A random sample of 27 individuals from a population of young men aged 18 was chosen and the length of the right foot of each individual was measured to the nearest centimetre. The data is as given in the table below.

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 $k > 27.0$.