Question

5. A manager wants to raise productivity at her company by increasing the speed at which her employees can use SPSS. She employs an external agency which provides SPSS training. The external agency offers 4 courses: Beginner, Intermediate I, Intermediate II, and Advanced. The manager is unsure which course is needed for the type of work being done at her company, so she sends 3 employees on the Beginner course, 3 employees on the Intermediate I course, 3 employees on the Intermediate Il course, and 3 employees on the Advanced course. When all employees return from their trainings, the manager gives them a problem to solve using SPSS and times how long it takes each employee to complete the problem. She then compares the times based on which course each employee attended. Are there differences in the average time to complete the problem at a 0.05 level of significance? Beginner 22 25 28 SPSS Course Intermediate Intermediate II Advanced 18 17 18 21 1914 26 20 16 Time to complete problem (minutes) Complete the ANOVA Table below (please use the blank page at the back of the quiz for your manual calculations - SHOW YOUR WORK!!), and below the ANOVA Table, write your conclusion and interpretation of the results of your ANOVA test. Source of Variation ANOVA Table: df sd MS Fstatistic Between Samples Within Samples Total (leave blank) Critical F (F): Conclusion & Interpretation:

Answer 1

We use one way ANOVA to solve this problem. A NOVA TABLE FOR ONE-WAY CLASSIFICATION IS: Source of Sum of Degrees of variation Mean I squarex (ss) freedomid square (ms) F-glatistic Between SSB SSB/K-1 Samples Enilyn-52 k-la. Within SS WE 6. MSWA EE(Yin-Y:.) N-k Samples IN-k = MSW SST=) Tolal F, SSB EMSB Sswl Esly;-)2 N-1 - 37. where k- Total No reatments & č.26 = 8 de autowered N - Total No. of observations "ni - No: of observations receiving ith treatment. Ji. = n Hace 3 & Jij de 31,2,.. } = &nihil & nig) 318 AT Arona Let us write the hypothesis : Hol Null hypothesis): There is no significant difference in the average time to write complete the problem, v/s 86.081 H. (Alternative hypothesis) There is a significant difference in the average time to s complete the problem. In the given design, K= 4 [ie. Beginner, Interinediate I and II, Advanced n = n₂ = n₃ = 114=3 laterdreht breo maintinos loped car eve tuer su NEM: 12 wolnie w ww verland ente ro of event nens ar

v;8, -ũ)? Time to com- -plete problem nigi Beginner 22/125 28 175 25 196/3 AT Ave Intermediate 182126 765 65/3 16/ 30 I. Intermediate 1/17 82 D 19 20 56 56/3 25/3 Advanced 18 14 16 48 16 169/3² Driftin WEM: &niji, &nilji. - 7 ) 2 elfneod. = 135.3333 P- 33 = 244 8 2129 20.353 35B = 135.33 zilveritat toon laotis weer 55T : LE Cyij - Z) 2 198.6667 Hij is theith obscuration breceiving ith treatment 55w=55T-55 B = 63.3334 ST ANOVA TABLE: (Refer above) NS 5 ш 9 Variation Start der IMSF -statistic Between samples t Within sampler Total Total 135.333 45.11 5.6982 63.3334 II 8 und 7.9167 Tilawala H. 198.6667 11 Hvad Critical F (F.): F(x:3, 8) = 4.066 (where &= 0.05) Conclusion and Interpretation: Ftable > For implies we reject the null hypothesis and we can say that there is a significant difference in the average time to complete the problem at 0.05 level of significance.