Question

5- To compare two programs for training industrial workers to perform a skilled job, 20 workers are included in an experiment. Of these, 10 are selected at random and trained by method 1; the remaining 10 workers are trained by method 2. After completion of training, all the workers are subjected to a time-and-motion test that records the speed of performance of a skilled job. The following data are obtained. Method 1 15 23.222 20 31.222 11 13.222 Time (minutes) 23 16 21 19.222 23.222 17.222 18 28.222 16 27 26.222 25.222 24 28.222 Method 2 a (a Can you conclude from the data that the mean job time is significantly less after training with method 1 than after training with method 2? (Test with a = .05.) (b) State the assumptions you make for the population distributions. Construct a 95% confidence interval for the population mean difference in job times between the two methods.

Answer 1

alven, n=10 * = 19.) SI = 4.82 n = 10 Fra 23:52 Az = 5.56 df = (silom + kulma) mt 12- df = (4.827110 + 5.56/162 (4.8270)² + (5:56/10) df ~ let los 18 be d= 0.05 Ho: M = M2 Hi Mi a la under Ho Estata XI-X2 t = 19.1- 23:52 t = -1.899 0.0369 p-value =

since p-value <a, we refect Ho. we can conclude that time for Methel I ps for method B. mean sob less than b) - The two sample data are continous. 7 The data follows normal destoibution two samples are Endependent → The 1 The samples are simple random samples taken from the population. het los be da 0.05 trap=2.101 df - 18 95).CI - (*) - x2) It = (1901-23.52) + 2.00 / 4.82% 5.562 10 = = -4.42 + 4.89 (-9.31, 0.47) interval spoce 'o ps included in the we fail to reject Ho.