### 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 $$\sigma^2_y = \sigma^2 + n \sigma^2_{bc(a)}$$