STUDENT NAME: CALDERON FLORES IVAN EDUARDO MAJOR IN STRUCTURES SUBJECT: DESIGN OF EXPERIMENTS ASSIGNMENT 1: CHAPTER3 COMPARING TWO ENTITIES: REFERENCE DISTRIBUTIONS, TESTS, AND CONFIDENCE INTERVALS PROBLEMS 1-8 August 7th, 2015 Considering the readings: A)Wthout s-142= 8,6 B)With s-142= 3,4 AssumingUA>UB, analyzing the data in R Studio we get the following results: Two Sample t-test data:A and B t = 3.1305, df = 2, p-value = 0.04434 alternative hypothesis: true difference in means is greater than 0 95 percent confidence interval: 0.2353569 Inf sample estimates: mean of x mean of y7.0 3.5 Since p-value = 0.04434 < 0.05 the hypothesis UA>UB is taken as true, thereadingsofasbestosfiberarehigherwithoutthes-142,thereforewe could assume the s-142 is working, but given that we only have two readings using the chemical, and checking the historical results we can seethisresultshavebeenpreviouslyobtainedwithouttheuseofthe chemical, further testing is necessary for an accurate answer. She might have asked this question so she can be sure that the results obtained are valid for the study and there are no other factors that might have altered the readings affecting the comparation between them. We begin with a null hypothesis UA=UB stated by one group who says thereisnodifferenceandanalternativehypothesisUAUBstatedby the other group saying there is a difference. Analyzing the data with R studio checking if A is greater than B we get the following results: Two Sample t-test data:a and b t = 0.93326, df = 6, p-value = 0.1934 alternative hypothesis: true difference in means is greater than 0 95 percent confidence interval: -0.2434838Inf sample estimates: mean of x mean of y3.075 2.850 With p-value = 0. 1934 > 0.05, the null hypothesis is accepted therefore there is not a significant difference between this readings. We begin stating a null hypothesis UA=UB and an alternative hypothesis UAUB Analyzing the data with R studio we get the following results: Two Sample t-test data:PLATEA and PLATEB t = 1.2111, df = 12, p-value = 0.1246 alternative hypothesis: true difference in means is greater than 0 95 percent confidence interval: -0.06286564 Inf sample estimates: mean of x mean of y 1.2087141.075429 With p-value = 0.1246> 0.05, the null hypothesis is accepted therefore there is not a significant difference between this readings. Thereisanapparentsystematicdifferencevisibleinsomeofthe readings but after making the t-test shows theres no difference.