7. Sampling Distributions & Confidence Intervals: Mean

Introduction to Confidence Intervals

7. Sampling Distributions & Confidence Intervals: Mean

Introduction to Confidence Intervals

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Multiple Choice

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?

$y hat equals dollar sign 600$ , $E equals dollar sign 20$

$y hat equals dollar sign 650$ , $E equals dollar sign 70$

$ŷ=\$600$ , $E equals dollar sign 40$

$ŷ=\$650$ , $E equals dollar sign 140$

Verified step by step guidance

Identify the given confidence interval, which is ($580, $720).

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} $.

Substitute the values into the formula: $ \hat{y} = \frac{580 + 720}{2} $.

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} $.

Substitute the values into the formula: $ E = 720 - \hat{y} $ or $ E = \hat{y} - 580 $.

6:33m

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