Answer:
Given,
Ho : u = 5
Ha : u != 5
a)
P(Type I error) = 1 - P(4.83 < xbar < 5.13)
= 1 - P((4.83 - 5)/(0.74/sqrt(6)) < (x-u)/(s/sqrt(n)) < (5.13 - 5)/(0.74/sqrt(6)))
= 1 - P(-0.56 < z < 0.43)
= 1 - [P(z < 0.43) - P(z < - 0.56)]
= 1 - 0.6664022 - 0.2877397 [since from z table]
= 0.0459
b)
P(Type II error) = P(4.83 < xbar < 5.13)
= P((4.83 - 5.1)/(0.74/sqrt(6)) < (x-u)/(s/sqrt(n)) < (5.13 - 5.1)/(0.74/sqrt(6)))
= P(-0.89 < z < 0.10)
= P(z < 0.10) - P(z < - 0.89)
= 0.5398278 - 0.1867329 [since from z table]
= 0.3531
Power = 1 - P(Type II error)
= 1 - 0.3531
= 0.6469
Question 8 A manufacturer is interested in the output voltage of a power supply used in...
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