Chapter 5

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107 line 16: \(= qE^2\) must be \(= 2E^2\)

107 line 13: from below under the first figure: Insert these lines: For the general case of this configuration we have \(\acute{E} = M\) and \(E_q = (E_{max} - M) = \frac{E_{max}-E_{min}}{2}, E_1 = (E_{min} - M) = \frac{-E_{max}-E_{min}}{2}\) and \( \sum\limits_{i=1}^{q} (E_i-\bar{E})^2 = 2(\frac{\delta}{2})^2=\frac{\delta^2}{2}\) with \( \delta = E_{max} - E_{min}\).

107 line 7: from below under the last figure: Insert these lines: (This corresponds with the D-optimal designs in linear regression). Hence for the general case of this configuration we have \( E = M, E_q = (E_{max} - M) = \frac{E_{max}-E_{min}}{2} \) and \(E_1 = (E_{min} - M ) = - \frac{E_{max}-E_{min}}{2} \) and = \(q(\frac{\delta}{2})^2 = \frac{q\delta^2}{4}\) .
Hence we have \(\lambda_{min}= \frac{\delta^2}{2\sigma^2}\) and \(\lambda_{max} = \frac{q\delta^2}{4\sigma^2}\) with \(\sigma = (Emax - Emin).\)

120 line 9: \( \lambda_a = \frac{bn}{\sigma^2} \sum (a_i – \acute{a})^2 \; \; \; \; \; \; \lambda_b = \frac{an}{\sigma^2} \sum (b_i – \acute{b})^2\)

120 line 11: \( \lambda_a = \frac{bn}{\sigma^2} \sum (a_i – \acute{a} )^2 \)

120 line 13: \( \lambda_b = \frac{an}{\sigma^2} \sum (b_i – \acute{b} )^2 \)

121 After second line of Problem 5.5 insert:
For a two-way classification without interaction and with equal subclass numbers the solution can be done as for a one-way classification with an example given in Problem 5.1. Only the calculated sample size for \(n\) for \(H_{0A}\) must then be divided by \(b\). As an example we choose \(\alpha = 0.05, \beta=0.1, a=4, b=2, \delta = 1\sigma\) and calculate the minimal and maximal subclass numbers for \(A\).

> size.anova(model="a",a=4,, alpha=0.05, beta=0.1, delta=1,case="minimin")
n
16


Hence minimal sample size \(n\) is \( \frac{16}{b} = \frac{16}{2} = 8\).

> size.anova(model="a",a=4,, alpha=0.05, beta=0.1,delta=1,case="maximin")
n
30


Hence maximal sample size \(n\) is \(\frac{30}{b} = \frac{30}{2} = 15\). For a two-way cross-classification with interaction and with equal subclass numbers , see Table 5.10 for the ANOVA table. The Solution is given below for this design.

121 line 17: from below: \(\beta=0.1, \delta = 1\sigma\) .

128 line 1: \(\lambda_a = (\frac{bn}{\sigma^2}) \sum … , \lambda_b = (\frac{an}{\sigma^2}) \sum … \lambda_{ab} = (\frac{n}{\sigma^2})\sum\sum … \)

128 line 5: \(\lambda_a = (\frac{bn}{\sigma^2})\sum …\)

128 line 5: \(\lambda_b = (\frac{an}{\sigma^2})\sum …\)

128 line 9: \(\lambda_{ab} = (\frac{n}{\sigma^2})\sum\sum … \)

131 in formula (5.2) insert at the end: \(N_i. .= \sum\limits_{j=1}^{b} n_{ij}\)

132 line 11: Insert at the end of the line after where: for the balanced case

132 line 12: \(\lambda_a = (\frac{bn}{\sigma^2}) \sum … ., … \lambda_{bin a} = \frac{n}{\sigma^2} \sum\sum …..\)

134 line 16: \(\delta = 1\sigma \).

134 line 14: from below: \(\delta = 1\sigma \).

138 In second caption of Table 5.15: (5.30) must be (5.31) [three times]

144 line 1: \(\delta = 0.5\sigma\)

147 line 17: \(\delta = 0.5\sigma\)

147 line 18: \(alpha = 0.01\)

147 line 14 from below: \(1\) must be \(11\) for \(n\)

152 line 20 from below: delete \(2\) (after beta =) and delete \(5\) (after (delta =)

152 line 16 from below: \(\delta = 0.5\sigma\)

156 line 7 from below: This line must be replaced by: \(H_{AxB 0}: (a,b)_{ij} = 0 \) (for all i and j).

157 line 2: "axc" must be "axb"

157 line 6: 5 must be 8




Chapter 7

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213 line 9 from below: \(\delta= 1.6\sigma\)

221 formula (7.17): \(F_B = \frac{MS_{BinA}}{MS_{Res}}\)

222 line 22: \(\delta = 0.1\) must be \(\beta=0.1\)

224 formula (7.19) \(F = \frac{MS_A}{MS_{AxC}}\)

224 line 4 from below: \(f_2 = (a - 1)(c - 1)\)

227 line 15 from below: \(\frac{\delta}{\sigma}\) must be \(\frac{\delta}{\sigma_y}\)

227 line11 from below: \(\frac{\delta}{\sigma}=0.5\) must be \(\frac{\delta}{\sigma_y} = 0.5\) with \(\sigma_y^2 = \sigma^2 + bn\sigma^2_{ac}\)

228 line 8: \(\frac{\delta}{\sigma} = 0.5\) must be \(\frac{\delta}{\sigma_y} = 0.5\) with \(\sigma_y^2 = \sigma^2 + bn\sigma^2_{ac}\)

228 line 12 from below: \(\frac{\delta}{\sigma} = 1\) must be \(\frac{\delta}{\sigma_y} = 1\) with \(\sigma_y^2 = \sigma^2 + n\sigma^2_{abc}\)

233 line 15 from below: \(\frac{\delta}{\sigma} = 0.4\) must be \(\frac{\delta}{\sigma_y} = 0.4\) with \(\sigma^2_y = \sigma^2 + cn\sigma^2_{b(a)}\)

235 line 14 from below: \(abc – 1\) must be \(ab(c – 1)\) degrees of freedom.

