SOLUTION: In a previous poll, 29% of adult Americans with children under the age of 18 reported that their family ate dinner together seven nights a week. Suppose that, in a more recent poll

Question 1181346: In a previous poll, 29% of adult Americans with children under the age of 18 reported that their family ate dinner together seven nights a week. Suppose that, in a more recent poll, 318 of 1157 adults with children under the age of 18 reported that their family ate dinner together seven nights a week. Is there sufficient evidence that the proportion of families with children under the age of 18 who eat dinner together seven nights a week has decreased? Use the α = 0.10 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 proportion:
**1. State the Hypotheses:**
* **Null Hypothesis (H0):** The proportion of families who eat dinner together seven nights a week is the same as the previous poll (p = 0.29).
* **Alternative Hypothesis (H1):** The proportion of families who eat dinner together seven nights a week has *decreased* (p < 0.29). This is a left-tailed test.
**2. Significance Level:** α = 0.10
**3. Calculate the Sample Proportion:**
* Sample proportion (p̂) = (Number of successes) / (Sample size)
* p̂ = 318 / 1157
* p̂ ≈ 0.275
**4. Calculate the Test Statistic (z-score):**
z = (p̂ - p) / √(p(1 - p) / n)
z = (0.275 - 0.29) / √(0.29 * (1 - 0.29) / 1157)
z = -0.015 / √(0.2059 / 1157)
z ≈ -0.015 / 0.0134
z ≈ -1.12
**5. Determine the P-value:**
Since this is a left-tailed test, the p-value is the probability of getting a z-score as extreme as -1.12 or *lower*. Using a z-table or calculator:
P(z < -1.12) ≈ 0.1314
**6. Make a Decision:**
Compare the p-value to the significance level (α):
* p-value (0.1314) > α (0.10)
Since the p-value is *greater* than the significance level, we *fail to reject* the null hypothesis.
**7. Conclusion:**
There is *not* sufficient evidence at the α = 0.10 level of significance to conclude that the proportion of families with children under the age of 18 who eat dinner together seven nights a week has decreased.

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