Question 1177450
**1. State the Hypotheses**

* Null Hypothesis (H0): The population mean rating is equal to 9.00 (µ = 9.00)
* Alternative Hypothesis (H1): The population mean rating is less than 9.00 (µ < 9.00)

**2. Determine the Test Statistic**

Since the population standard deviation is unknown, we'll use a t-test. The test statistic is calculated as:

```
t = (x̄ - µ) / (s / √n)
```

Where:

* x̄ = sample mean (8.33)
* µ = hypothesized population mean (9.00)
* s = sample standard deviation (1.96)
* n = sample size (16)

Plugging in the values:

```
t = (8.33 - 9.00) / (1.96 / √16) 
t ≈ -1.36
```

**3. Find the P-value**

We need to find the P-value associated with this t-statistic.  Since the alternative hypothesis is µ < 9.00 (a one-tailed test), we want the area to the left of t = -1.36 on a t-distribution with 15 degrees of freedom (df = n - 1 = 15).

Using a t-table or calculator, the P-value is approximately 0.096.

**4. Compare P-value to Significance Level (α)**

* α = 0.05 (given)
* P-value (0.096) > α (0.05)

**5. Decision**

Since the P-value is greater than α, we fail to reject the null hypothesis.

**6. Conclusion**

There is not enough evidence at the 0.05 level of significance to support the claim that the sample comes from a population with a mean rating of less than 9.00.

**Interpretation**

While the sample mean (8.33) is less than 9.00, the difference is not statistically significant. The observed difference could be due to random chance.