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Sagot :
Explanation:
To find the margin of error and the interval estimate of the number of eligible people under 20 years old who had a driver's license in 1995, we will follow these steps:
1. **Calculate the sample proportion (\( \hat{p} \)) for 1995:**
- Given: \( \hat{p} = 0.639 \)
- Sample size (\( n \)) = 1200
2. **Calculate the standard error (SE):**
\[
SE = \sqrt{\frac{\hat{p} (1 - \hat{p})}{n}}
\]
3. **Find the critical value (\( z \)) for a 95% confidence interval:**
- For a 95% confidence level, \( z \approx 1.96 \)
4. **Calculate the margin of error (ME):**
\[
ME = z \times SE
\]
5. **Calculate the interval estimate:**
\[
\text{Interval estimate} = \hat{p} \pm ME
\]
Let's calculate these step-by-step.
### Step 1: Sample Proportion
\[
\hat{p} = 0.639
\]
### Step 2: Standard Error
\[
SE = \sqrt{\frac{0.639 \times (1 - 0.639)}{1200}}
\]
### Step 3: Critical Value
\[
z = 1.96
\]
### Step 4: Margin of Error
\[
ME = 1.96 \times SE
\]
### Step 5: Interval Estimate
\[
\text{Interval estimate} = \hat{p} \pm ME
\]
Let's do the calculations:
#### Standard Error
\[
SE = \sqrt{\frac{0.639 \times 0.361}{1200}}
\]
\[
SE = \sqrt{\frac{0.230679}{1200}}
\]
\[
SE = \sqrt{0.0001922325}
\]
\[
SE \approx 0.01386
\]
#### Margin of Error
\[
ME = 1.96 \times 0.01386
\]
\[
ME \approx 0.02715
\]
#### Interval Estimate
\[
\text{Lower bound} = \hat{p} - ME = 0.639 - 0.02715 = 0.61185
\]
\[
\text{Upper bound} = \hat{p} + ME = 0.639 + 0.02715 = 0.66615
\]
### Summary
- **Margin of Error (ME):** \( 0.02715 \)
- **Interval Estimate:** \( 0.61185 \) to \( 0.66615 \)
These are the 95% confidence interval estimates for the proportion of eligible people under 20 years old who had a driver's license in 1995.
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