一家保险公司宣称，汽车保险理赔的平均支付金额为 $10 000。审计员检查了 25 个理赔案件的随机样本，并记录了每个案件的支付金额。支付金额 $ $x$ 汇总如下： $\sum(x - 10\ 000) = -6560 \quad \text{and} \quad \sum(x - 10\ 000)^2 = 7\ 493\ 500.$

1. Calculate the sample mean and the unbiased estimate of the population variance.

Let $y = x - 10\ 000$.
Given $n = 25$, $\sum y = -6560$ and $\sum y^2 = 7\ 493\ 500$.

Sample mean,
$\bar{x} = 10\ 000 + \frac{\sum y}{n} = 10\ 000 + \frac{-6560}{25} = 9737.6$

Unbiased estimate of population variance,
$s^2 = \frac{1}{n-1} \left( \sum y^2 - \frac{(\sum y)^2}{n} \right)$
$s^2 = \frac{1}{24} \left( 7\ 493\ 500 - \frac{(-6560)^2}{25} \right)$
$s^2 = 240\ 506.5$

2. State the necessary assumptions, as well as the null and alternative hypotheses.

Assumptions:

1. The population of payments made on motor insurance claims follows a normal distribution.
2. The population variance is assumed to be equal to the sample unbiased estimate $s^2$.

Let $\mu$ be the population mean payment.
Null hypothesis, $H_0: \mu = 10\ 000$
Alternative hypothesis, $H_1: \mu < 10\ 000$

3. Compute the test statistic and the $p$-value.

Level of significance: $1\%$

Since $n = 25$ is small and the population variance is unknown, under the assumptions made, we use the test statistic $Z \approx \frac{\bar{X} - 10\ 000}{s / \sqrt{n}} \sim N(0, 1)$.

$z = \frac{9737.6 - 10\ 000}{\sqrt{240\ 506.5} / \sqrt{25}} = -2.6753 \text{ (to 5 s.f.)}$

Using a graphing calculator,
$p\text{-value} = P(Z \le -2.6753) \approx 0.00373$

4. State the conclusion of the test in the context of the question.

Since $p\text{-value} = 0.00373 < 0.01$, we reject $H_0$.

There is sufficient evidence at the $1\%$ significance level to conclude that the mean payment made is less than $\char36{}10\ 000$.

1. State the meaning of the 1% significance level in the context of the given hypotheses

The $1\%$ significance level represents the probability of making a Type I error, which is rejecting the null hypothesis ($H_0: \mu = 10\,000$) when it is actually true.

In context: There is a $1\%$ probability of concluding that the mean payment made on motor insurance claims is less than $\char36{}10\,000$, when the true mean payment is actually exactly $\char36{}10\,000$.

1. Concisely calculate the sample mean, unbiased estimate of variance, and the $p$-value for the one-tailed test from part (i).

To state the reason without further tests, we rely on the results calculated in part (i). Let $y = x - 10\,000$.
$\bar{x} = 10\,000 + \frac{-6560}{25} = 9737.6$
$s^2 = \frac{1}{24}\left(7\,493\,500 - \frac{(-6560)^2}{25}\right) = 240\,506.5$
Under $H_0: \mu = 10\,000$, using a $t$-test ($n=25$, unknown population variance), the test statistic is:
$t = \frac{9737.6 - 10\,000}{\sqrt{240\,506.5/25}} \approx -2.6753$
With $24$ degrees of freedom, the $p$-value for the one-tailed test ($H_1: \mu < 10\,000$) is $0.006762$.

2. Deduce the $p$-value for the two-tailed test and determine if the conclusion changes.

For a two-tailed test, the new $p$-value is doubled:
$\text{New } p\text{-value} = 2 \times 0.006762 = 0.0135 \text{ (to 3 s.f.)}$
Since $0.0135 > 0.01$ (the $1\%$ significance level), we do not reject $H_0$.
This contradicts the conclusion in part (i), where $H_0$ was rejected ($0.006762 < 0.01$). Therefore, the conclusion would not be the same.

1. Calculate the sample mean $\bar{x}$ from the given summary statistics.

$\bar{x} = 10000 + \frac{\sum(x - 10000)}{25} = 10000 + \frac{-6560}{25} = 9737.6$

2. Set up the condition for the null hypothesis not to be rejected at the 5% significance level.

Let $\mu_0$ be the declared mean payment. The auditor tests $H_0: \mu = \mu_0$ against $H_1: \mu < \mu_0$ (testing for an overestimate).
For the claim not to be rejected at the 5% significance level, the test statistic $z$ must be greater than or equal to the critical value $-1.64485$ (since it is a left-tailed test).

$Z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} \ge -1.64485$

3. Substitute the known values to find the largest value of $\mu_0$.

Using $\sigma = 500$ and $n = 25$:
$\frac{9737.6 - \mu_0}{500 / \sqrt{25}} \ge -1.64485$
$\frac{9737.6 - \mu_0}{100} \ge -1.64485$
$9737.6 - \mu_0 \ge -164.485$
$\mu_0 \le 9737.6 + 164.485$
$\mu_0 \le 9902.085$

To the nearest dollar, the largest declared mean is $\char36{}9902$.