A sample of 220 families showed that 180 of them own more than 1 car. What is
the best point estimate of the percentage of all families that own more than 1 car?
Give the 90% confidence interval for this estimate.

To find the best point estimate of the percentage of all families that own more than one car, we use the sample proportion. Then, we'll calculate the 90% confidence interval for this estimate.

### 1. Point Estimate

The point estimate for the proportion of families owning more than one car is given by the sample proportion (\( \hat{p} \)).

\[
\hat{p} = \frac{\text{number of families owning more than 1 car}}{\text{total number of families in the sample}}
\]

\[
\hat{p} = \frac{180}{220} = 0.8182
\]

Thus, the best point estimate of the percentage of all families that own more than one car is 81.82%.

### 2. Confidence Interval

To construct a 90% confidence interval for the population proportion, we use the formula for the confidence interval of a proportion:

\[
\text{CI} = \hat{p} \pm Z_{\alpha/2} \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
\]

where:
- \(\hat{p}\) is the sample proportion,
- \(Z_{\alpha/2}\) is the critical value for the desired confidence level,
- \(n\) is the sample size.

For a 90% confidence level, \(Z_{\alpha/2}\) is approximately 1.645.

\[
\hat{p} = 0.8182
\]
\[
n = 220
\]

First, calculate the standard error (SE):

\[
\text{SE} = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.8182 \times (1 - 0.8182)}{220}} = \sqrt{\frac{0.8182 \times 0.1818}{220}} = \sqrt{\frac{0.1488}{220}} = \sqrt{0.000676} \approx 0.026

Now, calculate the margin of error (ME):

\[
\text{ME} = Z_{\alpha/2} \times \text{SE} = 1.645 \times 0.026 = 0.0428
\]

Finally, construct the confidence interval:

\[
\text{CI} = 0.8182 \pm 0.0428
\]

\[
\text{Lower bound} = 0.8182 - 0.0428 = 0.7754
\]
\[
\text{Upper bound} = 0.8182 + 0.0428 = 0.8610
\]

### Conclusion

The best point estimate of the percentage of all families that own more than one car is 81.82%. The 90% confidence interval for this estimate is (77.54%, 86.10%).

Would you like more details or have any questions about this process? Here are some related questions you might find interesting:

1. What is the critical value for a 95% confidence interval?
2. How does increasing the sample size affect the width of the confidence interval?
3. What assumptions are necessary for this confidence interval calculation?
4. How would the confidence interval change if the sample proportion were 0.70 instead of 0.8182?
5. What is the formula for calculating the confidence interval for a mean?
6. How do you interpret a 90% confidence interval in the context of this problem?
7. Can you explain the concept of margin of error in more detail?
8. How does the confidence level affect the critical value and the width of the confidence interval?

**Tip:** Always check the conditions for using the normal approximation in proportion confidence intervals: \(np \geq 5\) and \(n(1-p) \geq 5\).

Learn how to estimate the percentage of families owning more than 1 car with a detailed explanation and 90% confidence interval calculation. This problem involves statistics and confidence intervals, suitable for advanced high school students.

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