SOLUTION: A manufacturer of children's toys claim that less than 3% of his products are defective.when 500 toys were drawn from a large production rub 5% were found to be defective. Calculat
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Question 1178879: A manufacturer of children's toys claim that less than 3% of his products are defective.when 500 toys were drawn from a large production rub 5% were found to be defective. Calculate the population interest
You can put this solution on YOUR website! Let's break down this problem and calculate the population proportion of defective toys.
**1. Define the Variables:**
* **Claimed Proportion (p₀):** 0.03 (3%) - This is the manufacturer's claim.
* **Sample Size (n):** 500 toys
* **Sample Proportion (p̂):** 0.05 (5%) - This is what was found in the sample.
**2. Understanding the Problem:**
* The manufacturer claims that p < 0.03.
* We want to test if the sample proportion (p̂ = 0.05) provides enough evidence to reject the manufacturer's claim.
**3. Set up the Hypothesis Test:**
* **Null Hypothesis (H₀):** p ≤ 0.03 (The proportion of defective toys is less than or equal to 3%)
* **Alternative Hypothesis (H₁):** p > 0.03 (The proportion of defective toys is greater than 3%)
* This is a right-tailed test.
**4. Calculate the Test Statistic (z-score):**
* We'll use the z-test for proportions because we have a large sample size.
* The formula for the z-score is:
z = (p̂ - p₀) / √[p₀(1 - p₀) / n]
* Plug in the values:
z = (0.05 - 0.03) / √[0.03(1 - 0.03) / 500]
z = 0.02 / √[0.03(0.97) / 500]
z = 0.02 / √[0.0291 / 500]
z = 0.02 / √0.0000582
z = 0.02 / 0.0076289
z ≈ 2.62
**5. Find the P-value:**
* We need to find the probability of getting a z-score of 2.62 or higher in a standard normal distribution.
* Using a z-table or calculator, P(z > 2.62) ≈ 0.0044.
**6. Conclusion:**
* The p-value (0.0044) is very small.
* If we were to use a significance level of 0.05, since 0.0044 < 0.05, we would reject the null hypothesis.
* This means there is strong evidence to suggest that the true proportion of defective toys is greater than 3%.
* Therefore, the manufacturer's claim is likely incorrect.
