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 volt, and the manufacturer wishes to test HO: = 7 volts against H1: u* 7 volts. Calculate the probability of a type II error if the true mean output is 7.04 volts and round your answers to three decimal places (e.g. 98.765). and n = 13 (a) a = 0.01 B - i and n = 17 (b) a = 0.01 B = i

Answer 1

a) As this is a two tailed test, we have for 0.01 level of significance, from standard normal tables here:
P(-2.576 < Z < 2.576) = 0.99

Therefore the critical mean values here are obtained as:

$\bar X = \mu_0 \pm 2.576*\frac{\sigma}{\sqrt{n}} = 7 \pm 2.576*\frac{0.25}{\sqrt{13}}$

The probability of type II error is computed as the probability of not rejecting the null hypothesis given the true mean is 7.04 as:

7 – 2.576 * © 23 – 7.04 8 = P( - <Z< 7 + 2.576 * ° 73 – 7.04 0.25 V13 0.25 V13

B= P(-3.1529< Z < 1.9991)

$\beta = P( Z < 1.9991) - P(Z < -3.1529 )$

Getting it from the standard normal tables, we have here:

$\beta = 0.9772 - 0.0008 = 0.9764$

Therefore 0.9764 is the required probability of type II error here.

b) For sample size n = 17, the same probability of type II error here is computed as:

$\beta = P( \frac{7 - 2.576*\frac{0.25}{\sqrt{17}} - 7.04}{\frac{0.25}{\sqrt{17}}} < Z < \frac{7 + 2.576*\frac{0.25}{\sqrt{17}} - 7.04}{\frac{0.25}{\sqrt{17}}} )$

$\beta = P( -3.2357 < Z < 1.9163 )$

$\beta = P( Z < 1.9163 )- P(Z < -3.2357)$

Getting it from the standard normal tables, we have here:

$\beta = 0.9723 - 0.0006 = 0.9717$

Therefore 0.9717 is the required probability of type II error here.