Question

4. (30 pts.) Each clamp produced by a machine is OK with probability P, and faulty with probability 1- P, independent of all else. Unfortunately, P depends on things such as maintenance etc that we cannot know for sure, so that it is reasonable to model P as a random variable that is uniformly distributed on (0.5,0.75). The machine produces a single batch of 5 clamps each day, so it seems reasonable to assume that each clamp in a single batch sees the same value of P, but different batches see different values of P (a) Use simulation to compute the expected number of OK clamps in a batch of 5 clamps (call this g, say). Give a 95% confidence interval for g that is accurate to approximately 1 decimal place, choosing the runlength appropriately. (b) Use your simulation to give a 95% confidence interval for the probability, q say, that 4 or more of the 5 clamps in a batch are OK. You should choose the runlength so that your estimate of q is accurate to approximately 10% of the true value.

Answer 1

Kindly refer to the document attached below.

I have tried to show the formula that i have used to arrive at my results while solving in excel .Please note that the formula is only notional for your guidence of generating/simulating the sample .

95% confidence test OK probability of an independent clamp-P not OK probability 1-P Probablity distirbution (0.50,0.75) 1 :RAN D ) 0 75 0.50 0.50 2RAND() =RAND 2RAND 2RAND =RAND() = F C15c=SB 15,1,01:F(D15c=SB15,1,0 =FIE15c=SB15 1,0 = F(F15c=SB15,1,0) = F(G15c=SB15,1,0 =5UM H15 L15) = F(M15 =4,1,0 2RANDO 0.75-.50)+0.50RANDORANDORANDORANDRANDOIFC16K $B16,1,IF(D16-$B16,1,0)IF(E16-$B16,1,0)F(F16 B16,1,0)IG16 $B16,1,0)SUM(H16:16)F(M16 4,10) 3 RAN D ) () /5-0.50) 0.50 RAND() RANIX) RAND() RAND() RAND() IF(C1/c SB1,1,0) ㅖ(Dİ/c SB11,1,0) ir(1/c SB11,1,0) r(1/ SB1 / 1,0) -r(G1/ SB17,10) SUM(H17:17) ir(M1/> 4,1,0) RAND) 0.75-0.50)+0.S0-RANDORANDORANDO RANDRANDO-IF(C18-SB18,1,0)-IFID18-SB18,1,0)-IF(E18-SB18,1,0)IF(F18-$818,1,0)IFIG18-$B18,1,0-SUM(H18:L18) -IF(M18-4,1,0) RAND O 75-50 +0.50 RAND RAND RAND RAND RAND -F( 19 -SB19,101 FID19-SB19,10 FIE19t-SB19 1,0) -IF(F19e-SB19,1,0) -I (51ge SB19,1,0) -SUM(H19:L19 F(M19 4,1,0) б -RANDO O 75 0.50 0.50 -RAND -RANDO -RAND -RANDO -RAND() = F C20-SB20 1,0 -FD20<-SB2010 -FIE20<-SB2010 -FIF20€-SB20.10 -F G20<-$B20, 101-5 M(H20:L20 -F M20>=4,1,0 7 -RANDO 0,75 0.50 0.50 -RAND -RANDO -RAND -RAND -RAND -F(C21-=SB21,1,0 -FD21 -SB21,1,0 =FIE21-=SB21,1,0 = FF21-=$821,1,0) = F G21-5B2 1,1,0 =5UM(H21:L21) =IF M21>=4,1,0 8 =RAND O 75 0.50 +0.50 2RAND() = AND 2RAND 2RAND 2RAND = F(C22 =SB22,1 -IF(D22 =SB22,1,0 = F(E22 =SB22 1,0 = r(F22 =SB22,1,0 = F(G22 =SB22,1,0 =5UM H22:L22) =IF(M22 =4,1,0 10 RANDO (0.75-0.50)+0.50-RANDRAND)-RANDORANDRANDOIF(C24-$B24,1,0IF(D2-SB24,1,0)IF(E24-B24,1,0)IF(F24 B24,1,IFG24-$B21,1,0)SUM(H24:L24) -IF(M24-1,1,0) 50)0.50 -RANDO -RAND-RANDRAND-RAND-IF(C25-SB25,1,0-IFID25SB25,10)-IFIE25-SB25,1,0) -IFF25-SB25,1,0-IFG25-SB25,1,0-SUMH25:125)-IFM25-4,1 RANDO 0.700.50 -RANDO-RAND RANDRAND -RANDIIFIC26-SB26,1,0) IFID26 SB26,10) IFIE2G SB26,1,0) IF(F26-B26,1,0 IF G26-SB26,1,0 -SUMH26:L26) IFM26>-4,10 13 -RAND()"(0,75 0.50) 0 50 -RAND()-RANDO «RAND() -RANDO «RAND() =IF C27<=SB27,1,0) s|FID27-SB27,1,0) = FIE27<=SB27,1,0) = F(F27-=SB27,1,01 IF(G27-=$B27,1,0} -SUM(H27:L27) = F(M27>-4,1,0)

To have a 95% confidence interval accurate to approximately 1 decimal place , we need a confidence interval of half width 0.05 We first run 100 trials to estimate the sample size required to obtain a confidence interval of half-width 0.05 with confidence 95%. From the 100 trial runs, the sample standard deviation is 1.1315. The standard normal quantile for a 95% confidence interval can be computed with NORMSINV(0.975), which is 1.96. In order to find the number of replications, we want 0.05 = 1-868 1.96 × 1.1315) 2 You may observe a different value, due to randomness. Then, using a similar setup, 1.96s Therefore, -1968 0.05 we obtain the confidence interval [3.07, 3.17

b).

Again, we use 100 trial runs to find the appropriate runlength. Using 100 trial runs, we see that the mean of q is 0.43, and the standard deviation of q is 0.4975. If we want to have the estimate accurate to approximately 10% of the true value, we choose 1.96515 0.1q | = Using the same setup, we obtain the confidence interval (0.38,0.46)