Math

8.46.) A random render of 10 miniature Tootsie Rolls was taken from a bag. Each piece was weighed on a significant 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) Construct a 90 sight confidence breakup for the true mean cargo. 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 exemplification sizing would be necessary to estimate the true weight down 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 promoters which skill cause fun in the weight of Tootsie Rolls during manufacture. The equipment that the Tootsie Rolls are make on could be a factor if they are non calibrated properly or as intended. The chroma of the equipment or the speed of the candy that is fed into the cutter likewise influences the weight. thither could als o be a weight variation if in that location is a fluctuation in the temperature. 8.62.

) In 1992, the FAA conducted 86,991 pre-employment do drugs tests on job applicants who were to be engaged in safety and security-related jobs, and found that 1,143 were positive. (a) Construct a 95 percent confidence interval 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 progeny 7 surmise not a problem, despite the very small time value of p? The normality assumption really is not a problem because the sample size is very lar ge. Even though the small value of p, p is n! ormally distributed by the Central Limit Theorem.If you necessity to tie a full essay, order it on our website: BestEssayCheap.com

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