EastHam 是一支处于竞争激烈联赛中的足球俱乐部。研究发现，EastHam 在比赛中打入首粒进球所需的平均时间为 50 分钟。教练试用了一种新的战术，并记录了本赛季随机抽取的 35 场比赛中，打入首粒进球所需的时间 $x$ 分钟。结果显示，总用时为 1650 分钟，时间的方差为 250 minutes$^2$。 教练想要检验：随着新战术的实施，打入首粒进球的平均时间是否有所减少。

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).