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$^2$.

1. State the relevant theorem and condition that allows the test to be carried out

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.

$\hat{\mu} = \bar{x} = \frac{\sum x}{n} = \frac{1650}{35} \approx 47.143$ (to 3 d.p.)

2. Calculate the unbiased estimate of the population variance.

$\hat{\sigma}^2 = s^2 = \frac{n}{n-1} \times \text{Sample Variance} = \frac{35}{34} \times 250 \approx 257.353$ (to 3 d.p.)

1. State the hypotheses and define the population mean.

Let $\mu$ be the population mean time taken by Eastham to score their first goal under the new tactic.

$H_0: \mu = 50$
$H_1: \mu \neq 50$
(Note: A 1-tailed test with $H_1: \mu < 50$ is also acceptable if interpreting the tactic's aim as specifically wanting to reduce the time.)

2. Identify the appropriate test and state the test statistic.

Since the sample size $n = 35 \ge 30$, by the Central Limit Theorem, the sample mean $\bar{X}$ is approximately normally distributed. We use a $Z$-test.

Under $H_0$, the test statistic is $Z = \frac{\bar{X} - \mu}{s / \sqrt{n}} \sim N(0, 1)$ approximately.

3. Calculate the value of the test statistic and the $p$-value.

Using the unbiased estimates calculated in part (b), $\bar{x} = \frac{1650}{35} = \frac{330}{7} \approx 47.143$ and $s^2 = \frac{35}{34} \times 250 = \frac{4375}{17} \approx 257.353$.

$z = \frac{\frac{330}{7} - 50}{\sqrt{\frac{4375/17}{35}}} = \frac{-20/7}{\sqrt{125/17}} \approx -1.0537$

The $p$-value for a two-tailed test is $2 \times P(Z < -1.0537) \approx 0.292$.
(If a 1-tailed test was used, $p$-value $\approx 0.146$)

4. Compare the $p$-value with the significance level and conclude in context.

Since the $p$-value $= 0.292 > 0.10$ (the $10\%$ level of significance), 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 changed from $50$ minutes.

1. State the null and alternative hypotheses.

Let $\mu$ be the population mean time taken by NewPalace to score their first goal.
$H_0: \mu = 40$
$H_1: \mu \neq 40$
Level of significance $\alpha = 0.10$.

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

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

3. Find the critical $Z$-values for a two-tailed test at the 10% significance level.

For a two-tailed test at $10\%$ significance, the critical probability in each tail is $5\%$ ($0.05$).
Using the standard normal distribution, the critical $z$-values are $z = \pm 1.6449$.
The test statistic is given by:
$Z = \frac{\bar{X} - 40}{\sqrt{5.6}}$
We reject $H_0$ if $Z \leq -1.6449$ or $Z \geq 1.6449$.

4. Calculate the corresponding range of values for the sample mean.

For the lower tail:
$\frac{\bar{x} - 40}{\sqrt{5.6}} \leq -1.6449 \implies \bar{x} \leq 40 - 1.6449\sqrt{5.6}$
$\bar{x} \leq 36.108$
For the upper tail:
$\frac{\bar{x} - 40}{\sqrt{5.6}} \geq 1.6449 \implies \bar{x} \geq 40 + 1.6449\sqrt{5.6}$
$\bar{x} \geq 43.892$
Thus, the fan's claim is rejected if $\bar{x} \leq 36.1$ or $\bar{x} \geq 43.9$ (to 3 s.f.).