Question

CAN ANYONE PLEASE HELP ME WITH THIS STAT QUESTION?? Thumbs up will be given if done correctly. THANK YOU!

Many furniture pieces are built with veneer, a thin layer of fine wood that is fastened to less expensive wood products underneath. Face checks are cracks that sometimes develop in the veneer. When face checks appear, the furniture needs to be reconstructed. Because this is a fairly expensive process, researchers seek ways to minimize the occurrence of face checks by controlling the manufacturing process. In one study, the type of adhesive used was one of the factors examined. Because of the way that the veneer is cut, it has two different sides, called loose and tight, either of which can face out. Here are tables giving the numbers of veneer panels with and without face checks for two different adhesives, PVA and UF. Separate tables are given for the loose side and the tight side. Loose side Tight side Face checks Face checks Adhesive No Yes No Yes PVA 10 54 44 20 UF 21 43 37 27 Analyze the data. Write a summary of your results concerning the relationship between the adhesive and the occurrence of face checks. Be sure to include numerical and graphical summaries as well as the details of your significance tests.

I know I have to perform the Chi-squared statistic test, but I am having trouble interpreting the given table.

Answer 1

Saution Gaven Tght Side Face checks Total Laoke Side Face chechs No yes Adhesive Yes No 44 o PVA 31 43 UF 56 47 Total 81 3 I Value of Cach cell exfected Noa to find the We Knoa that The nuda of exPetee Value is Fiven by x Colum total Voc total Exfedted Value Gramd total Side face checks Fa ce checks TIjht Side Loe Total yes Adlesve yes No 28x 3 PVA -48.5| s6 =40.5 3.5 IS. 5 256 128 x 4 0.5 OF 15.5 23.5 56 256 256 8 31 Total

find t we knoa that Nos to Chi-s2 uane test statistic statistic is Jiven_ by The fomula of Chi- S2uae test 3-)3 Loe side Fa ce checks TTht side Fa ce checks Total No yes yes Ad hesve No |he-15.5)2 (54-48.5) 40.5)0-3-s)2 PVA 5-5 48-5 405 3.5 3.399 t516 0.623 o.3625 ニ0.5213 (i43-48-5) (1-4s-5)27-23,5)- 21-5-5) UF IS-5 o.5 48-5 3.5 3.391 0.623 0-Sa13 6.7982 -ol26 O.605 3.9032 Total 6.1982 Noc to find the dejoecs of fneedom (-) (o) C2-) Noa to find the f-value teae using the Values of chi-s2uase test Statistic df (3) then Cue get th Pvalue and fvalue .0786 --

Decision P-Value (ou078 6) Thevrefae the dlo0s) inu y Pothe sis Ho) s ince Ge fail to eject the PValue eate than the 0.05 level of Sianitiance.