Solution to Exercise 8.30.

  Again, I have to criticize the author's language. Since $\bar{x} = 11.3 mg.$, the data do ``indicate'' that the average daily zinc intake falls below the recommended amount of 15 mg. Do they offer strong evidence that it falls below 15? Now that's a question for statistical hypothesis testing. So, letting $\mu$ denote the population mean of zinc intake (in mg./day) for men 65-74, we will test

$$H_0 : \mu \ge 15 , \quad vs. \quad H_1 : \mu < 15 .$$

The sample size n = 115 is quite sufficient to allow us to use a z-test (generally $n \ge 30$is considered acceptable), so

$$z \; = \; \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \; = \; \frac{11.3 -15}{6.43/\sqrt{115}} \; = \; -6.170774 .$$

Clearly signficant. We reject for small values of z. At the .05 level, we reject for z < -1.645, so clearly we can reject at the .05 level of significance. From the table in the book, we have P[Z < -3.49] = 0.0002, so we know that we can reject at any level of significance above 0.0002. In fact, the P-value is about $3.4 \times 10^{-10}$.