A company is interested in estimating lunch costs in their cafeteria, so they survey employees asking for their monthly lunch expenses. The staff calculated a 95% confidence interval for the mean amount of money spent on lunch to be (\$580, \$720). What are the point estimate and margin of error for this sample data?

• A. $\hat{y}=\mathrm{\backslash (}600$, $E=\mathrm{\backslash )}20$
• B. $\hat{y}=\mathrm{\backslash (}650$, $E=\mathrm{\backslash )}70$
• C. $ŷ=\$600$, $E=\$40$
• D. $ŷ=\$650$, $E=\$140$

Answer: (B)

Verified step by step guidance

1. Identify the given confidence interval, which is (\$580, \$720).
2. The point estimate (mean) is the midpoint of the confidence interval. Calculate it using the formula: \( \hat{y} = \frac{\text{Lower Limit} + \text{Upper Limit}}{2} \).
3. Substitute the values into the formula: \( \hat{y} = \frac{580 + 720}{2} \).
4. Calculate the margin of error (E) by finding the difference between the point estimate and one of the interval limits. Use the formula: \( E = \text{Upper Limit} - \hat{y} \) or \( E = \hat{y} - \text{Lower Limit} \).
5. Substitute the values into the formula: \( E = 720 - \hat{y} \) or \( E = \hat{y} - 580 \).