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.
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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