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Math Essay

566 words - 3 pages

Chapter Review questions 5 and 6 & Chapter Exercises 8.46 and 8.62

Chapter Review
5. State the main points of the Central Limit Theorem for a mean.

The most important result from sample mean is the Central Limit Theorem. The Central Limit Theorem states that a large sample size n, the sampling distribution of the mean approaches a normal distribution.

6. Why is population shape of concern when estimating a mean? What does sample size have to do with it?

Population shape is in concern when estimating a mean because if the shape relates to the frequency of values that can be achieved by complete sampling. Smaller samples can greatly differ from the population and give a false outcome or reading to how the population actually feels. As the sample size increases the ...view middle of the document...

Confidence interval =[pic]+ z*[pic]

sample mean = 3.3048 = μ
3.087 + 3.131 + 3.241 + 3.241 + 3.270 + 3.353 + 3.400 + 3.411 + 3.437 + 3.477 = 33.048 / 10 = 3.3048
Std. Deviation= 0.1320
z(90) = 1.645
Sample Size= N = 10

Confidence Interval=
μ + 1.645*σ = 3.3048 + 1.645*0.1320 = (3.236, 3.373)
√10 √10

(b) What sample size would be necessary to estimate the true weight with an error of ± 0.03 grams with 90 percent confidence?

1.645*σ = 0.03
1.645*σ = √N
N = (1.645*σ/0.03)2 = (1.645*0.1320/0.03)2 = 52.4 = 53 Samples

(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.)

To manufacture a Tootsie Roll there are several steps to production. Calibrating machines yearly will allow a time of conduciveness. Precision of the amount of ingredients in the mix, humidity and temperature control is in total effect and adds up to a variation in the weight.

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.
95% confidence interval:
E = z *√ [[pic](1-[pic])]

z = 1.96, x = 1,143, n = 86,991
[pic] = x/n = 1,143/86,991 = 0.0131
E = z *√ [[pic](1-[pic])]
E = 0.0131± 1.96 *√ [0.0131*(1-0.0131)]
E = 0.0131± 0.00076
E = (0.01234, 0.01386)

b) Why is the normality assumption not a problem, despite the very small value of p? (Data are from Flying 120, no. 11 [November 1993], p. 31.)
Normality assumption is not a problem because the sample size (n) is a large value of 86,991.

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