# 1 Chi-Square Test -- X 2 Test of Goodness of Fit.

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1 Chi-Square Test -- X 2 Test of Goodness of Fit

2 (Pseudo) Random Numbers  Uniform: values conform to a uniform distribution  Independent: probability of observing a particular value is independent of the previous values  Should always test uniformity

3 Test for Independence  Autocorrelation Test Tests the correlation between successive numbers and compares to the expected correlation of zero e.g. 2 3 2 3 2 4 2 3  There is a correlation between 2 & 3 We won’t do this test  software available

4 Hypotheses & Significance Level  Null Hypotheses – Ho Numbers are distributed uniformly Failure to reject Ho shows that evidence of non-uniformity has not been detected  Level of Significance – α (alpha) α = P(reject Ho|Ho is true)

5 Frequency Tests (Uniformity)  Kolmogorov-Smirnov More powerful Can be applied to small samples  Chi Square Large Sample size >50 or 100 Simpler test

6 Overview  Not 100% accurate  Formalizes the idea of comparing histograms to candidate probability functions  Valid for large samples  Valid for Discrete & Continuous

7 Chi-Square Steps - #1  Arrange the n observations into k classes  Test Statistic: X 2 = Σ (i=0..k) ( O i – E i ) 2 / E i O i = observed # in i th class E i = expected # in i th class  Approximates a X 2 distribution with (k-s-1) degrees of freedom

8 Degrees of Freedom  Approximates a X 2 distribution with (k-s-1) degrees of freedom  s = # of parameters for the dist.  Ho: RV X conforms to ?? distribution with parameters ??  H1: RV X does not conform  Critical value: X 2 (alpha,dof) from table  Ho reject if X 2 > X 2 (alpha,dof)

9 X 2 Rules  Each Ei > 5  If discrete, each value should be separate group  If group too small, can combine adjacent, then reduce dof by 1  Suggested values n = 50, k = 5 – 10 n = 100, k = 10 – 20 n > 100, k = sqrt(n) – n/5

10 Degrees of Freedom  k – s – 1  Normal: s=2  Exponential: s = 1  Uniform: s = 0

11 X 2 Example  Ho: Ages of MSU students conform to a normal distribution with mean 25 and standard deviation 4.  Calculate the expected % for 8 ranges of width 5 from the mean.

12 X 2 Example  Expected percentages & values <10-15 = 2.5% 5 15-20 = 13.5%27 20-25 = 34%68 25-30 = 34%68 30-35 = 13.5%27 35-40> = 2.5%5

13 X 2 Example  Consider 200 observations with the following results: 10-15 = 1 15-19 = 70 20-24 = 68 25-29 = 41 30-34 = 10 35-40+ = 10

14 Graph of Data 10 15 20 25 30 35 70 60 50 40 30 20 10 0

15 X 2 Example  X 2 Values – (O-E) 2 /E 10-15 = (5-1) 2 /53.2 15-20 = (27-70) 2 /2768.4 20-25 = (68-68) 2 /680 25-30 = (68-41) 2 /6810.7 30-35 = (27-10) 2 /2710.7 35-40+ = (5-10) 2 /45 Total98

16 X 2 Example  DOF = 6-3 = 3  Alpha = 0.05  X 2 table value = 7.81  X 2 calculated = 98  Reject Hypothesis

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