In auction bidding, the "winner's curse" is the phenomenon of the winning (or highest) bid price being above the expected value of the item being auctioned. A study was conducted to see if less-experienced bidders were more likely to be impacted by the curse than super-experienced bidders. The study showed that of the 172 bids by super-experienced bidders, 24 winning bids were above the item's expected value, and of the bids by the 138 less-experienced bidders, 36 win bids were above the item's expected value. Complete parts a through d. a. Find an estimate of p,, the true proportion of super-experienced bidders who fall prey to the winner's curse. P1 = 0.140 (Type an integer or a decimal. Round to three decimal places as needed.) b. Find an estimate of p2, the true proportion of less-experienced bidders who fall prey to the winner's curse. P2 = 0.261 (Type an integer or a decimal. Round to three decimal places as needed.) c. Construct a 90% confidence interval for (p, - P2). (Round to three decimal places as needed.)

MATLAB: An Introduction with Applications
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Author:Amos Gilat
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17) part C
In auction bidding, the "winner's curse" is the phenomenon of the winning (or highest) bid price being above the expected value of the item being auctioned. A study was conducted to see if less-experienced bidders were more likely to be impacted by the curse than super-experienced bidders. The study showed that of the 172 bids by super-experienced bidders, 24 winning bids were above the item’s expected value, and of the bids by the 138 less-experienced bidders, 36 winning bids were above the item’s expected value. Complete parts a through d.

a. Find an estimate of \( p_1 \), the true proportion of super-experienced bidders who fall prey to the winner’s curse.

\[
\hat{p}_1 = \frac{24}{172} = 0.140 \quad \text{(Type an integer or a decimal. Round to three decimal places as needed.)}
\]

b. Find an estimate of \( p_2 \), the true proportion of less-experienced bidders who fall prey to the winner’s curse.

\[
\hat{p}_2 = \frac{36}{138} = 0.261 \quad \text{(Type an integer or a decimal. Round to three decimal places as needed.)}
\]

c. Construct a 90% confidence interval for \( (\hat{p}_1 - \hat{p}_2) \).

\[
\quad \square \quad \square \quad \boxed{\boxed{\boxed{\boxed{\frown}}}}
\]

\[
\text{(Round to three decimal places as needed.)}
\]
Transcribed Image Text:In auction bidding, the "winner's curse" is the phenomenon of the winning (or highest) bid price being above the expected value of the item being auctioned. A study was conducted to see if less-experienced bidders were more likely to be impacted by the curse than super-experienced bidders. The study showed that of the 172 bids by super-experienced bidders, 24 winning bids were above the item’s expected value, and of the bids by the 138 less-experienced bidders, 36 winning bids were above the item’s expected value. Complete parts a through d. a. Find an estimate of \( p_1 \), the true proportion of super-experienced bidders who fall prey to the winner’s curse. \[ \hat{p}_1 = \frac{24}{172} = 0.140 \quad \text{(Type an integer or a decimal. Round to three decimal places as needed.)} \] b. Find an estimate of \( p_2 \), the true proportion of less-experienced bidders who fall prey to the winner’s curse. \[ \hat{p}_2 = \frac{36}{138} = 0.261 \quad \text{(Type an integer or a decimal. Round to three decimal places as needed.)} \] c. Construct a 90% confidence interval for \( (\hat{p}_1 - \hat{p}_2) \). \[ \quad \square \quad \square \quad \boxed{\boxed{\boxed{\boxed{\frown}}}} \] \[ \text{(Round to three decimal places as needed.)} \]
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