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) 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

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 The number of homes sold of each model for 100 sales over the past two years is shown below. k = 4

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

Goodness of Fit Test  =.05 (column) 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

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 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

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 m = 2

45 Price Colonial Log Split-Level A-Frame Total < $99K > $99K Total Price Colonial Log Split-Level A-Frame Total < $99K > $99K Total Expected Frequencies ( e i ) Observed Frequencies ( f i ) Independence Test

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

 =.05 (column) 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

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

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 Frequency  otal Arrivals = 0(0) + 1(1) + 2(4) + 3(10) (1) = 600 Estimate of  = 600/100 = 6 Total one-minute intervals = n = = 100

x f ( x ) n∙ f ( x ) x f ( x ) n∙ f ( x ) For x = Goodness of Fit Test: Poisson Distribution The hypothesized probability of x cars arriving during the time period is

i f i e i f i - e i or 1 or or more Goodness of Fit Test: Poisson Distribution

With  = 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

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

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 n = 33, x = 71.76, s = Goodness of Fit Test: Normal Distribution

z.z. k = 33/5 = 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

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

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

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

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

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

Observed and Expected Frequencies f i e i f i – e i (f i – e i ) 2 /e i Total LL UL ∞ Data Table   ∞ Goodness of Fit Test: Normal Distribution

 =.05 (column) 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