During training, the time in seconds for a soldier to dismantle a certain type of equipment is a normally distributed continuous random variable $T$. The standard deviation of $T$ is 3.5 and the expected value of $T$ is 35.1. After a new set of instructional material is introduced, $n$ soldiers are selected at random and the mean time taken for this sample of soldiers to dismantle the equipment is found to be $\bar{t}$ seconds. A test is carried out, at 5% significance level, to determine whether the mean time taken to dismantle the equipment has been reduced.

1. State the null and alternative hypotheses, defining the population mean $\mu$.

Let $\mu$ be the population mean time taken for a soldier to dismantle the equipment.
$H_0: \mu = 35.1$
$H_1: \mu < 35.1$

1. State the distribution of the sample mean under the null hypothesis and identify the rejection criterion.

Under $H_0$, the population mean is $\mu = 35.1$. Since the population is normally distributed with known standard deviation $\sigma = 3.5$, the sample mean for $n=20$ follows:
$\bar{T} \sim \text{N}\left(35.1, \frac{3.5^2}{20}\right)$
For a left-tailed test at the $5\%$ level of significance, we reject $H_0$ if the test statistic falls in the lower $5\%$ tail.

2. Standardise the sample mean and set up the inequality using the critical $z$-value.

The critical $z$-value for a lower tail area of $0.05$ is $z = -1.64485$.
Standardising the sample mean $\bar{t}$, the condition to reject $H_0$ is:
$\frac{\bar{t} - 35.1}{\frac{3.5}{\sqrt{20}}} \le -1.64485$

3. Solve the inequality to find the critical region for $\bar{t}$.

Solving for $\bar{t}$:
$\bar{t} \le 35.1 - 1.64485 \left(\frac{3.5}{\sqrt{20}}\right)$
$\bar{t} \le 35.1 - 1.2872$
$\bar{t} \le 33.813$
Rounding to 3 significant figures, we get $\bar{t} \le 33.8$.

1. Set up the inequality for the test statistic given that $H_0$ is not rejected.

The test statistic for the sample mean is $z = \frac{\bar{t} - \mu}{\frac{\sigma}{\sqrt{n}}}$.

For a 1-tailed test (testing for a reduction) at the 5% significance level, the critical value is $z \approx -1.64485$.

Since $H_0$ is not rejected, the test statistic must fall outside the critical region, meaning $z \ge -1.64485$:
$\frac{33.2 - 35.1}{\frac{3.5}{\sqrt{n}}} \ge -1.64485$

2. Solve the inequality to find the range of possible values for $n$.

$\frac{-1.9}{3.5} \sqrt{n} \ge -1.64485$
$\sqrt{n} \le \frac{-1.64485 \times 3.5}{-1.9}$
$\sqrt{n} \le 3.02998$
$n \le 9.1808$

3. Identify the largest integer value for $n$.

Since the sample size $n$ must be an integer, the largest possible value is:
$n = 9$