Math

8.46.) A random represent of 10 miniature Tootsie Rolls was interpret from a bag. Each scrap was weighed on a square accurate scale. The results in grams were 3.087 3.131 3.241 3.241 3.270 3.353 3.400 3.411 3.437 3.477 (a) effect a 90 set confidence breakup for the consecutive mean cantiness. E = 1.96(s/sqrt(n)) = 1.96[0.131989/sqrt(10)]=1.96*0.41739 =0.081808 C.I. = (x-bar-E,x-bar+E) = (3.3048-0.0818,3.3048+0.0818) (b) What representative size would be requisite to estimate the true weight with an error of ± 0.03 grams with 90 pct confidence? n=[z*s/E]^2 n=[1.645*0.131989/0.03]^2 = 52.38; rounding up, n=53 (c) Discuss the factors which adroitness cause variation in the weight of Tootsie Rolls during manufacture. The equipment that the Tootsie Rolls are let on could be a factor if they are non correct properly or as intended. The volume of the equipment or the step on it of the candy that is fed into the sculptor alike influences the weight. thither could also be a weight variation if in that localization of function is a fluctuation in the temperature. 8.62.

) In 1992, the FAA conducted 86,991 pre-employment drug tests on job applicants who were to be meshed in safety and security-related jobs, and lay down that 1,143 were compulsory. (a) Construct a 95 percent confidence breakup for the population proportion of positive drug tests.     E = 1.96*sqrt [0.01314*(1-0.01314)/86,991] = 0.0007567.. 95% CI: 0.79615, 83671 (b) Why is the atomic number 7 presumption non a problem, in spite of the very small succession hold dear of p? The normalcy assumption really is not a problem because the savor size is very large. yet though the small value of p, p is normally distributed by the Central Limit Theorem.If you indispensability to get a secure essay, order it on our website: Ordercustompaper.com

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