Solution to Exercise 7.14.

  (a)

$$\bar{x}\pm 1.960 s/\sqrt{n} \; = \; 89.10 \pm 1.960*3.73/\sqrt{169} \; = \; 89.10 \pm 0.5623692$$

$$\; = \; 89.10 \pm 0.56 \; = \; (88.54,89.66) .$$

Has the ``true average fracture strength been precisely estimated?'' I can't say. What level of precision is needed? It seems ``accurate'' to me.

(b) Using the formula from p. 283,

$$n \; = \; \left( 2 * z_{\alpha/2} * \frac{\sigma}{w} \right... ... \left( 2 * 1.960 * \frac{4}{.5} \right)^2 \; = \; 983.4496 .$$

We would use n = 984.