Res 341 Wk 5 E-Text

8.46 A random sample of 10 miniature Tootsie Rolls was interpreted from a bag. Each piece was weighed on a in truth 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) frame a 90 pct faith detachment for the true mean slant. The standard mistake is 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 sample size would be necessary to promise the true cant over with an error of ± 0.03 grams with 90 share confidence? n=[z*s/E]^2 n=[1.645*0.131989/0.03]^2 = 52.38; rounding up, n=53 (c) Discuss the ingredients which might let variation in the weight of Tootsie Rolls during manufacture. (Data are from a view by MBA student hydrogen Scussel.) There are legion(predicate) factors that can influence weight variation in the yield of Tootsie Rolls. The equipment that the Tootsie Rolls are manufactured on could be a factor if they ar e no. adjust properly or as required. The speed of the equipment or the al-Quran of the candy that is fed into the sculptor could also influence the weight. Additionally, in that respect could also be a weight variation if there is a fluctuation in the temperature. 8.

62 In 1992, the FAA conducted 86,991 pre-employment drug tests on stock applicants who were to be engaged in safety and security-related jobs, and effectuate that 1,143 were positive. (a) Construct a 95 percent confidence interval for the nation 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 normality assumption not a wor ry, patronage the very nice value of p? (Da! ta is from trajectory 120, no. 11 [November 1993], p. 31.) The normality assumption is not a problem because the sample size is very large. Additionally, even though the small value of p, p is normally distributed by the fundamental settle Theorem.If you want to get a full essay, piece it on our website: OrderEssay.net

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