Etext 8.46

8.46-A random sample of 10 miniature tootsie rolls was taken from a bag. Each piece was weighed on a rattling diminutive scale. The results in grams were: 3.087, 3.131, 3.241, 3.270, 3.353, 3.400, 3.411, 3.437, 3.477 (a) Construct a 90% sanction detachment for the true mean weight. Sample mean = 3.3048 Sample commonplace deviation = 0.13199 Standard error: 1.645*0.13199/Sqrt(10) = 0.06866 90% CI- 3.3048-0.06866 < u < 3.3048 + 0.06866 90% CI- 3.23614 < u < 3.37346 (b) What sample size would be necessary to retrieve the true weight with an error rate of +/- 0.03 grams with 90% potency? n = [1.645*0.13199/0.03]^2 = 7.42^2 = 52.38 Rounding up to merry-go-round for n = 53 (c) Discuss the factors which might cause disagreement in the weight of the tootsie rolls during manufacturing. Factors such as the amount of ingredients apply to pick out the tootsie rolls along with the humidity and temperature control. Also the political machine tolerances may be a factor in make variation during manufacturing. 8.

62- In 1992, the FAA conducted 86,991 pre-employment drug tests on the job applicants who were to be engaged in safety and security related jobs, and tack together that 1,143 were absolute. (a) Construct a 95% confidence interval for the creation proportion of positive drug tests. p-hat = 1143/86991 = 0.013139... E = 1.96*sqrt[(0.013139)*(0.98686)/86991] = 0.002393 CI is (0.013139 0.002393, 0.013139 + 0.00239) (b) why is the normality speculation not a line of work, notwithstanding the very small care for of p? The normality boldness was not a problem because the random sample was very ! large.If you want to get a full essay, order it on our website: BestEssayCheap.com

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