Energy bars are a convenient source of nutrition for athletes. A typical energy bar weighs between 30-50 g and is likely to supply about 200-300 calories. An American company sells energy bars and a nutritionist working in the company claims that the mean number of calories in a product, Perfect Protein bars, is 350 calories. A random sample of 30 bars is collected and the number of calories, $x$ cal, in the 30 bars is summarised as follows. $\sum x = 10446, \; \sum x^2 = 3638000.$

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:
$\sqrt{\frac{v}{45}} \ge \frac{2.2}{1.750686}$
$\sqrt{\frac{v}{45}} \ge 1.25665$
$\frac{v}{45} \ge 1.57917$
$v \ge 71.063$
$v \ge 71.1 \text{ (to 3 s.f.)}$