Lecture 42 Section 14.3 Mon, Nov 19, 2007 Test of Goodness of Fit Lecture 42 Section 14.3 Mon, Nov 19, 2007

The Effect of the Sample Size What if the previous sample distribution persisted in a much larger sample, say n = 600? Would it be significant? 1 2 3 4 5 6 80 (100) 100 140 120 90 70

TI-83 – Goodness of Fit Test The TI-83 will not automatically do a goodness-of-fit test, but the TI-84 may.

TI-83 – Goodness of Fit Test The following procedure will compute 2. Enter the observed counts into list L1. Enter the expected counts into list L2. Evaluate the expression (L1 – L2)2/L2. Select LIST > MATH > sum and apply the sum function to the previous result, i.e., sum(Ans). The result is the value of 2.

The List of Expected Counts To get the list of expected counts, you may Store the list of hypothetical probabilities in L3. Multiply L3 by the sample size and store in L2.

The List of Expected Counts For example, if the probabilities for 3 categories are p1 = 0.25 , p2 = 0.44, and p3 = 0.31 and the sample size is n = 184, then Store {0.25, 0.44, 0.31} in L3. Compute L3*184 and store in L2. Do not round off. {0.25, 0.44, 0.31}*184 L2.

Case Study 8 (revisited) Who gets pulled over. Test the hypothesis that the distribution of the race of people who get pulled over matches the distribution of race in the population.

The Hypotheses Let H0 : p1 = 0.73, p2 = 0.20, p3 = 0.06. p1 = proportion of whites pulled over. p2 = proportion of blacks pulled over. p3 = proportion of Hispanics pulled over. H0 : p1 = 0.73, p2 = 0.20, p3 = 0.06. H1 : H0 is not true.  = 0.05.

The Observed Counts We have p1^ = 0.65, p2^ = 0.24, p3^ = 0.07. The sample size is n = 520,079. The observed counts are O1 = 0.65  520079 = 338051. O2 = 0.24  520079 = 124819. O3 = 0.07  520079 = 36406.

The Expected Counts Hypothetically, we have p1 = 0.73, p2 = 0.20, p3 = 0.06. In a sample size of size n = 520,079, the expected counts are E1 = 0.73  520079 = 379658. E2 = 0.20  520079 = 104016. E3 = 0.06  520079 = 31205.

The Counts We get the table Whites Blacks Hispanics 338051 (379658) 124819 (104016) 36406 (31205)

The Test Statistic The test statistic is Its value is

Compute the p-value In this example, df = 2. To find the p-value, use the TI-83 to calculate the probability that 2(2) would be at least as large as 9587.16. p-value = 2cdf(9587.16, E99, 2) = 0 (Too small to measure).

Conclusion We reject H0. We conclude that the rates at which the different races are pulled over are not in proportion to their representation in the population.

Conclusion Does mean that the police are practicing racial profiling? State demographics.