A particular streaming service platform claims that the average time a user spends watching content on the platform per visit is more than 15 minutes. A team of data analysts for the platform wants to verify this claim. They randomly select a sample of 80 users and record the amount of time each user spends watching content in a single visit. The sample mean of time spent is found to be 16 minutes. The population standard deviation of time spent is assumed to be $\sigma$ minutes.

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).