Let’s solve the problem step by step.

Step 1: Calculate the sample proportion ($\hat{p}$)

The sample proportion is given by:

$\hat{p} = \frac{x}{n}$

where:

• $x = 114$ (number of adults with kids),
• $n = 300$ (total adults sampled).

$\hat{p} = \frac{114}{300} = 0.38$

Step 2: Calculate the standard error (SE)

The standard error for a proportion is calculated as:

$SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$

$SE = \sqrt{\frac{0.38(1 - 0.38)}{300}} = \sqrt{\frac{0.38 \cdot 0.62}{300}} = \sqrt{\frac{0.2356}{300}} \approx 0.0280$

Step 3: Determine the critical value for a 97% confidence level

For a 97% confidence level, the critical value ($z^*$) is approximately 2.17 (obtained from the standard normal table).

Step 4: Calculate the margin of error (ME)

The margin of error is given by:

$ME = z^* \cdot SE$

$ME = 2.17 \cdot 0.0280 \approx 0.0608$

Step 5: Construct the confidence interval

The confidence interval is:

$\hat{p} - ME < p < \hat{p} + ME$

Substituting the values:

$0.38 - 0.0608 < p < 0.38 + 0.0608$

$0.319 < p < 0.441$

Step 6: Write in point estimate ± margin of error form

The point estimate ± margin of error form is:

$p = 0.380 \pm 0.061$

Final Answer:

1. Tri-inequality form:
   $0.319 < p < 0.441$

2. Point estimate ± margin of error:
   $p = 0.380 \pm 0.061$

Would you like a detailed explanation of any specific step?

5 Related Questions:

1. How is the critical value $z^*$ determined for different confidence levels?
2. Why is the formula $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ used for standard error in proportion problems?
3. What are the implications of using a smaller or larger sample size for this confidence interval?
4. How do confidence intervals change with a different confidence level (e.g., 90% or 99%)?
5. What is the difference between a confidence interval and a hypothesis test?

Tip:

When constructing a confidence interval, ensure that $n\hat{p}$ and $n(1 - \hat{p})$ are both greater than 5 to justify using the normal approximation.