Question 1181347
Here's how to conduct a hypothesis test for this scenario:

**1. State the Hypotheses:**

*   **Null Hypothesis (H0):** Internet users spend the same amount of time watching television as the average American (μ = 154.8 minutes).
*   **Alternative Hypothesis (H1):** Internet users spend less time watching television than the average American (μ < 154.8 minutes).  This is a one-tailed test (left-tailed).

**2. Significance Level:** α = 0.05

**3. Test Statistic:**

Since we have a sample size greater than 30, we can use a z-test. The formula for the z-statistic is:

z = (sample mean - population mean) / (standard deviation / √sample size)

z = (128.7 - 154.8) / (46.5 / √50)
z = -26.1 / (46.5 / 7.07)
z ≈ -26.1 / 6.58
z ≈ -3.97

**4. P-value:**

Because this is a left-tailed test, the p-value is the probability of getting a z-score as extreme as -3.97 or *lower*.  Using a z-table or calculator:

P(z < -3.97) ≈ 0.0000 (very close to zero)

**5. Decision:**

Since the p-value (≈ 0.0000) is *less than* the significance level (α = 0.05), we *reject* the null hypothesis.

**6. Conclusion:**

There is sufficient evidence at the α = 0.05 level of significance to conclude that Internet users spend less time watching television than the average American.