The recommended dietary allowance (RDA) of iron for adult fernales is 18 milligrams ( $m g$ ) per day. The given iron intakes $(\mathrm{mg})$ were obtained for 45 random adult females. At the $1 \%$ significance level, do the data suggest that adult females are, on average, getting less than the RDA of $18 \mathrm{mg}$ of iron? Assume that the population standard deviation is $4.6 \mathrm{mg}$. Preliminary data analyses indicate that applying the z-test is reasonable, (Note $x=14.65 \mathrm{mg}$ ) Click here lo vew the iron intake data. Click here to view Page 1 of the table of areas under the standard normat curve. Click here to view Pene 2 of the table of areas under the standard normal curve. State the hypotheses for the one-mean z-test: \[ \begin{array}{l} H_{0} \mu=18 \mathrm{mg} \\ H_{\mathrm{a}} \mu<18 \mathrm{mg} \end{array} \] (Type integers or decimals. Do not round) Compute the value of the test statistic \[ z=\square \] (Round to two decimal places as needed)
Solution
Step 1 ：State the hypotheses for the one-mean z-test: \(H_{0}: \mu = 18 \, mg\) and \(H_{a}: \mu < 18 \, mg\).
Step 2 ：Given values are: sample mean (\(x_{bar}\)) = 14.65 mg, population mean (\(\mu\)) = 18 mg, population standard deviation (\(\sigma\)) = 4.6 mg, and sample size (n) = 45.
Step 3 ：Compute the value of the test statistic using the formula \(z = \frac{x_{bar} - \mu}{\sigma / \sqrt{n}}\).
Step 4 ：Substitute the given values into the formula to get \(z = \frac{14.65 - 18}{4.6 / \sqrt{45}}\).
Step 5 ：The calculated z-test statistic is approximately -4.89. This value is negative, which indicates that the sample mean is less than the population mean. This is consistent with our alternative hypothesis \(H_{a}: \mu < 18 \, mg\).
Step 6 ：The final answer is: The value of the test statistic is approximately \(\boxed{-4.89}\).
