永初农场（Eng Choo Farm）的业主声称，其农场生产的鸡蛋平均每只含有 $\mu_0$ 毫克的胆固醇。 健康促进局从该农场随机抽取了 16 只鸡蛋，并测量了每只鸡蛋的胆固醇含量 $x$ 毫克。结果汇总如下： $\sum x = 2848$ 且 $\sum x^2 = 509884$。

1. Find the sample mean and unbiased estimate of variance

$\bar{x} = \frac{\sum x}{n} = \frac{2848}{16} = 178$
$s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right) = \frac{1}{15}\left(509884 - \frac{2848^2}{16}\right) = \frac{2940}{15} = 196$
So the estimated standard deviation is $s = 14$ and $\dfrac{s}{\sqrt{n}} = \dfrac{14}{\sqrt{16}} = 3.5$.

2. Set up the two-tail acceptance condition

The owner claims the mean is $\mu_0$, so we test $H_0: \mu = \mu_0$ against $H_1: \mu \neq \mu_0$ (a 2-tail test) at the 5% level. Assuming the population is normal and using $Z = \dfrac{\bar{x} - \mu_0}{s/\sqrt{n}}$, the claim is not rejected when
$-1.95996 < \frac{178 - \mu_0}{3.5} < 1.95996.$

3. Solve the inequality for μ₀

$|178 - \mu_0| < 1.95996 \times 3.5 = 6.8599$
$178 - 6.8599 < \mu_0 < 178 + 6.8599$
$171.14 < \mu_0 < 184.86$
Hence the set of values is $\{\mu_0 : 171 < \mu_0 < 185\}$.

1. Find the sum and sum of squares of $x$ for the second sample.

For the second sample, $n_2 = 36$ and $\bar{x}_2 = 185$.
$\sum x_2 = 36 \times 185 = 6660$

Using the biased sample variance formula $\frac{1}{n_2}\left(\sum x_2^2 - \frac{(\sum x_2)^2}{n_2}\right) = 120$:
$\frac{1}{36}\left(\sum x_2^2 - \frac{6660^2}{36}\right) = 120$
$\sum x_2^2 = 36(120) + 36(185^2) = 4320 + 1232100 = 1236420$

2. Calculate the combined unbiased estimates for the population mean and variance.

For the combined sample, $n = 16 + 36 = 52$.
$\sum x = 2848 + 6660 = 9508$
$\sum x^2 = 509884 + 1236420 = 1746304$

Unbiased estimate of population mean, $\hat{\mu}$:
$\bar{x} = \frac{9508}{52} = \frac{2377}{13} \approx 182.85 \text{ mg}$

Unbiased estimate of population variance, $s^2$:
$s^2 = \frac{1}{n-1} \left( \sum x^2 - \frac{(\sum x)^2}{n} \right)$
$s^2 = \frac{1}{51} \left( 1746304 - \frac{9508^2}{52} \right) = \frac{1}{51} \left( 1746304 - \frac{90402064}{52} \right)$
$s^2 = \frac{33812}{221} \approx 153.00 \text{ mg}^2$

3. State the hypotheses and find the $p$-value using a test statistic.

$H_0: \mu = 180$
$H_1: \mu > 180$

Under $H_0$, since the sample size $n = 52$ is large, by the Central Limit Theorem the sample mean is approximately normally distributed. The test statistic is:
$Z = \frac{\bar{X} - 180}{s / \sqrt{52}} \sim N(0, 1) \text{ approximately.}$

Using a graphing calculator (Z-Test) with $\mu_0 = 180$, $\bar{x} = \frac{2377}{13}$, $s_x = \sqrt{\frac{33812}{221}}$, and $n = 52$:
$z \approx 1.659$
$p\text{-value} \approx 0.0485$ (to 3 s.f.)

4. Explain the meaning of the $p$-value in the context of the question.

The $p$-value of $0.0485$ is the probability of obtaining a sample mean cholesterol content of $\frac{2377}{13}$ mg (approx. $182.8$ mg) or more, assuming that the true population mean cholesterol content of the eggs is $180$ mg.