某一特定的流媒体服务平台声称，用户每次访问该平台观看内容的时间平均超过 15 分钟。该平台的一个数据分析团队想要验证这一说法。他们随机抽取了 80 名用户作为样本，并记录了每位用户在单次访问中观看内容的时间。结果发现观看时间的样本均值为 16 分钟。假设观看时间的总体标准差为 $\sigma$ 分钟。

1. Identify the directional nature of the claim to justify a 1-tail test

The claim specifically states that the average time spent is more than 15 minutes. This corresponds to an alternative hypothesis of $H_1: \mu > 15$, which is directional. Hence, a 1-tail test is appropriate.

1. State the condition and the theorem that justifies the validity of the hypothesis test

Since the sample size $n = 80$ is large ($\ge 30$), by the Central Limit Theorem, the sample mean is approximately normally distributed.

1. Formulate the test statistic under the null hypothesis.

Let $\mu$ be the population mean of time spent watching content.
$H_0: \mu = 15$
$H_1: \mu > 15$

Under $H_0$, the test statistic is given by:
$Z = \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}} = \frac{16 - 15}{\frac{\sigma}{\sqrt{80}}} = \frac{\sqrt{80}}{\sigma}$

2. Determine the critical value and the condition for rejecting $H_0$.

For the platform's claim to be valid, $H_0$ must be rejected at the $5\%$ significance level.
The critical value for an upper 1-tail test at the $5\%$ significance level is $z_{0.05} \approx 1.6449$.
Therefore, the test statistic must fall in the critical region:
$Z > 1.6449$

3. Solve the inequality to find the range of $\sigma$.

$\frac{\sqrt{80}}{\sigma} > 1.6449$

Since population standard deviation $\sigma$ must be positive ($\sigma > 0$):
$\sigma < \frac{\sqrt{80}}{1.6449}$
$\sigma < 5.4377$

4. State the final valid range of values for $\sigma$.

Combining this with the fact that $\sigma > 0$, the range of values is:
$0 < \sigma < 5.44 \text{ (to 3 s.f.)}$

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

The significance level is the probability of rejecting the null hypothesis when it is true. In this context, it means there is a probability of $0.05$ of concluding that the average time a user spends watching content per visit is more than $15$ minutes (rejecting $H_0$), when the true average time is actually $15$ minutes ($H_0$ is true).