Question

The “winner’s curse” in transaction bidding. In transaction bidding, the “winner’s curse” is the miracle of the winning (or loftiest) shot price being above the anticipated value of the item being auctioned. The Review of Economics and Statistics (Aug. 2001) published a study on whether shot experience impacts the liability of the winner’s curse being. Two groups of a stab in a sealed-shot transaction were compared (1)super-experienced stab and (2) less educated stab. In the super-experienced group, 29 of 189 winning flings were above the item’s anticipated value; 32 of 149 winning flings were above the item’s anticipated value in the less-educated group.

1. Find an estimate of p1, the true proportion of super educated stab who fell prey to the winner’s curse
2. Find an estimate of p2, the true proportion of less-educated stab who fell prey to the winner’s curse.
3. Construct a 90 confidence interval for p1-p2.
4. d. Give a practical interpretation of the confidence interval, part c. Make a statement about whether shot experience impacts the liability of the winner’s curse being.

Short answer

There is no evidence to conclude that the bid experience impacts the likelihood of the winner’s curse occurring.

Step-by-step solution

Step-by-Step Solution Step 1: Let x1 and x2

x₁ = 29, n₁ 189, x₂=32, andn₂=149.

Let x₁ be the number of winning bids above the item’s anticipated value by the super experienced group and x₂ be the number of winning bids above the item’s anticipated value by the less-experienced group.

Here we have informed that 29 of 189 winning bids were above the item’s anticipated value by the super-experienced group, and 32 of the 149 winning bids were above the item’s expected value by the less-experienced group.

that means we have x₁ = 29, n₁ 189, x₂=32, andn₂=149.

(a) Estimate the true proportion of super-experienced bidders

By definitions, the estimate of the true proportion of super-experienced bidders who fall prey to the winner’s curse (p₁) is given as

$\underline{P_1}=\frac{x_1}{n_1}$

$=\frac{29}{189}$

$\underline{P_1}=0.153$

(b) Estimating the true proportion of less-experienced bidders

By definitions, the estimate of the true proportion of less-experienced stab who fall prey to the winner’s curse (p₂) is given as

$\underline{P_2}=\frac{x_2}{n_2}$

$=\frac{32}{149}$

$\underline{P_2}=0.215$

(c) Estimating the difference between the population proportions (P1 – P2)

We know that a (1-α)% confidence interval to estimate the difference between the population proportions (P₁ – P₂) is defined as

(P₁ – P₂)$+z_{\frac{a}{2}}\sqrt{\frac{p_1{q}_1}{n_1}+\frac{p_2{q}_2}{n_2}}$

$z_{\frac{a}{2}=}{z}_{\frac{0.1}{2}}$

role="math" localid="1652709934887" $=z_{0.05}$

$z_{\frac{a}{2}}=1.645$

Thus the 90% confidence interval estimates the difference between the population proportions.

= (P₁ – P₂) ± 1.645 ×$\sqrt{\frac{p_1{q}_1}{n_1}+\frac{p_2{q}_2}{n_2}}$

(P₁ – P₂) is defined as = ( 0.153 – 0.215) ± 1.645 ×role="math" localid="1652710155075" $\sqrt{\frac{0.153\times \left(1-0.153\right)}{189}+\frac{0.215\times \left(1-0.215\right)}{149}}$

= - 0.061 ± 1.645 ×

= - 0.061 ± 0.070

= ( - 0.131, 0.009 )

Therefore the 90% confidence interval to estimate the difference between the population proportions (p₁ – p₂ ) is obtained as (-0,131, 0.009).

(d) Estimate the difference between population proportions

From part (C), we are 90% confident that we estimate the difference between the population proportions. (P₁-P₂) the between -0.131 and 0.009.

Here we observe that zero passes through the 90% confidence interval.

That means there is no evidence of the difference between the population proportions.

Final answer

So, there is no evidence to conclude that the bid experience impacts the likelihood of the winner’s curse occurring.