AGENDA Hypotheses concerning two means with small samples INFERENCES CONCERNING TWO MEANS SMALL SAMPLES Population mean and variance are unknown Summary statistics are taken from samples Samples are <30 Critical value and test statistic will involve the t distribution GENERAL HYPOTHESES PROCEDURE Identify Ho and Ha Determine level of significance (generally 0.05 or 0.01) Determine “critical value” criterion from level of significance Calculate “test statistic” Make decision Fail to reject Ho Reject Ho DIFFERENT VERSIONS OF THE T-TEST Two sample t-test assuming equal variances Two sample t-test assuming unequal variances AKA Smith-Satterthwaite Test We will learn how to determine if the variance is equal or unequal in chapter 8 Paired t-test TWO SAMPLE T-TEST ASSUMING EQUAL VARIANCES Use n1, n2 or both < 30 each Population means and variances are unknown Sample variances are assumed to be equal Assuming normality and independence of populations Calculate summary statistics xbar1, var1 xbar2, var2 TWO SAMPLE T-TEST EQUAL VARIANCES HYPOTHESES Null u1-u2=gamma, gamma can be = 0 Alternate u1-u2gamma, one sided u1-u2 not equal to gamma, two sided test EXAMPLE 1 T-TEST ASSUMING EQUAL VARIANCES Heat producing capacity of coal from two different mines Use alpha = 0.01 to test if the difference between the means of the mines are significant Summary statistics xbar1=8,230, var1=15,750 xbar2=7,940, var2=10,920 EXAMPLE 1 T-TEST ASSUMING EQUAL VARIANCES Hypotheses Ho: u1-u2=0 Ha: u1-u2 not equal to 0 Level of significance = 0.01 Criterion 9 degrees of freedom Reject if t> 3.250 or t<-3.250 Calculations t=4.19 EXAMPLE 1 T-TEST ASSUMING EQUAL VARIANCES Decision Test statistic of 4.19 is > critical value of 3.250 Reject Ho (coal is not the same between mines) TWO SAMPLE T-TEST ASSUMING UNEQUAL VARIANCES Use n1, n2 or both < 30 each Population variances are unknown Sample variances are assumed to be unequal Assuming normality and independence of populations Calculate summary statistics xbar1, var1 xbar2, var2 This test manipulates the degrees of freedom for the critical value to account for the unequal variances Note that this test is not described in the text, problem 7.70. TWO SAMPLE T-TEST UNEQUAL VARIANCES EXAMPLE 2 T-TEST ASSUMING UNEQUAL VARIANCES Damage to two different bumper guards on cars. Use alpha = 0.01 to test if the difference in cost to repair is significant Summary statistics xbar1=127.33, var1=597.87, n1=6 xbar2=129.00, var2=202.00, n2=6 EXAMPLE 2 T-TEST ASSUMING UNEQUAL VARIANCES Hypotheses Ho: u1-u2=0 Ha: u1-u2 not = 0 Level of significance = 0.01 Criterion Degrees of freedom=8.03 Round up or round down? Reject if t> 3.355 or t<-3.355 Calculations t= -0.144 EXAMPLE 2 T-TEST ASSUMING UNEQUAL VARIANCES Decision Test statistic of -0.144 is between +- 3.355 Cannot Reject Ho (bumper damage is the same) PAIRED T-TEST Use Compare “before” and “after” paired results n1 and n2 must be equal Assuming normality of populations Data is not independent, pairs are correlated Examples Worker knowledge before and after a training program Productivity before and after the installation of new equipment Measurement of samples between two instruments Calculate summary statistics Calculate the difference between the “before” and “after” values for each observation D bar is the average of the differences between obs varD is the variance of the differences between obs PAIRED T-TEST EXAMPLE 3 PAIRED T-TEST Weekly losses of worker hours due to accidents in 10 different industrial plants before and after a safety program Use alpha=0.05 to determine if the program is effective Difference in accidents Plant 1 2 3 4 5 6 7 8 9 10 Before 45, 73, 46, 124, 33, 57, 83, 34, 26, 17 After 36, 60, 44, 119, 35, 51, 77, 29, 24, 11 D is the difference in accidents for each plant before and after Dbar=5.2 VarD=4.08 EXAMPLE 3 PAIRED T-TEST Hypotheses Ho: uD=0 Ha: uD > 0 Level of significance = 0.05 Criterion 9 degrees of freedom Reject if t> 1.833 Calculations t=4.03 EXAMPLE 3 PAIRED T-TEST Decision Test statistic of 4.03 is > critical value of 1.833 Reject Ho (safety program is effective)