EGR 252 S10 JMB Ch.10 Part 3 Slide 1 Statistical Hypothesis Testing - Part 3  A statistical hypothesis is an assertion concerning one or more populations.  In statistics, a hypothesis test is conducted on a set of two mutually exclusive statements: H 0 : null hypothesis H 1 : alternate hypothesis  New test statistic of interest:

EGR 252 S10 JMB Ch.10 Part 3 Slide 2 Goodness-of-Fit Tests  Procedures for confirming or refuting hypotheses about the distributions of random variables.  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 S10 JMB Ch.10 Part 3 Slide 3 Goodness of Fit Tests (cont.)  Test statistic is χ 2  Draw the picture  Determine the critical value for goodness of fit test χ 2 with parameters α, ν = k – 1  Calculate χ 2 from the sample  Compare χ 2 calc to χ 2 crit  Make a decision about H 0  State your conclusion.  Discussion: Look at Table 10.4 in text.

EGR 252 S10 JMB Ch.10 Part 3 Slide 4 Tests of Independence (without computer)  Example:Worker type and Choice of pension plan  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 Worker Type Pension Plan Total #1#2#3 Salaried Hourly Total

EGR 252 S10 JMB Ch.10 Part 3 Slide 5 Worker vs. Pension Plan Example 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)= *500 = 136 P(#1 ∩ H) = P(#1)*P(H) = (200/500)*(160/500)= *500 = 64 Worker Type Pension Plan Total #1#2#3 Salaried Hourly Total #1#2#3 S (exp.)136 H (exp.) 64

EGR 252 S10 JMB Ch.10 Part 3 Slide 6 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 ( )^2/136 + ( )^2/136 + (40-68)^2/68 + (40-64)^2/64 + (60-64)^2/64 + (60-32)^2/32 = 49.63

EGR 252 S10 JMB Ch.10 Part 3 Slide 7 The Chi-Squared Test of Independence 4. Compare to the critical statistic for a test of independence, χ 2 α, r where r = (a – 1)(b – 1) a = # of columns b = # of rows For our example, let’s use α = 0.01 _ χ ,2 = (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.

Chi-Square Test Using Minitab plan1plan2plan3 salaried hourly EGR 252 S10 JMB Ch.10 Part 3 Slide 8 Minitab Output: Chi-Square Test: plan1, plan2, plan3 Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts plan1 plan2 plan3 Total Total Chi-Sq = , DF = 2, P-Value = Minitab Input: