We are tasked with constructing a 90% confidence interval for the difference in the proportion of residents in the North and South regions who support the construction of the new high school. Specifically, we want to estimate $p_S - p_N$, where:

• $p_S$ is the proportion of residents in the South who support the construction.
• $p_N$ is the proportion of residents in the North who support the construction.

Step 1: Calculate the sample proportions

• Proportion in the North: $p_\text{N} = \frac{\text{Yes in North}}{\text{Total in North}} = \frac{54}{120} = 0.45$
• Proportion in the South: $p_\text{S} = \frac{\text{Yes in South}}{\text{Total in South}} = \frac{77}{140} \approx 0.55$

Step 2: Calculate the standard error for the difference in proportions

The standard error (SE) for the difference in proportions is given by the formula: $SE = \sqrt{\frac{p_\text{N}(1 - p_\text{N})}{n_\text{N}} + \frac{p_\text{S}(1 - p_\text{S})}{n_\text{S}}}$ where:

• $n_\text{N} = 120$ (sample size for the North),
• $n_\text{S} = 140$ (sample size for the South),
• $p_\text{N} = 0.45$,
• $p_\text{S} = 0.55$.

Substituting the values: $SE = \sqrt{\frac{0.45(1 - 0.45)}{120} + \frac{0.55(1 - 0.55)}{140}} = \sqrt{\frac{0.45 \times 0.55}{120} + \frac{0.55 \times 0.45}{140}}$ $SE = \sqrt{\frac{0.2475}{120} + \frac{0.2475}{140}} \approx \sqrt{0.0020625 + 0.0017679} \approx \sqrt{0.0038304} \approx 0.0619$

Step 3: Calculate the margin of error

For a 90% confidence interval, the critical value $z^*$ is approximately 1.645 (from the standard normal distribution).

The margin of error (ME) is: $ME = z^* \times SE = 1.645 \times 0.0619 \approx 0.1016$

Step 4: Construct the confidence interval

The difference in sample proportions is: $p_\text{S} - p_\text{N} = 0.55 - 0.45 = 0.10$

The 90% confidence interval is: $(p_\text{S} - p_\text{N}) \pm ME = 0.10 \pm 0.1016$ This gives the interval: $(0.10 - 0.1016, 0.10 + 0.1016) = (-0.0016, 0.2016)$

Step 5: Match with the answer choices

The closest interval to our calculation is: $(-0.002, 0.202)$

Thus, the correct answer is: Choice C: $(-0.002, 0.202)$

Do you have any questions or need further details on this calculation?

Here are 5 related questions:

1. How does changing the confidence level (e.g., from 90% to 95%) affect the margin of error?
2. What assumptions need to be checked when using confidence intervals for proportions?
3. How do you calculate the sample size required for a given margin of error in a confidence interval for proportions?
4. What is the interpretation of a confidence interval for the difference in proportions?
5. How can you determine if there is a statistically significant difference between two proportions?

Tip: When constructing confidence intervals for differences in proportions, always double-check that the sample sizes are sufficiently large (usually $n \cdot p$ and $n \cdot (1 - p)$ should both be greater than 10) to ensure that the normal approximation is valid.