © 2008 Thomson South-Western. All Rights Reserved Normal Distribution Goodness of Fit Test n Example: IQ Computers IQIQ IQ Computers (one better than HP?) 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.05 significance level, if the annual sales volume (number of units sold by a salesperson) follows a normal probability distribution.

36 Slide © 2008 Thomson South-Western. All Rights Reserved A simple random sample of 30 of the salespeople was taken and their numbers of units sold are below. Normal Distribution Goodness of Fit Test n Example: IQ Computers (mean = 71, standard deviation = 18.54) IQIQ

37 Slide © 2008 Thomson South-Western. All Rights Reserved n Hypotheses Normal Distribution Goodness of Fit Test H a : The population of number of units sold does not have a normal distribution with does not have a normal distribution with mean 71 and standard deviation mean 71 and standard deviation H 0 : The population of number of units sold has a normal distribution with mean 71 has a normal distribution with mean 71 and standard deviation and standard deviation

38 Slide © 2008 Thomson South-Western. All Rights Reserved n Interval Definition Normal Distribution Goodness of Fit Test To satisfy the requirement of an expected To satisfy the requirement of an expected frequency of at least 5 in each interval we will divide the normal distribution into 30/5 = 6 equal probability intervals.

39 Slide © 2008 Thomson South-Western. All Rights Reserved n Interval Definition Areas = 1.00/6 =.1667 Areas = 1.00/6 = .43(18.54) = = (18.54) Normal Distribution Goodness of Fit Test

40 Slide © 2008 Thomson South-Western. All Rights Reserved n Observed and Expected Frequencies Normal Distribution Goodness of Fit Test Less than to to to to to to to to More than i f i e i f i - e i i f i e i f i - e i Total

41 Slide © 2008 Thomson South-Western. All Rights Reserved n Test Statistic With  =.05 and k - p - 1 = = 3 d.f. With  =.05 and k - p - 1 = = 3 d.f. (where k = number of categories and p = number of population parameters estimated), Reject H 0 if p -value n Rejection Rule Normal Distribution Goodness of Fit Test

42 Slide © 2008 Thomson South-Western. All Rights Reserved Normal Distribution Goodness of Fit Test n Conclusion Using the p -Value Approach The p -value > . We cannot reject the null hypothesis. There is little evidence to support rejecting the assumption the population is normally distributed with  = 71 and  = The p -value > . We cannot reject the null hypothesis. There is little evidence to support rejecting the assumption the population is normally distributed with  = 71 and  = Because  2 = is between.584 and in the Chi-Square Distribution Table, the area in the upper tail Because  2 = is between.584 and in the Chi-Square Distribution Table, the area in the upper tail of the distribution is between.90 and.10. Area in Upper Tail  2 Value (df = 3) A precise p -value can be found A precise p -value can be found using Minitab or Excel. using Minitab or Excel. A precise p -value can be found A precise p -value can be found using Minitab or Excel. using Minitab or Excel.

43 Slide © 2008 Thomson South-Western. All Rights Reserved End of Chapter 12