A manufacturer is interested in the output voltage of a power supply used in a P

Question

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 V, and the manufacturer wished to test H 0: = 8 V against H 1: * 8 V , using n = 10 units. Statistical Tables and Char (a) The critical region is 7.83 or R> 8. 17 . Find the value of Round the answer to 3 decimal places. = the tolerance is +/-2% (b) Find the power of the test for detecting a true mean output voltage of 8.08 . Round the answer to 3 decimal places. The power of the test is

Explanation / Answer

Here standard deviation = 0.25 V

standard error of sample mean se0 = /sqrt(n) = 0.25/sqrt(10) = 0.079

Here we are rejection null hypothesis when x < 7.83 or x > 8.17

Here test statistic

Here critical test statistic Zcritical = (8.17 - 8)/0.079 = 2.152

here type I error here = 2 * Pr(Z > 2.152) = 0.032

Here True mean output voltage = 8.08 V

Power of the test = Pr(x > 8.17 ; 8.08 ; 0.079)

Z = (8.17 - 8.08)/ 0.079 = 1.13924

so Power of the test = Pr(x > 8.17 ; 8.08 ; 0.079) = 1- NORM (Z < 1.13924) = 1 - 0.8727 = 0.1273

Question 2

Standard deviation = 0.23

standard error of sample mean se0 = /sqrt(n) =   0.23/ sqrt(8) = 0.0813

Here critical test statistic Zcritical = (5.16- 5)/0.0813 = 1.968

here type I error here = 2 * Pr(Z > 1.968) = 0.050

(b) True mean output voltage = 5.1 volts

Power = Pr(x > 5.16 ; 5.10 volts; 0.0813)

Z = (5.16 - 5.10)/ 0.0813 = 0.738

Power = Pr(x > 5.16 ; 5.10 volts; 0.0813) = Pr(Z > 0.738) = 1 - Pr(Z < 0.738) = 1 - 0.7697 = 0.2303

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