Question

Chapter 8, Section 2, Exercise 042 Consider the data with analysis shown in the following computer output: Level Mean z in StDev 3.061 2.168 2.387 10.000 16.800 10.800 u u Source DF SS MS F P Groups 2 138.13 69.07 10.48 0.002 12 79.07 6.59 Error Total 14 217.20 What is the pooled standard deviation? What degrees of freedom are used in doing inferences for these means and differences in means? Round your answer for the pooled standard deviation to two decimal places. The pooled standard deviation is degrees of freedom=

Answer 1

following own in the aven Consider the data with analysis shown in computer output : Level N Mean StDev A $ 10.000 3.061 16-800 2.168 C 5 10-800 2.387 Here, nA = Mg = m = 5 and $A= 3.061, 88=2.168 and yo = 2.387. where $AJ $p and we are standard deviations of levels A, B and C respectively and magha and ne are the rrumber of levels of AgB and C respectively. Now we have to find the pooled standard deviation. Thus, pooled standard deviation = 18 = (14-1) 4+ (-4) $+(me=1);- V (4-1)+(re-1)+(re-4) = (6-1)(3-061) +(6-1) 2.468)2+(5-12-287) v (5-1)+(-4) +(5-1) | 4x (3-061+ 4x(Q.168)2 + 4x (2-387) 3-061.82 +(2-168)+(2.2872 2.66696 2-67. Thus, pooled standard deviation $$ = 2.57. Now we have to find out the degrees of freedom which are used in deing inferences for these means and differences in means.

Hereg another information is given which is as follows: Source DF $$ Ms F P. - Groups & 138-13 69-07 10-48 0.002, Error 12 74.07 6-59 Total 14 217-20 Heren from the given information, we can say that the method of analysis of variance or A NOVA has been used to do inferences for the means and differences in means. From the earlier given information, we get MA = 10.000 Mo = 16-800 He = 10-800 where Mag Me and the are means of levels A, B and e respectively. This to make inferences for the means and differences in means the test statistie used is F= MS (Greups) which follone MS(Erpor) F distribution with degrees of freedom (k-1) and (mk), where kis number of groups of differences in means and nis total number of levele. Here, k=3[ sirice ("A-16), ("A-1) and (148-16) are the 3 groups of differences. and n = mat ng the = 15. Thus, degrees of freedom that are used are (3-1)=2 and (n-k) = (15-3) = 12.

Thus, the pooled standard deviation is 2.57. and the degrees of freedom thast are used are 2 and 12.