从 18 岁年轻男性的总体中随机抽取了 27 个个体，测量了每个个体的右脚长度，精确到厘米。数据如以下表格所示。

1. State the null and alternative hypotheses.

Let $\mu$ be the population mean foot length of an 18 year old man.
$H_0: \mu = 27$
$H_1: \mu > 27$

2. Calculate the sample mean and the unbiased estimate of the population variance.

From the frequency table, $n = 27$.
$\sum x = 24(1) + 25(2) + 26(3) + 27(9) + 28(6) + 29(5) + 30(1) = 738$
$\sum x^2 = 24^2(1) + 25^2(2) + 26^2(3) + 27^2(9) + 28^2(6) + 29^2(5) + 30^2(1) = 20224$

Sample mean, $\bar{x} = \frac{738}{27} = \frac{82}{3} \approx 27.333$

Unbiased estimate of population variance, $s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right)$
$s^2 = \frac{1}{26}\left(20224 - \frac{738^2}{27}\right) = \frac{1}{26}(20224 - 20172) = \frac{52}{26} = 2$

3. State the necessary assumptions for the hypothesis test.

Since the sample size $n = 27$ is small ($n < 30$), we must assume that the population of foot lengths of 18 year old men is normally distributed.
(Additionally, we assume the unknown population variance $\sigma^2$ is approximately equal to the sample variance $s^2 = 2$).

4. Calculate the test statistic and $p$-value using a graphing calculator.

Under $H_0$, the test statistic is $Z = \frac{\bar{X} - \mu}{s/\sqrt{n}} \sim N(0,1)$ approximately.
Using a graphing calculator (Z-Test):
$z \approx 1.22$
$p\text{-value} = P(Z > 1.2247) \approx 0.110$

5. State the conclusion of the test in the context of the question.

Since $p\text{-value} = 0.110 > 0.02$, we do not reject $H_0$.
There is insufficient evidence at the $2\%$ level of significance to conclude that the mean foot length of an 18 year old man is more than $27\text{ cm}$.

1. Calculate the unbiased estimate of the population variance, $s^2$.

$s^2 = \frac{1}{n-1}\sum (x-\overline{x})^2 = \frac{1}{60-1}(123.20) = \frac{123.20}{59} = \frac{616}{295} \approx 2.0881$

2. State the hypotheses and determine the test statistic.

Let $\mu$ be the population mean foot length of an 18-year-old woman.
$H_0: \mu = k$
$H_1: \mu \neq k$
Since $n = 60 \ge 30$, by the Central Limit Theorem, the sample mean $\overline{X}$ is approximately normally distributed. Under $H_0$, the test statistic is:
$Z = \frac{\overline{X} - k}{s/\sqrt{n}} \sim N(0,1) \text{ approximately.}$

3. Identify the critical region for a 4% level of significance.

For a two-tailed test at the 4% level of significance, the critical regions are $Z < -z_{0.02}$ or $Z > z_{0.02}$.
From the standard normal distribution, $z_{0.02} \approx 2.0537$.
Thus, we reject $H_0$ if $|Z| > 2.0537$.

4. Formulate the inequality and solve for the range of values for $k$.

$\left| \frac{26.6 - k}{\sqrt{\frac{616}{295 \times 60}}} \right| > 2.0537$
$\frac{|26.6 - k|}{0.18655} > 2.0537$
$|26.6 - k| > 0.3831$
This gives $26.6 - k > 0.3831$ or $26.6 - k < -0.3831$.
$k < 26.2169 \quad \text{or} \quad k > 26.9831$
Rounding to 3 significant figures, $k < 26.2$ or $k > 27.0$.