Question

To test whethoer the mean time needed to mix a batch of materlal is the same for machines produced by three manufacturers, a chemical company obtained the follewing data on the time (is misutes) needed to mix the material Manufacturer 1 2 3 19 28 19 27 26 20 23 32 23 23 30 (a) Use these data to test whether the population mean times for mbing a batch of material differ for the three manufacturers. Use e0.05 State the null and alternative hypotheses O Mo O Hoi At least two of the population means are equal. H At least two of the population means are different O Mo H Not all the population means are equal. O Ho O Ha Not all the population means are equal Find the value of the test statistic (Round your answer to two decimal places.) 957 AM

Answer 1

MI SSE SSW= 23 29 Mean: 23 29 20. And . To test. Ho: Mi= Me= M3 vs. Hi Not Source of Variation all pobed a means are equal. Between To test the above hypothesis the test sortistic is Within F M SB MSW Total wheres, MSB SSB K-1 þ-value MSW= SSH nou • p-value SSB: between Surn of squares . . {"cãi - 55) Conclusion: S&W= Within Sum of squares .. A.M Reject Ho the no Eni is not Hare, n= 12 , K= 3 . S8 e 23+29 +20 24

SSB= 4(23-24)2 + 4 ( 29 -24)² + 4(20-24) 2 168 SSW (19-23) + (22 - 23) + ... + (1920) ² + (18-20) 66 : ANOVA Source of Variation of ss MSF MS Between 162 168 84F284 7.33 2 11.45 Within 66 7.33 . Total 1 234 0.003 the meell bypotheses. þ-value & P[F> Fo] - " p-value < 0.05; so he seeject Conclession: He: There is sufficient the mean time needed to is not the same for each evidence to conclude that mix a leatch of material manufacturer.

6). LSD = JMS (t + ) X F.08; 2,9 +- 33 x 2 x 39 142 √3 665 x39.4 60914ectas 4 - 2 75.41 The pairwise absolute difference b/w sample means for manufacturer 1 and 3 is - 18-231+ 23-201 - 3. : u. (3 F1541) does not contain o. There is not a significant differene 6/10 the means for manufacturer 1 23.