Question

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:

248 Circuit 1: 251 Circuit 2: 250 255 253 258 249 257 256 250 259 251 252 254 260 250 251

Test for evidence that the 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.

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Answer 1

Let denote mean current in first circuit and second circuit respectively.

111, 112

Hypotheses : To test null hypothesis HV - = 0 against alternative hypothesis H1 H1-H2 > 0 This is a one tailed right sided test as the alternative hypothesis contains '>' sign. Here, 21 = 252.67,81 = 3.46, n = 9,22 =253.75, S2 = 4.13, n2 = 8 The test statistic can be written as: (21-22-0 ta + V m T12 which under H, follows at distribution with 13 df. where + (3.402 +45°) 2 df= 13 (1,012) (4.172, 9 1 + + 8-1 T-1 12-1 and standard error for the difference between means si SE + 3.462 9 + 4.132 8 = 1.862059 ni n2 Decision rule / Rejection region : We reject H, at 0.05 level of significance if P-value < 0.05 or if tstat > 0.05,df 0.4- 0.3- 0.2- 0.1- 0.0- -2 t Now Value of the test statistic : The value of the test statistic is tstat = (252.67 -253.75) 3.462 V9 4.132 + -0.581793 -0.582 associated degrees of freedom = 13 The critical value = tcritical = t0.05,13 =1.770933 1.771 P-value = P(t13 > tstat) = P(t13 > -0.581793) = 1- P(t13 -0.581793) = 1-0.285329 0.714671 0.7147 Conclusion : Since p-value > 0.05 and tstat * tcritical = 1.771, so We fail to reject Ho at 0.05 level of significance Hence, we can conclude that mean for first population is not significantly greater than the mean for second population.

There is sufficient evidence to support the claim that the first circuit can deliver a higher current that circuit 2.