SOLUTION: In 2008, there were 507 children in Arizona out of 32,601 who were diagnosed with Autism Spectrum Disorder (ASD) ("Autism and developmental," 2008). Find the proportion of ASD in
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Question 1186541: In 2008, there were 507 children in Arizona out of 32,601 who were diagnosed with Autism Spectrum Disorder (ASD) ("Autism and developmental," 2008). Find the proportion of ASD in Arizona with a confidence level of 99%.
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Here's how to calculate the confidence interval for the proportion of ASD in Arizona:
1. **Calculate the sample proportion (p̂):**
p̂ = (Number of children with ASD) / (Total number of children)
p̂ = 507 / 32601
p̂ ≈ 0.01555
2. **Find the critical z-score:**
* For a 99% confidence level, alpha (α) is 1 - 0.99 = 0.01.
* Since confidence intervals are two-tailed, divide alpha by 2: 0.01 / 2 = 0.005.
* Find the z-score that corresponds to 0.005 in each tail (or 0.995 in the center). Using a z-table or calculator, the z-score is approximately 2.576.
3. **Calculate the standard error (SE):**
SE = sqrt[ (p̂ * (1 - p̂)) / n ]
SE = sqrt[ (0.01555 * (1 - 0.01555)) / 32601 ]
SE ≈ sqrt(0.00000472)
SE ≈ 0.00217
4. **Calculate the margin of error (ME):**
ME = z * SE
ME = 2.576 * 0.00217
ME ≈ 0.00559
5. **Calculate the confidence interval:**
Lower Bound = p̂ - ME = 0.01555 - 0.00559 ≈ 0.00996
Upper Bound = p̂ + ME = 0.01555 + 0.00559 ≈ 0.02114
6. **Express the confidence interval:**
The 99% confidence interval for the proportion of ASD in Arizona is approximately (0.00996, 0.02114).
**Interpretation:**
We are 99% confident that the true proportion of children in Arizona with ASD in 2008 was between 0.996% and 2.114%.
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