能量棒是运动员便利的营养来源。一根典型的能量棒重 30-50 克，可能提供大约 200-300 卡路里的热量。一家美国公司销售能量棒，该公司的一位营养学家声称，其产品“完美蛋白质”（Perfect Protein）能量棒的平均热量为 350 卡路里。 收集了 30 根能量棒的随机样本，这 30 根能量棒的热量 $x$ （单位：卡路里）汇总如下： $\sum x = 10446, \; \sum x^2 = 3638000.$

1. Define the population parameter and state the null and alternative hypotheses.

Let $\mu$ be the population mean number of calories in a Perfect Protein bar.

$H_0: \mu = 350$
$H_1: \mu < 350$

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

Sample mean, $\bar{x} = \frac{\sum x}{n} = \frac{10446}{30} = 348.2$

Unbiased estimate of population variance, $s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right)$
$s^2 = \frac{1}{29}\left(3638000 - \frac{10446^2}{30}\right)$
$s^2 = \frac{702.8}{29} \approx 24.234$ (5 s.f.)

3. State the test statistic and calculate the $p$-value using a graphing calculator.

Since $n = 30$ is sufficiently large, by the Central Limit Theorem, the sample mean $\bar{X}$ is approximately normally distributed.
Under $H_0$, the test statistic is $Z = \frac{\bar{X} - 350}{s / \sqrt{30}} \sim N(0, 1)$ approximately.

Using a graphing calculator with $\mu_0 = 350$, $\bar{x} = 348.2$, $s_x = \sqrt{24.234}$, and $n = 30$:
$p\text{-value} = 0.022604 \approx 0.0226$ (3 s.f.)

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

Since the $p$-value ($0.0226$) is less than the level of significance ($\alpha = 0.03$), we reject $H_0$.
There is sufficient evidence at the $3\%$ level of significance to conclude that the mean number of calories in Perfect Protein bars is less than $350$ calories.

1. Explain why the test is inadmissible for a small sample size without further assumptions.

Since the sample size of $15$ is small ($<30$) and the population distribution of the number of calories is unknown, the Central Limit Theorem cannot be applied to assume that the sample mean is normally distributed.

2. State the assumption required to proceed with the test.

It must be assumed that the number of calories in the Perfect Protein bars is normally distributed.

1. Calculate the sample mean and the unbiased estimate of the population variance (from part i).

$\bar{x} = \frac{\sum x}{n} = \frac{10446}{30} = 348.2$

$s^2 = \frac{1}{n-1} \left( \sum x^2 - \frac{(\sum x)^2}{n} \right)$
$s^2 = \frac{1}{29} \left( 3638000 - \frac{10446^2}{30} \right) = \frac{702.8}{29} \approx 24.2345$

2. Calculate the test statistic using the Central Limit Theorem.

Under the null hypothesis $H_0: \mu = 350$, we use a $Z$-test since the sample size $n = 30$ is sufficiently large.

$z = \frac{\bar{x} - \mu}{\sqrt{s^2 / n}} = \frac{348.2 - 350}{\sqrt{24.2345 / 30}} \approx -2.0027$

3. Identify the test as 2-tailed and compute the $p$-value.

The test is for whether the mean is $350$ calories, meaning $H_1: \mu \neq 350$. This makes it a 2-tailed test.

$p\text{-value} = 2 \times \mathrm{P}(Z < -2.0027) = 2 \times 0.022604 \approx 0.045208$

4. State the smallest level of significance for which the claim will be rejected.

The nutritionist's claim ($H_0$) is rejected if the level of significance is at least the $p$-value.
Smallest level of significance $= 0.0452 \text{ or } 4.52\% \text{ (to 3 s.f.)}$

1. State the hypotheses and the condition for the null hypothesis not to be rejected.

Let $\mu$ be the population mean number of calories of Dynamic Protein bars.
$H_0: \mu = 350$
$H_1: \mu > 350$
Since there is no reason to believe $\mu > 350$, $H_0$ is not rejected at the $4\%$ level of significance. This means the test statistic $z$ must be less than or equal to the critical value.

2. Find the critical value and set up the inequality involving the unknown variance.

For a $4\%$ right-tailed test, the critical value is $z_{0.04} \approx 1.750686$.
Let $v$ be the variance used in the calculation. The test statistic is given by $z = \frac{\bar{x} - \mu}{\sqrt{\frac{v}{n}}}$.
$\frac{352.2 - 350}{\sqrt{\frac{v}{45}}} \le 1.750686$

3. Solve the inequality to find the range of values for $v$.

$\frac{2.2}{\sqrt{\frac{v}{45}}} \le 1.750686$
Since $v > 0$, we can rearrange the inequality:
$\sqrt{\frac{v}{45}} \ge \frac{2.2}{1.750686}$
$\sqrt{\frac{v}{45}} \ge 1.25665$
$\frac{v}{45} \ge 1.57917$
$v \ge 71.063$
$v \ge 71.1 \text{ (to 3 s.f.)}$