Question 138258: Hello again, need assistance please!! THANKS A BUNCH - UR a LIFESAVER!!
1. 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.
(b) What sample size would be necessary to estimate the true weight with an error of ± 0.03 grams with 90 percent confidence?
(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.)
2. 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.
(b) Why is the normality assumption not a problem, despite the very small value of p?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1. 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.
(b) What sample size would be necessary to estimate the true weight with an error of ± 0.03 grams with 90 percent confidence?
(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.)
Go to www.algebra.com and search for 136270
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2. 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.
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p-hat = 1143/86991 = 0.01314...
(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
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(b) Why is the normality assumption not a problem, despite the very small value of p?
The sample is random
The sample size is less than 10% of the population
etc.
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Cheers,
Stan H.
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