Question 1181347: According to a survey, the typical American spends 154.8 minutes (2.58 hours) per day watching television. A survey of 50 Internet users results in a mean time watching television per day of 128.7 minutes per day with a standard deviation of 46.5 minutes. Conduct the appropriate test to determine if Internet users spend less time watching television than the average American at α = 0.05 level of significance.
[Remember, you should include your H0 , H1 , your p-value, and a conclusion statement explaining whether you accept or reject the claim.]
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! 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.
|
|
|
