Question 1181362: 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(1987)   (Show Source):
You can put this solution on YOUR website! Here's how to conduct a hypothesis test for a proportion:
**1. Hypotheses:**
* H₀: p ≥ 0.29 (The proportion has *not* decreased, it's the same or higher than the previous poll.)
* H₁: p < 0.29 (The proportion *has* decreased.) This is a left-tailed test.
**2. Level of Significance:** α = 0.10
**3. Test Statistic:**
We'll use a z-test for proportions because the sample size is large enough.
* Sample proportion (p̂) = 318/1157 ≈ 0.275
* Population proportion (p) = 0.29
* Sample size (n) = 1157
z = (p̂ - p) / sqrt[p(1-p)/n]
z = (0.275 - 0.29) / sqrt[0.29 * 0.71 / 1157]
z = -0.015 / sqrt(0.000177)
z ≈ -0.015 / 0.0133
z ≈ -1.13
**4. P-value:**
Since this is a left-tailed test, the p-value is the area to the left of our z-statistic (-1.13) on the standard normal distribution. Using a z-table or calculator:
P-value ≈ 0.129
**5. Decision:**
Our p-value (0.129) is *greater* than our significance level (0.10). Therefore, we *fail to reject* the null hypothesis.
**6. 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. While the sample proportion is lower, the difference is not statistically significant.
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