236 line 17: \(ab(c - 1)\) must be \(a(b – 1)\) [ twice ]

237 line 1: \(\frac{\delta}{\sigma} = 1\) must be \(\frac{\delta}{\sigma_y} = 1\) with \(\sigma^2_y = \sigma^2 + n\sigma^2_{c(ab)} + cn\sigma^2_{b(a)}\)

238 line 5: \(\frac{\delta}{\sigma} = 1\) must be \(\frac{\delta}{\sigma_y}\)

238 line 8: \(\frac{\delta}{\sigma} = 1\) must be \(\frac{\delta}{\sigma_y} = 1\) with \(\sigma^2_y = \sigma^2 + n\sigma^2_{c(ab)}\)

241 line 13 from below: \(\frac{\delta}{\sigma}\) must be \(\frac{\delta}{\sigma_y}\)

241 line 9 from below: \(\frac{\delta}{\sigma} = 0.5\) must be \(\frac{\delta}{\sigma_y} = 0.5\) with \(\sigma^2_y = \sigma^2 + cn\sigma^2_{ab}\)

244 line 15 from below: \(\frac{\delta}{\sigma}\) must be \(\frac{\delta}{\sigma_y}\)

244 line 12 from below: \(\frac{\delta}{\sigma} = 0.5\) must be \(\frac{\delta}{\sigma_y} = 0.5\) with \(\sigma^2_y = \sigma^2 + n \sigma^2_{c(ab)}\)

245 line 2: \(\frac{\delta}{\sigma}\) must be \(\frac{\delta}{\sigma_y}\)

245 line 5 : \(\frac{\delta}{\sigma} = 0.5\) must be \(\frac{\delta}{\sigma_y} = 0.5\) with \(\sigma^2_y = \sigma^2 + n \sigma^2_{c(ab)} \)

245 line 17 from below: \(\frac{\delta}{\sigma}\) must be \(\frac{\delta}{\sigma_y}\)

245 line 14 from below: \(\frac{\delta}{\sigma} = 0.5\) must be \(\frac{\delta}{\sigma_y} = 0.5\) with \(\sigma^2_y = \sigma^2 + n \sigma^2_{c(ab)}\)

246 line 12 from below: \(\frac{\delta}{\sigma}\) must be \(\frac{\delta}{\sigma_y}\)

246 line 9 from below : \(\frac{\delta}{\sigma} = 0.5\) must be \(\frac{\delta}{\sigma_y} = 0.5\) with \(\sigma^2_y = n \sigma^2_{c(ab)} + cn\sigma^2_{ab}\)

247 line 2 from below: \(c_{k(ij)}\) must be \(c_k\) [ twice ]

247 line 2 from below: delete \(j,i,\)

249 line 2 below Table 7.23: \(\frac{\delta}{\sigma}\) must be \(\frac{\delta}{\sigma_y}\)

249 line 5 below Table 7.23: \(\frac{\delta}{\sigma} = 0.5\) must be \(\frac{\delta}{\sigma_y} = 0.5\) with \(\sigma^2_y = \sigma^2 + bn \sigma^2_{ac}\)

250 line 6 below Table 7.24: \(\frac{\delta}{\sigma}\) must be \(\frac{\delta}{\sigma_y}\) .

250 line 9 below Table 7.24: \(\frac{\delta}{\sigma} = 0.5\) must be \(\frac{\delta}{\sigma_y} = 0.5\) with \(\sigma^2_y = \sigma^2 + cn \sigma^2_{b (a)} + n\sigma^2_{bc(a)}\)

250 line 2 from below: \(\frac{\delta}{\sigma}\) must be \(\frac{\delta}{\sigma_y}\) .

251 line 1: \(\frac{\delta}{\sigma} = 0.5\) must be \(\frac{\delta}{\sigma_y} = 0.5\) with \(\sigma^2_y = \sigma^2 + n \sigma^2_{bc(a)}\)

252 line 15 from below: \(\frac{\delta}{\sigma}\) must be \(\frac{\delta}{\sigma_y}\) .

252 line 12 from below: \(\frac{\delta}{\sigma} = 0.5\) must be \(\frac{\delta}{\sigma_y} = 0.5\) with \(\sigma^2_y = \sigma^2 + n \sigma^2_{bc(a)} + bn \sigma^2_{ac}\)

253 line 2: \(\frac{\delta}{\sigma}\) must be \(\frac{\delta}{\sigma_y}\) .

253 line 5: \(\frac{\delta}{\sigma} = 0.5\) must be \(\frac{\delta}{\sigma_y} = 0.5\) with \(\sigma^2_y = \sigma^2 + n \sigma^2_{bc(a)}\)

255 line 5: \(\frac{\delta}{\sigma}\) must be \(\frac{\delta}{\sigma_y}\) .

255 line 8: \(\frac{\delta}{\sigma} = 0.5\) must be \(\frac{\delta}{\sigma_y} = 0.5\) with \(\sigma^2_y = \sigma^2 + n \sigma^2_{bc(a)}\)