3. Confidence Interval: You operate a quality control company that evaluates the
ID: 3363302 • Letter: 3
Question
3. Confidence Interval: You operate a quality control company that evaluates the operational quality of slot machines that are distributed through your partner vendor. For each slot machine that you test, the expected loss (gain for the casino) is 10 cents per $1 bet. In order to test each machine, you run 1000 bets and calculate the amount that a hypothetical player would receive as a payout. For each bet, the player either wins $5 or loses a. Compute a 95% confidence interval for the average loss per bet for the outcomes below (i-iii), which slot machines (if any) would you suspect as being defective? If you suspect a machine to be defective, state your level of confidence. Note: A machine is defective if it deviates significantly from an expected loss of 10 cents per game. Payouts $1250 (net winnings = $250) Payouts $1100 (net winnings = $100) ii. Payout-$700 (net winnings300) b. Repeat (a) using a 95% confidence interval of the proportion of winning games. Compare your results with those obtained in (a) and (b). Note: The proportion of games that a player can expect to win for a non-defective slot machine is 18%.Explanation / Answer
Given the proportion of games win is 18% P = 0.18
Q=1-P= 1-0.18 = 0.82
case 1 : number of bets (n) = 1000
Let X be the number of games expected to win ; Expected number of games win is nP = 1000*0.18 =180 ; Variance = nPQ = 180*0.82 = 147.6; SD = 12.149
In case 1 : X =250 ; p =250/1000 =0.25 ; SD(x) =sqrt(npq) = sqrt(1000*0.25*0.75) =13.69
The 95% confidence interval for the number of wins = X +- z* SD(X)= X+- (z* sqrt(npq))
The Z value at 5% level is 1.96
The 95% confidence interval for the number of wins = 250 +- (1.96* 13.69) =(223.2 , 276.8)
Case 2: X= 100 ; p=0.1 ; SD(X) = sqrt(1000*0.1*0.9) =9.49
The 95% confidence interval for the number of wins = 100 +- (1.96* 9.49) =(81.4 , 118.6)
Case 3:X = 300 ; p =300/1000 = 0.3 ; SD(X) =sqrt(1000*0.3*0.7)=14.49
The 95% confidence interval for the number of wins = 300 +- (1.96* 14.149) =(271.2 , 328.4)
If the machine is not defective then the 95% confidence interval would be
The 95% confidence interval for the number of wins = np+- (z* sqrt(nPQ) =180 +- (1.96* 12.149) =(203.8 , 276.2)
There is a 95% confidence that the number of wins would between 203.8 to 276.2 for a non defective machine.
Therefore case 2 is a non defective machine.
For 95% confidence interval , the case 1 and 3 the machines are defective
95% confidence interval for proportions for cases 1 to 3
proportion p =250/1000 = 0.25 ; SD(p) = sqrt(p*q/n) = 0.014
The 95% CI of proportion of winning games is = p+- (z*sqrt(p*q/n) = 0.25+- (1.96*0.014) = (0.223, 0.277)
Case 2: p =100/1000=0.1 ;SD(p) = sqrt(0.1*0.9/1000) = 0.0095
The 95% CI is = p+- (z*sqrt(p*q/n) = 0.1+- (1.96*0.0095) = (0.081, 0.119)
Case 3: p =300/1000=0.3 ;SD(p) = sqrt(0.3*0.7/1000) = 0.015
The 95% CI is = p+- (z*sqrt(p*q/n) = 0.3+- (1.96*0.015) = (0.272, 0.328)
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