Question

(17 Points) An electrical engineer must design a circuit to deliver the maximum amount of current to a display tube to achieve sufficient image brightness. Within her allowable design constraints, she has developed two candidate circuits and test prototypes of each. The resulting data (in microamperes) are as follows: Circuit 1: 251 255 258 257 250 251 254 250 248 Circuit 2: 250 253 249 256 252 260 251 259 Test for evidence that first circuit can deliver a higher current that circuit 2 using a 5% level. You may assume that the populations are normal for this problem. Show all work for this problem! Step 1 Step 2: Step 3: Step 4: Step 5: P-value: Step 6 Step 7: E o IDI

Answer 1

Solution - We have to find mean and standard deviations for both circuit I & circuit 2 step I for circuit 1 Exi 2274 X2 = 574660 exi 2274 18 X = 252.667 S na mi zxp2 - 1 (3x:)] 574660 (2274) 2 9-1 L S = 3.464 Shop II :- for circuit 2. Ex2i = 2 03 0 EXzi =515232 x EX21 2030 12 8 X2 = 253.75 2 2 Exz 1 Sz (sx=;)] 12-1 12 545992 (2030127 S2-4.132 na = 8

step II :- Hypothesis Ho li 시그 VS Ha lle ela This earro usp ends to right tailed t test for two samples- Step II :- critical value. Here d = 0.05 nitha - 2 g + 8 -2 dif-15 : critical value for this right tailed tast is 1.753 The rejection region is, R = { t t > 1.753] step Test statistic since it assumed that population variance are equal, the t-statistic as computed as follows: + x X2 (n,r) sp +(12-1) S₂² ni tn 2-2 no ni 252.667 253.75 (9-1) (3.464)2+(8-1) (4.132)2 9+8-2 6 9 مسن 0.588

P-valve :- P-value = P-value = Pret > -0.588 0.7174 Step I :- Decision about neell hypothesis :- We tailed to reject the null hypothesis because P-value = 0.7174 > a = 0.05. OR t = -0.588 < te= 1.753 it is concluded that the null hypothesis is not rejected. step i conclusion :- It is concluded that the null hypothesis Ho is not rejected. " There is not enough eviendi evidence to claim that first circuit can deliver higher current that circuit 2 aut 5% level of significance