EastHam 是一家处于竞争激烈联赛中的足球俱乐部。研究发现，EastHam 在比赛中攻入首球的平均时间为 50 分钟。教练试验了一种新的战术，并在该赛季中随机抽取了 35 场比赛，记录了攻入首球所需的时间 $x$（分钟）。结果发现总用时为 1650 分钟，且时间的方差为 $250\ \text{minutes}^2$。 教练想要检验在实施新战术后，攻入首球的平均用时是否有所减少。

1. State the relevant condition and theorem that justifies the hypothesis test

Since the sample size $n = 35$ is large ($n \ge 30$), by the Central Limit Theorem, the sample mean $\bar{X}$ is approximately normally distributed.

1. Calculate the unbiased estimate of the population mean.

$\bar{x} = \frac{\sum x}{n} = \frac{1650}{35}$
$\bar{x} \approx 47.143$

2. Calculate the unbiased estimate of the population variance using the given sample variance.

The given variance is the biased sample variance ($250$). The unbiased estimate $s^2$ is found by multiplying by $\frac{n}{n-1}$:
$s^2 = \frac{n}{n-1} \times \text{sample variance}$
$s^2 = \frac{35}{34} \times 250 = \frac{4375}{17}$
$s^2 \approx 257.353$

1. State the hypotheses and define the population mean.

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 and its distribution under $H_0$.

Under $H_0$, since the sample size $n = 35$ is large ($n \ge 30$), by the Central Limit Theorem, the sample mean is approximately normally distributed. The test statistic is:
$Z = \frac{\bar{X} - 50}{S / \sqrt{n}} \sim N(0, 1) \text{ approximately.}$

3. Calculate the sample statistics and use the graphing calculator to find the $z$-value and $p$-value.

From the given data (and worked in part (b)):
$\bar{x} = \frac{1650}{35} = \frac{330}{7} \approx 47.143$
$s^2 = \frac{35}{34} \times 250 = \frac{4375}{17} \approx 257.353$

Using a 1-tailed $Z$-test on the graphing calculator with $\mu_0 = 50$, $\bar{x} = \frac{330}{7}$, $s = \sqrt{\frac{4375}{17}}$ and $n = 35$:
$z = -1.05 \text{ (to 3 s.f.)}$
$p\text{-value} = 0.146 \text{ (to 3 s.f.)}$

4. Compare the $p$-value against the significance level and state the conclusion in context.

Level of significance, $\alpha = 0.10$.
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 with the implementation of the new play tactic.

1. State the hypotheses for the test.

Let $\mu$ be the population mean time taken by NewPalace to score the first goal.
$H_0: \mu = 40$
$H_1: \mu \neq 40$

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

Under $H_0$, the sample mean $\bar{X}$ is normally distributed with mean $\mu = 40$ and variance $\frac{\sigma^2}{n} = \frac{280}{50} = 5.6$.
$\bar{X} \sim \text{N}(40, 5.6)$

3. Determine the critical value and formulate the rejection region in terms of the test statistic.

For a two-tailed test at $10\%$ level of significance, the critical $z$-values are $\pm 1.6449$.
We reject $H_0$ if the test statistic $Z = \frac{\bar{x} - 40}{\sqrt{5.6}}$ falls in the critical region:
$Z \le -1.6449 \quad \text{or} \quad Z \ge 1.6449$

4. Solve for the range of values of the sample mean $\bar{x}$.

Substituting the expression for $Z$:
$\frac{\bar{x} - 40}{\sqrt{5.6}} \le -1.6449 \implies \bar{x} \le 40 - 1.6449\sqrt{5.6}$
$\bar{x} \le 36.108$
$\frac{\bar{x} - 40}{\sqrt{5.6}} \ge 1.6449 \implies \bar{x} \ge 40 + 1.6449\sqrt{5.6}$
$\bar{x} \ge 43.892$
To 3 significant figures, the fan's claim would be rejected if $\bar{x} \le 36.1 \text{ or } \bar{x} \ge 43.9$.