Test review for Data eval
Test review for Data eval BME 3721
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This 3 page Study Guide was uploaded by Kathleen Quijada on Friday February 13, 2015. The Study Guide belongs to BME 3721 at Florida International University taught by Wei-Chiang Lin in Spring2015. Since its upload, it has received 114 views. For similar materials see Data Evaluation Principals in Biomedical Sciences at Florida International University.
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Date Created: 02/13/15
Exam I hymn Qucsh39bn L In order to estimate the mean diameter of a particular variety of orange a sample of 64 oranges were selected and the sample mean was found to be 75cm with a sample standard deviation of 15cm 1 What is a 95 confidence interval for the population mean Please specify the assumptions used in yo ur calculation SaMplt mm 3 9539 CI For a popula on mans M d is a 015 chance l lxatJht population mun Cam b fennel limp 1M in rm CalcuttaPeel va dim Sample mam Ya 951 CI mean mi SE Trims 8 39 39 0quot ISL 1 conelth39on Samplc man dis lyibeth is n M69940 Mm a normal Alaninan 8 sample 3sz is orgc 715mi AA 727m 71300 873 2 If a 90 confidence interval is constructed it will be he 95 confidence interval A wider than C the same as 7 7 1 lt3 77 7 7 15 a a it l ml l Ll ml 3 If the 95 confidence interval is constructed using only 9 samples it would be the one constructed with 64 samples iihi i un B narrower than C the same as Quash on Z A researcher carried out a study to test the efficiency of a new drug X in treating high LDL cholestrol He recruited 16 patients in this study and measured their LDL cholestrol levels before and after 1 month of usage of drug X The data shows the average decrease in LDL cholesterol level after 1 month consumption of drug X is 20 and the corresponding standard deviation is 12 1 What type of significance test would you use to analysis this data set Paired 39t te t 2 What are the assumptions used in your significance test SamPle mean eliS lYilouh39on me a normal alludinan Small Popula 39oo n5 3 ab D n lt 3 8 12 3 What is the nu hypothesis of your significance test nquot 1 usage 0F olruj X overa mm will haw no impact ond e LDL cholesvol low 01ka lawnHi bondmp39h39bn of clnrvg X 4 What is the test results X o 05 e given Dr nl 1 gt for 2 171 to39ogj39 21753 we W u hull h iPDJWILS US94 ohlrvg X Dvorame OF usagn 9 m5 gtlt appwr 1 Aunasc 392th LDL chbleslwl lml ner MO 39l39h 9F Consumpan OFelvvgx Questh 3 1 What is the purpose of a one way ANOVA test 3 or more Ihdmem groups 1 QIinan 1 h kcoPquot of W null HyprbaSiS 2 What is the null hypothesis of a oneway ANOVA test gr 0 C s 71quot damplf mums ourL equal 061 P lt 005 CI 11m xedHuang had39i nc mums art104 e val 4 we reject 1 mm hypoHwSiS 3 Is normal distribution of data a required assumption in oneway ANOVA quotl U 4 You received the outcome of an ANOVA anaysis shown in the table below from your employee Source SS Di Mun SS T P W vavps libb 1 DD web 400 within GWWC L00 bi TO W loco 2L Would you accept this results Why or why not 123 became 1 MM in mm mm 4 Pym am good
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