here,
a) alpha = 1 - P(acceptance region)
= 1 - P(4.85 < X < 5.15)
we know that,
Z = (X - mean) / (standard deviation/sqrt(n)
= 1 - P((4.85-5) / (0.25/sqrt(8)) < Z < (5.15-5) / (0.25/sqrt(8)))
= 1 - P(-1.7 < Z < 1.7)
= 1 - {P(Z < 1.7) - P(Z < -1.7)}
= 1 - {0.9554-0.0446}
= 0.0892
b) calculation for the power of the test for detecting a true mean output voltage of 5.1 volts
beta = P(type II error)
= P(4.85 < X < 5.15)
= P((4.85-5.1) / (0.25/sqrt(8)) < Z < (5.15-5.1) / (0.25/sqrt(8)))
= P(-2.83 < Z < 0.57)
= 0.7157 - 0.0023
= 0.7134
so, power = 1 - beta = 1 - 0.7134 = 0.2866
c) for alpha = 0.01
Z = 2.58
critical region = mean +/- Z*standard deviation/sqrt(n)
= 5 +/- 2.58*0.25/sqrt(8)
= 5 +/- 0.228
= (4.772 , 5.228)
so the critical region would be, Xbar < 4.772 and Xbar > 5.228
d) here,
n = 16, alpha = 0.01, so , Z = 2.58
true mean output = 5.05
Xbar > 5 + Z*standard deviation/sqrt(16)
Xbar > 5 + 2.58*0.25/sqrt(16)
Xbar > 5.161
Type II error = P(X < 5.161) = P(Z < (5.161-5.05)/(0.25/sqrt(16))) = P(Z < 1.776) = 0.9621
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