Problem 6 (8 points) A manufacturer is interested in the output voltage of a pow
ID: 3324030 • Letter: P
Question
Problem 6 (8 points) A manufacturer is interested in the output voltage of a power supply used in a PC. Output voltage is assumed to be normally distributed, with standard deviation 0.25 Volts, and the manufacturer wishes to test using n-8 units. a) The acceptance region is 4.85 5.15. Find the value of. b) Find the power of the test for detecting a true mean output voltage of 5.1 Volts. c) Find the boundary of the critical region if the type 1 error probability is -001 d) Find the probability of a type II error if the true mean output is 5.05 volts and =0.01 and n=16.Explanation / Answer
(a) Here acceptance region is given.
Standard deviation = 0.25 volts
Margin of error = Critical test statistic * Standard error of sample mean
Margin of errro = 5.15 - 5.00 = 0.15
standard error of sample mean se = / sqrt(n) = 0.25/ sqrt(8) = 0.0884
so,
0.15 = 0.0884 * Critical test statistic (Z*)
Z* = 1.697
so alpha= 2 * Pr(Z > 1.697) = 2* (1- 0.9552) = 0.09
(b) Here true mean output voltage = 5.1 volts
so Here Power means that we will reject the null hypothesis when it is false
Pr(x > 5.15 ) = NORM(x > 5.15; 5.1; 0.0884)
Z = (5.15 - 5.1)/ 0.0884 = 0.5656
Pr(x > 5.15 ) = Pr(Z > 0.5656) = 1 - Pr(Z < 0.5656) = 1 - 0.7142 = 0.2858
(c) if type I error probbility is alpha = 0.01
then boundary of critical region are to be calculated by 99% confidence interval.
99% confidenceinterval around mean = x +- Z99% SE
= 5.00 + 2.575 * 0.0884
= (4.772, 5.228)
so critical region would be x < 4.772 & x> 5.228
(d) If true mean output is 5.05 volts
n = 16
standard erro of sample mean = 0.25/ sqrt(16) = 0.0625
and alpha = 0.01
so we have to find the sample above which we will be able to reject the null hypothesis.
x > 5.00 + 2.575 * 0.0625 = 5.161
So Pr (Type II error) = Pr(x < 5.161; 5.05 ; 0.0625)
Z = (5.161 - 5.05)/ 0.0625 = 1.775
Pr(Type II error ) = Pr(Z < 1.775) = 0.9621