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 $250\ \text{minutes}^2$. 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 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 $\bar{x} \le 36.1 \text{ or } \bar{x} \ge 43.9$.