Etext 8.46

8.46-A random sample of 10 miniature tootsie rolls was taken from a bag. Each piece was weighed on a genuinely lead 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 breakup for the true mean weight. Sample mean = 3.3048 Sample well-worn 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 surmise the true weight with an error rate of +/- 0.03 grams with 90% assurance? n = [1.645*0.13199/0.03]^2 = 7.42^2 = 52.38 Rounding up to conduct for n = 53 (c) Discuss the factors which might cause variate 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 railway car tolerances may be a factor in create va riation 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 implant that 1,143 were absolute. (a) Construct a 95% confidence interval for the macrocosm 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, in spite of the very small care for of p? The normality assumption was not a problem because the random sa mple was very large.If you want to get a ful! l essay, order it on our website: BestEssayCheap.com

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