Korean dramas, or Kdramas, have been immensely popular among students, captivating their attention during leisure hours. Mr. Kim, a high school Math teacher, wants to investigate the average amount of time that students from junior colleges (JC) in Singapore spend watching Kdrama per day.

1. State the condition for a random sample and explain how the method fails to meet it.

For a sample to be a random sample, every JC student in the target population (all JC students in Singapore) must have an equal probability of being selected. By posting the survey on his own social media and taking the first $100$ responses, only students who see his social media post and are quick to respond are able to be chosen. Thus, not every student has an equal chance of being selected.

1. Calculate the unbiased estimate of the population mean.

$\bar{x} = \frac{\sum x}{n} = \frac{1707}{50} = 34.14$

2. Calculate the unbiased estimate of the population variance.

$s^2 = \frac{1}{n-1} \left( \sum x^2 - \frac{(\sum x)^2}{n} \right)$
$s^2 = \frac{1}{49} \left( 71695 - \frac{1707^2}{50} \right)$
$s^2 = \frac{1}{49} (71695 - 58276.98)$
$s^2 = \frac{13418.02}{49} \approx 273.837 \approx 274$ (to 3 s.f.)

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

Let $\mu$ be the population mean time, in minutes, spent by a KJC student watching Kdrama per day.
$H_0: \mu = 37$
$H_1: \mu < 37$

2. Identify the test distribution and calculate the test statistic and $p$-value.

Since the sample size $n = 50$ is large, by the Central Limit Theorem, the sample mean $\bar{X}$ is approximately normally distributed. We carry out a $Z$-test.
From (b), $\bar{x} = 34.14$ and the unbiased estimate of variance $s^2 \approx 273.837$.
Under $H_0$, the test statistic is:
$Z = \frac{\bar{X} - 37}{S/\sqrt{50}} \sim N(0, 1) \text{ approximately}$
Using a graphing calculator:
Test statistic $z \approx -1.222$
$p\text{-value} \approx 0.1108 > 0.05$

3. Compare the $p$-value with the level of significance and draw a conclusion in context.

Since the $p$-value of $0.111$ is greater than the level of significance $\alpha = 0.05$, we do not reject $H_0$.
There is insufficient evidence at the $5\%$ level of significance to conclude that the population mean time spent is less than $37$ minutes. Therefore, we cannot conclude that Ms. Shin has overstated the average time spent watching Kdrama.

1. State the meaning of the level of significance in the context of the problem.

The 5% level of significance is the probability of rejecting the null hypothesis when it is true. In this context, it means there is a $0.05$ probability of concluding that the average time spent watching Kdrama per day by the students is less than 37 minutes, when in fact the true average time spent is exactly 37 minutes.

1. State the hypotheses and the distribution of the sample mean under the null hypothesis.

Let $\mu$ be the population mean time spent watching Kdrama per day.
$H_0: \mu = 37$
$H_1: \mu \neq 37$ (since we test if it is "different from" 37)

Given the population variance $\sigma^2 = 285$ and sample size $n = 50$.
Under $H_0$, the sample mean $\bar{X} \sim \text{N}\left(37, \frac{285}{50}\right)$, which simplifies to $\bar{X} \sim \text{N}(37, 5.7)$.

2. Determine the critical $z$-values for a 2-tailed test at the 5% level of significance.

For a 2-tailed test at $\alpha = 0.05$, the probability in each tail is $0.025$.
The critical $z$-values are $z = \pm 1.95996$.
$H_0$ will be rejected if the test statistic $z < -1.95996$ or $z > 1.95996$.

3. Set up the test statistic inequalities and solve for the set of values of $k$.

The test statistic is given by $z = \frac{k - 37}{\sqrt{5.7}}$.

For there to be sufficient evidence to reject $H_0$, $k$ must fall in the critical region:
$\frac{k - 37}{\sqrt{5.7}} < -1.95996 \quad \text{or} \quad \frac{k - 37}{\sqrt{5.7}} > 1.95996$

Solving for $k$:
$k < 37 - 1.95996\sqrt{5.7} \quad \text{or} \quad k > 37 + 1.95996\sqrt{5.7}$
$k < 32.321 \quad \text{or} \quad k > 41.679$

To 3 significant figures, $k < 32.3 \text{ or } k > 41.7$.