A nutritionist claims that the mean number of calories in an energy bar is 350 cal. The nutritionist collected and measured the number of calories of a random sample of 15 energy bars. The mean and variance of the sample was 347.2 cal and 20.74 $\text{cal}^2$ respectively.

1. Explain why a $t$-test is used instead of a $z$-test.

A $t$-test is used instead of a $z$-test because the population variance is unknown and the sample size $n = 15$ is small (typically $n < 30$).

2. State the necessary assumption for a small-sample test.

For a small sample test to be valid when the population variance is unknown, it must be assumed that the population (the number of calories in an energy bar) is normally distributed.

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

Let $\mu$ be the population mean number of calories in an energy bar.
$H_0 : \mu = 350$
$H_1 : \mu \neq 350$

2. Calculate the unbiased estimate of the population variance.

Given the sample size $n = 15$ and the sample variance $= 20.74$, the unbiased estimate of the population variance is:
$s^2 = \frac{n}{n-1} \times (\text{sample variance}) = \frac{15}{14} \times 20.74 = \frac{155.55}{7} \approx 22.221$

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

Under $H_0$, the test statistic is $T = \frac{\bar{X} - \mu}{s / \sqrt{n}} \sim t(14)$.

Using a graphing calculator to perform a $t$-test with:
$\mu_0 = 350$
$\bar{x} = 347.2$
$s_x = \sqrt{22.221} \approx 4.714$
$n = 15$

$p\text{-value} = 0.03717 \approx 0.0372 \text{ (to 3 s.f.)}$

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

Level of significance $\alpha = 0.05$.
Since $p\text{-value} = 0.0372 < 0.05$, we reject $H_0$.
There is sufficient evidence at the $5\%$ level of significance to conclude that the mean number of calories in an energy bar is not $350$ cal.

1. Identify the different test and explain its effect on the $p$-value or critical region.

The 'different test' refers to the $z$-test mentioned in part (i). For the same test statistic, the standard normal distribution has thinner tails than the $t$-distribution, meaning the $p$-value for the $z$-test will be smaller than the $p$-value for the $t$-test (alternatively, the $z$-test critical region is wider since $1.960 < 2.145$).

2. State whether the conclusion will change based on the previous decision.

Since the null hypothesis $H_0$ was already rejected in part (ii) at the 5% level of significance, a smaller $p$-value (or wider critical region) means $H_0$ will still firmly be rejected. Thus, the conclusion will not be different.

1. State the hypotheses and identify the appropriate test statistic.

From part (ii), the hypotheses are:
$H_0: \mu = 350$
$H_1: \mu \neq 350$

For the new sample, the population variance is known ($\sigma^2 = 20.74$) and the population follows a normal distribution. Thus, we use a $z$-test instead of a $t$-test.
Under $H_0$, the test statistic is $Z = \frac{\bar{X} - 350}{\sqrt{20.74 / 15}} \sim \mathrm{N}(0, 1)$.

2. Set up the non-rejection region for a 2-tailed test at the 5% level of significance.

For a 2-tailed test at the 5% level of significance, $H_0$ is not rejected if the test statistic falls within the acceptance region:
$-1.95996 \le Z \le 1.95996$

3. Substitute the expression for $Z$ and solve for $\bar{x}$.

$-1.95996 \le \frac{\bar{x} - 350}{\sqrt{20.74 / 15}} \le 1.95996$

$350 - 1.95996\left(\sqrt{\frac{20.74}{15}}\right) \le \bar{x} \le 350 + 1.95996\left(\sqrt{\frac{20.74}{15}}\right)$

$350 - 2.3047 \le \bar{x} \le 350 + 2.3047$

$347.695 \le \bar{x} \le 352.305$

Correct to one decimal place, the range is:
$347.7 \le \bar{x} \le 352.3$