368 Words2 Pages

8.46 A random sample of 10 miniature Tootsie Rolls was taken from a bag. Each piece was weighed on a very 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 percent confidence interval for the true mean weight.
The standard error 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 estimate the true weight with an error of ± 0.03 grams with 90 percent 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 might cause variation in the weight of Tootsie Rolls during manufacture. (Data are from a project by MBA student Henry Scussel.)
There are many factors that can influence weight variation in the production of Tootsie Rolls. The equipment that the Tootsie Rolls are manufactured on could be a factor if they are not calibrated properly or as required. The speed of the equipment or the volume of the candy that is fed into the cutter could also influence the weight. Additionally, there 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 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 normality assumption not a problem, despite the very small value of p? (Data is from Flying 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

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