EastHam is a football club in a competitive league. It is found that the average time taken by EastHam to score their first goal in a match is 50 minutes. The coach trialled a new play tactic and the time taken, $x$ minutes, to score the first goal is recorded for a random sample of 35 matches in the season. The total time taken is found to be 1650 minutes and the variance of the time is 250 minutes$^2$. The coach wants to test if the average time taken to score the first goal has decreased with the implementation of the new play tactic.

1. State the condition and invoke the Central Limit Theorem.

Since the sample size $n = 35$ is large ($n \ge 30$), by the Central Limit Theorem, the distribution of the sample mean is approximately normal regardless of the population's distribution.

1. Calculate the unbiased estimate of the population mean.

Unbiased estimate of population mean, $\bar{x} = \frac{\sum x}{n}$

$\bar{x} = \frac{1650}{35}$
$\bar{x} = 47.1428... \approx 47.143$ (to 3 d.p.)

2. Calculate the unbiased estimate of the population variance.

Unbiased estimate of population variance, $s^2 = \frac{n}{n-1} \times \text{sample variance}$

$s^2 = \frac{35}{35-1} \times 250$
$s^2 = \frac{35}{34} \times 250$
$s^2 = 257.3529... \approx 257.353$ (to 3 d.p.)

1. Define the necessary symbol and state the null and alternative hypotheses.

Let $\mu$ be the population mean time taken by EastHam to score their first goal.
$H_0: \mu = 50$
$H_1: \mu < 50$

2. State the test statistic distribution and calculate the $z$-statistic and $p$-value.

From part (b), the unbiased estimate of the sample mean is $\bar{x} = \frac{1650}{35} = \frac{330}{7}$ and the unbiased estimate of the population variance is $s^2 = \frac{35}{34}(250) = \frac{4375}{17}$.

Under $H_0$, since the sample size $n = 35$ is large ($n \ge 30$), the Central Limit Theorem applies and the test statistic is approximately standard normal:
$Z = \frac{\bar{X} - 50}{s/\sqrt{35}} \sim N(0, 1)$

Using a graphing calculator (1-tail Z-Test):
Test statistic, $z \approx -1.0537$
$p\text{-value} \approx 0.14602 \approx 0.146$ (to 3 s.f.)

3. State the statistical conclusion and interpret it in the context of the problem.

Since $p\text{-value} = 0.146 > 0.10$, we do not reject $H_0$.

There is insufficient evidence at the $10\%$ level of significance to conclude that the average time taken by EastHam to score their first goal has decreased.

1. State the null and alternative hypotheses.

Let $\mu$ be the population mean time taken by NewPalace to score the first goal.
The fan claims the average time is exactly $40$ minutes. Thus, this is a two-tailed test.
$H_0: \mu = 40$
$H_1: \mu \neq 40$

2. Identify the distribution of the sample mean under the null hypothesis.

Since the time taken is normally distributed with population variance $\sigma^2 = 280$, and the sample size is $n = 50$, the sample mean $\bar{X}$ is normally distributed.
Under $H_0$:
$\bar{X} \sim \text{N}\left(40, \frac{280}{50}\right)$
$\bar{X} \sim \text{N}(40, 5.6)$

3. Calculate the critical boundaries for the test statistic.

For a $10\%$ level of significance in a two-tailed test, the critical regions are the lower and upper $5\%$ tails.
Using the standard normal distribution, the critical $z$-values are $\pm 1.64485$.
To reject $H_0$, the sample mean must fall in the critical regions:
$\frac{\bar{x} - 40}{\sqrt{5.6}} < -1.64485 \quad \text{or} \quad \frac{\bar{x} - 40}{\sqrt{5.6}} > 1.64485$

4. Solve for the range of values of the sample mean.

Solving the inequalities yields:
$\bar{x} < 40 - 1.64485\sqrt{5.6} \implies \bar{x} < 36.108$
$\bar{x} > 40 + 1.64485\sqrt{5.6} \implies \bar{x} > 43.892$
Therefore, the fan's claim is rejected if $\bar{x} < 36.1$ or $\bar{x} > 43.9$ (correct to 3 significant figures).