ISM_chapter12

# ISM_chapter12 - Chapter 12 Considerations in Designing...

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257 Chapter 12: Considerations in Designing Experiments 12.1 (See Example 12.1) Let n 1 = ( ) ( ) 90 5 3 3 2 1 1 + σ + σ σ = n = 33.75 or 34 and n 2 = 90 34 = 56. 12.2 (See Ex. 12.1). If n 1 = 34 and n 2 = 56, then 7111 . 56 25 34 9 2 1 = + = σ Y Y In order to achieve this same bound with equal sample sizes, we must have 9 n + 25 n = .7111 The solution is n = 47.8 or 48. Thus, it is necessary to have n 1 = n 2 = 48 so that the same amount of information is implied. 12.3 The length of a 95% CI is twice the margin of error: 2(1.96) 9 n 1 + 25 n 2 , and this is required to be equal to two. In Ex. 12.1, we found n 1 = (3/8) n and n 1 = (5/8) n , so substituting these values into the above and equating it to two, the solution is found to be n = 245.9. Thus, n 1 = 93 and n 2 = 154. 12.4 (Similar to Ex. 12.3) Here, the equation to solve is 2(1.96) 9 n 1 + 25 n 1 = 2. The solution is n 1 = 130.6 or 131, and the total sample size required is 131 + 131 = 262. 12.5 Refer to Section 12.2. The variance of the slope estimate is minimized (maximum information) when S xx is as large as possible. This occurs when the data are as far away from x as possible. So, with n = 6, three rats should receive x = 2 units and three rats should receive x = 5 units. 12.6 When σ is known, a 95% CI for β is given by xx S z σ ± β α 2 / 1 ˆ . Under the two methods, we calculate that S xx = 13.5 for Method 1 and S xx = 6.3 for Method 2. Thus, Method 2 will produce the longer interval. By computing the ratio of the margins of error for the two methods (Method 2 to Method 1), we obtain 3 . 6 5 . 13 = 1.464; thus Method 2 produces an interval that is 1.464 times as large as Method 1. Under Method 2, suppose we take n measurements at each of the six dose levels. It is not difficult to show that now S xx = 6.3 n . So, in order for the intervals to be equivalent, we must have that 6.3 n = 13.5, and so n = 2.14. So, roughly twice as many observations are required. 12.7 Although it was assumed that the response variable Y is truly linear over the range of x , the experimenter has no way to verify this using Method 2. By assigning a few points at x = 3.5, the experimenter could check for curvature in the response function.

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258 Chapter 12: Considerations in Designing Experiments Instructor’s Solutions Manual 12.8 Checking for true linearity and constant error variance cannot be performed if the data points are spread out as far as possible. 12.9 a. Each half of the iron ore sample should be reasonably similar, and assuming the two methods are similar, the data pairs should be positively correlated. b. Either analysis compares means. However, the paired analysis requires fewer ore samples and reduces the sample to sample variability. 12.10 The sample statistics are: d = .0217, s D 2 = .0008967. a. To test H 0 : μ D = 0 vs. H a : μ D 0, the test statistic is | t | = 6 0008967 . 0217 . = 1.773 with 5 degrees of freedom. Since t .025 = 2.571, H 0 is not rejected. b. From Table 5, .10 < p value < .20. c. The 95% CI is 6 0008967 . 571 . 2 0217 . ± = .0217 ± .0314.
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## This note was uploaded on 03/18/2009 for the course STA 4321 taught by Professor Staff during the Spring '08 term at University of Florida.

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ISM_chapter12 - Chapter 12 Considerations in Designing...

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