An online auction website allows sellers to place a reserve on their items up for auction. Bids on an item with a reserve must exceed a predetermined price before the seller commits to the transaction. In a sample of 2,617 auctions, the mean reserve was $112 with a standard deviation of $560. Find an approximate 95% confidence interval for the true mean reserve price, µ.
A. 116 ± 18.46
B. 110 ± 20.61
C. 112 ± 25.56
D. 112 ± 21.46
To find the approximate 95% confidence interval for the true mean reserve price (µ), we can use the formula:
Confidence Interval = Sample Mean ± (Critical Value * Standard Error)
First, calculate the standard error, which is the standard deviation divided by the square root of the sample size:
Standard Error = Standard Deviation / √(Sample Size)
Standard Error = $560 / √(2,617)
Next, find the critical value associated with a 95% confidence level. This can be determined using a normal distribution table or a statistical calculator. Since the sample size is large (n > 30), we can assume a normal distribution for the sampling distribution of the mean.
For a 95% confidence level, the critical value is approximately 1.96.
Now, substitute the values into the formula:
Confidence Interval = $112 ± (1.96 * Standard Error)
Calculate the Standard Error:
Standard Error ≈ $560 / √(2,617)
Confidence Interval ≈ $112 ± (1.96 * Standard Error)
Now, calculate the confidence interval by substituting the values:
Confidence Interval ≈ $112 ± (1.96 * Standard Error)
Confidence Interval ≈ $112 ± (1.96 * $11.14)
Calculating the expression:
Confidence Interval ≈ $112 ± $21.84
So, the approximate 95% confidence interval for the true mean reserve price is:
112 ± 21.84
Rounded to the nearest two decimal places, the answer is:
112 ± 21.46
Therefore, the correct option is D. 112 ± 21.46.