Question 1170230: 3. Among 10,000 school children, 500 of them are randomly selected for a survey on nutritional status among school age children. Of these 500 school children, 91 are found to be moderately and severely underweight children. At 5% level of significance, is their evidence that the proportion of moderately and severely underweight children is 20%?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! We've already solved this problem in a previous response. Let's recap the steps and the answer.
**1. Define the Hypotheses**
* **Null Hypothesis (H0):** The proportion of moderately and severely underweight children is 20% (p = 0.20).
* **Alternative Hypothesis (H1):** The proportion of moderately and severely underweight children is not 20% (p ≠ 0.20).
**2. Set the Significance Level**
* Significance level (α) = 0.05
**3. Calculate the Sample Proportion**
* Sample size (n) = 500
* Number of underweight children in the sample (x) = 91
* Sample proportion (p̂) = x / n = 91 / 500 = 0.182
**4. Calculate the Test Statistic (z-score)**
* Standard error (SE) = √[p(1-p) / n] = √[0.20(0.80) / 500] = √(0.16 / 500) = √0.00032 = 0.0178885
* z = (p̂ - p) / SE = (0.182 - 0.20) / 0.0178885 = -0.018 / 0.0178885 ≈ -1.006
**5. Determine the Critical Value**
* Since this is a two-tailed test (p ≠ 0.20) and α = 0.05, the critical z-values are approximately ±1.96.
**6. Make a Decision**
* The calculated z-score (-1.006) falls within the non-rejection region (-1.96 < -1.006 < 1.96).
* Therefore, we fail to reject the null hypothesis.
**7. Draw a Conclusion**
* There is not enough evidence at the 5% significance level to conclude that the proportion of moderately and severely underweight children is different from 20%.
**Answer**
No, there is not sufficient evidence to say that the proportion of moderately and severely underweight children is different from 20% at the 5% level of significance.
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