Statistical Hypothesis Testing Part 3 q A statistical

Goodness-of-Fit Tests q Procedures for confirming or refuting hypotheses about the distributions of random

Goodness of Fit Tests (cont. ) q Test statistic is χ2 Ø Draw the

Tests of Independence (without computer) q Example: Worker type and Choice of pension plan

Worker vs. Pension Plan Example Worker Type Salaried Hourly Total Pension Plan #1 #2

Hypotheses Recall the general format of the hypotheses H 0: the categories (worker &

The Chi-Squared Test of Independence 4. Compare to the critical statistic for a test

Chi-Square Test Using Minitab salaried Minitab Input: hourly plan 1 plan 2 plan 3

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Statistical Hypothesis Testing - Part 3 q A statistical hypothesis is an assertion concerning one or more populations. q In statistics, a hypothesis test is conducted on a set of two mutually exclusive statements: H 0 : null hypothesis H 1 : alternate hypothesis q New test statistic of interest: EGR 252 S 10 JMB Ch. 10 Part 3 Slide 1

Goodness-of-Fit Tests q Procedures for confirming or refuting hypotheses about the distributions of random variables. q Hypotheses: H 0: The population follows a particular distribution. H 1: The population does not follow the distribution. Example: H 0: The data come from a normal distribution. H 1: The data do not come from a normal distribution. EGR 252 S 10 JMB Ch. 10 Part 3 Slide 2

Goodness of Fit Tests (cont. ) q Test statistic is χ2 Ø Draw the picture Ø Determine the critical value for goodness of fit test χ2 with parameters α, ν = k – 1 q Calculate χ2 from the sample q Compare χ2 calc to χ2 crit q Make a decision about H 0 q State your conclusion. q Discussion: Look at Table 10. 4 in text. EGR 252 S 10 JMB Ch. 10 Part 3 Slide 3

Tests of Independence (without computer) q Example: Worker type and Choice of pension plan q Hypotheses H 0: Pension Plan Choice and Worker Type are independent H 1: Pension Plan Choice and Worker Type are not independent 1. Develop a Contingency Table of Observed Values Pension Plan Worker Type Salaried Hourly Total #1 #2 #3 160 40 200 140 60 200 40 60 100 EGR 252 S 10 JMB Ch. 10 Part 3 Total 340 160 500 Slide 4

Worker vs. Pension Plan Example Worker Type Salaried Hourly Total Pension Plan #1 #2 160 140 40 60 200 #3 40 60 100 Total 340 160 500 2. Calculate expected probabilities. Multiply by total observations to determine expected values for each cell. P(#1 ∩ S) = P(#1)*P(S) = (200/500)*(340/500)=0. 272 P(#1 ∩ H) = P(#1)*P(H) = (200/500)*(160/500)=0. 128 #1 S (exp. ) 136 H (exp. ) 64 EGR 252 S 10 JMB Ch. 10 Part 3 #2 0. 272*500 = 136 0. 128*500 = 64 #3 136 64 Slide 5

Hypotheses Recall the general format of the hypotheses H 0: the categories (worker & plan) are independent H 1: the categories are not independent 3. Calculate the sample-based statistic (160 -136)^2/136 + (140 -136)^2/136 + (40 -68)^2/68 + (40 -64)^2/64 + (60 -32)^2/32 = 49. 63 EGR 252 S 10 JMB Ch. 10 Part 3 Slide 6

The Chi-Squared Test of Independence 4. Compare to the critical statistic for a test of independence, χ2α, r where r = (a – 1)(b – 1) a = # of columns b = # of rows For our example, let’s use α = 0. 01 _ χ20. 01, 2 = 9. 210 (from Table A. 5, pp 756) Comparison: χ2 calc > χ2 crit Decision: Reject the null hypothesis Conclusion: Worker and plan are not independent. ****Note that we are stating that there is an association between the two categories. We are not claiming a cause and effect relationship. EGR 252 S 10 JMB Ch. 10 Part 3 Slide 7

Chi-Square Test Using Minitab salaried Minitab Input: hourly plan 1 plan 2 plan 3 160 140 40 40 60 60 Minitab Output: Chi-Square Test: plan 1, plan 2, plan 3 Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts 1 plan 2 plan 3 160 140 40 136. 00 68. 00 4. 235 0. 118 11. 529 Total 340 2 40 64. 00 9. 000 60 64. 00 0. 250 60 32. 00 24. 500 160 Total 200 100 500 Chi-Sq = 49. 632, DF = 2, P-Value = 0. 000 EGR 252 S 10 JMB Ch. 10 Part 3 Slide 8