Question 1181346
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.