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 0.78 volt, and the manufacturer wishes to test H0:μ=5 volts against H1:μ≠5 volts, using n=7 units. Round your answers to four decimal places (e.g. 98.7654). (a) The acceptance region is 4.81≤x¯≤5.14 . Find the value of α . (b) Find the power of the test for detecting a true mean output voltage of 5.1 volts.

Answer #1

Given that,

mean = = 5

standard deviation = = 0.78

n = 7

= 5

= ( /n) = (0.78 / 7 ) = 0.2948

P ( 4.81 5.14 )

P(4.81 - 5 / 0.2948)
( - /_{})
(5.14 - 5 / 0.2948)

P (-0.19 / 0.2948 z 0.14 / 0.2948)

P ( -0.64 z 0.47)

P ( z 0.47 ) - P ( z -0.64 )

Using z table

= 0.6808 - 0.2611

= 0.4197

- Probability = 0.4197

b ) x = 5.1

Using z-score formula,

z = - /_{}

_{=} 5.1 -5 / 0.2948

= 0.1 / 0.2948

= 0.34

Using standard normal table,

z = 0.6331

Given that,

mean = = 5

standard deviation = = 0.78

n = 7

= 5

= ( /n) = (0.78 / 7 ) = 0.2948

P ( 4.81 5.14 )

P(4.81 - 5 / 0.2948)
( - /_{})
(5.14 - 5 / 0.2948)

P (-0.19 / 0.2948 z 0.14 / 0.2948)

P ( -0.64 z 0.47)

P ( z 0.47 ) - P ( z -0.64 )

Using z table

= 0.6808 - 0.2611

= 0.4197

- Probability = 0.4197

b ) x = 5.1

Using z-score formula,

z = - /_{}

_{=} 5.1 -5 / 0.2948

= 0.1 / 0.2948

= 0.34

Using standard normal table,

z = 0.6331

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