Solution. Dividing by the reliability, the corrected slope is \(1.5/0.6 = 2.5\).

No sample size removes the bias. By Theorem 14.1.1, the naive slope converges to \(0.6\beta _1\), not to \(\beta _1\), so larger samples only make the wrong answer more precise. The remedy is a better measurement or a correction based on an estimate of \(\sigma _\delta ^2\).

Solution. (a) Berkson: the recorded setting is fixed, and the truth varies around it. Unbiased. (b) Classical: the recorded value is the truth plus error. The slope is attenuated. (c) Berkson: everyone in the room receives the same recorded value, and each worker’s true exposure varies around it. Unbiased, though less precise.

Solution. The slope is the positive root of \(S_{xy}\beta ^2 + (\kappa S_{xx} - S_{yy})\beta - \kappa S_{xy} = 0\). Dividing by \(\kappa \) and letting \(\kappa \to \infty \) leaves \(S_{xx}\beta - S_{xy} = 0\), so \(\beta \to S_{xy}/S_{xx}\). Setting \(\kappa = 0\) leaves \(S_{xy}\beta ^2 - S_{yy}\beta = 0\), whose positive root is \(S_{yy}/S_{xy}\); by continuity of the roots in \(\kappa \), the positive root tends to it.

Solution. \(\hat x_0 = (9 - 1)/0.8 = 10\). Then \(g = 2.447^2(0.36)/(0.64 \times 42) = 0.080\), not negligible. The approximate standard error is \[ \frac {0.6}{0.8}\sqrt {1 + \frac 18 + \frac {25}{42}} = 0.75(1.312) = 0.984 , \] giving \(10 \pm 2.447(0.984)\), from \(7.59\) to \(12.41\). The exact interval has centre \(5 + 5/0.920 = 10.436\) and half-width \[ \frac {2.447(0.6)}{0.8(0.920)}\sqrt {0.920(1.125) + \frac {25}{42}} = 1.996(1.276) = 2.547 , \] so it runs from \(7.89\) to \(12.98\). It is shifted away from \(\bar x\) and is not symmetric about \(\hat x_0\), because the band widens as \(x\) moves away from \(\bar x\) (Figure 31). The symmetric approximation is poor here because \(\hat x_0\) is far from the centre of the standards and the line is not very precise.

Solution. The differences are \(-2\), \(1\), \(-1\), \(3\), \(0\), with mean \(0.2\) and sample variance \([(-2.2)^2 + 0.8^2 + (-1.2)^2 + 2.8^2 + (-0.2)^2]/4 = 14.8/4 = 3.7\). Since a difference has variance \(2\sigma _\delta ^2\), \(\hat \sigma _\delta ^2 = 3.7/2 = 1.85\).