Let's solve this step by step.

(a) Point Estimate

The point estimate for $p$, the proportion of shoppers who remained loyal, is calculated as: $\hat{p} = \frac{\text{Number of loyal shoppers}}{\text{Total number of shoppers}} = \frac{622}{728}$

$\hat{p} = 0.8544 \text{ (rounded to four decimal places)}.$

(b) 95% Confidence Interval for $p$

The formula for the confidence interval is: $\hat{p} \pm Z \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ Where:

• $\hat{p} = 0.8544$,
• $n = 728$,
• $Z$ (for 95% confidence) $= 1.96$.

First, calculate the standard error ($SE$): $SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.8544(1-0.8544)}{728}} = \sqrt{\frac{0.8544 \cdot 0.1456}{728}} = \sqrt{\frac{0.1243}{728}} = 0.0132 \text{ (rounded to four decimal places)}.$

Next, calculate the margin of error ($ME$): $ME = Z \cdot SE = 1.96 \cdot 0.0132 = 0.0259 \text{ (rounded to three decimal places)}.$

The confidence interval is: $\hat{p} \pm ME = 0.8544 \pm 0.0259.$

Lower limit: $0.8544 - 0.0259 = 0.8285.$

Upper limit: $0.8544 + 0.0259 = 0.8803.$

95% Confidence Interval: $0.829 \text{ to } 0.880 \text{ (rounded to three decimal places)}.$

Explanation of the Confidence Interval: The correct statement is:

We are 95% confident that the true proportion of loyal shoppers falls within this interval.

(c) How to Report the Survey Results

As a news writer, the best way to report the survey results is to:

Report $\hat{p}$ along with the margin of error.

This gives readers both the estimate and an idea of its precision.

Margin of Error

From part (b), the margin of error is: $0.0259 \text{ (rounded to three decimal places)}.$

Let me know if you need further clarification or additional steps explained!

5 Related Questions:

1. What does the margin of error represent in a confidence interval?
2. How is the $Z$-score determined for different confidence levels (e.g., 90%, 99%)?
3. Why is the confidence interval wider for smaller sample sizes?
4. What assumptions must hold for this confidence interval calculation to be valid?
5. How would the interval change if the confidence level was increased to 99%?

Tip: Always verify the sample size is large enough ($n \geq 30$) to ensure the normal approximation applies for proportions.