Biometrics BIO272 Final Exam (spring 2017) Total 105pts the following informatio
ID: 86582 • Letter: B
Question
Biometrics BIO272 Final Exam (spring 2017) Total 105pts the following information to get full creditel am, be sure to incl Use SPss for this why you chose 5%, expl using a 0.05 significance). Assume that the following data randomly sampled from anormal population o s02, and o 649. body mase (ngl each my o soa, 642, 0661 maes 500 10 pts) a nes drawn from a of mean body 0 e53.0 684, 0693. is Another randomly and asured (in 707, med 0.712, and 0.638 Test whether these two samples were selected from the populations ONE and explain why, your decision is based on some analysis, show that resut to support your 2 The following table lists tota hot cocoa sales for five stores dring four times of the year 85 3.0 94 92 10.1 7.9 (a) Appy an analysis of variance to this data, report your complete ANOVA table. (10 pes) cocoa sales are the same for all four months surveyed Assume that the March is a control group an te Use appropriate to compare the control mean hot cocoa consumption with the other means. (10 prs) 294 80 170 185 220 10.6 574 6.8 065 306 8.9 35.6 11.1 208 748 06 13 1 108 59 98 0.5 10.8 5.5 (a) Fit a simple 29 (b) and linear model of Y on one independent variable xaonly, report (1) Y intercept Fit a regression coefficient (2) test Ho (10 pts) the linear model of Y on ALL Xs report (1) the Y intercept and partial regression coefficients, (2) testif there is a significant multiple regression relationship between Y and Xs (3) Bs 0 (for X1) (15 pts)
Explanation / Answer
1) a) The mean body mass of the given sample of flies is 0.612 mg and standard deviation is 0.037 mg.The 95% confidence intreval is decsribed by mean + 2SD. This 95% confidence intreval ranges from 0.538 mg and 0.684 mg.
95% confidence intreval does not include 0.500 mg. So,this sample fly population is not drawn froma population with mean body mass of 0.500 mg.
b) The mean and standard deviation of the given sample of flies are 0.671 mg and 0.038 mg. The sample1 and 2 are compared by unpaired t test and p-value was found to be 0.01 which is significant. Therefore, the two samples are not selected from the population with equal means.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.