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
H0:μ=5
volts against
H1:μ≠5
, using n = 8 units. a. The acceptance region is 4.85 ≤
x¯¯¯≤5.15
. 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

Answer:

a= 0.0897

b= 0.71186

Power of the test for detecting a true mean output voltage of 5.1 volts is 0.28814.

Step-by-step explanation:

See attached pictures.

Answer 2

The value of $\alpha$ and $\beta$ are 0.0897 and 0.71186 respectively.

Probability

It is the ratio of favorable events to the total events.

Given

Standard deviation ($\sigma$) = 0.25

$\mu$ = 5

n = 8

How to calculate ?

The acceptance region as

4.85 ≤ $\rm \bar{x}$ ≤5.15

a.  Then type I error of probability,

$\alpha = P(4.85> \bar{x}\ when\ \mu =5) + P(\bar{x}\ > 5.15\ when\ \mu =5)\\\alpha = P(\dfrac{4.85-5}{0.25\sqrt{8} } > \dfrac{\bar{x} - \mu}{\sigma / \sqrt{n} } ) + (\dfrac{\bar{x} - \mu}{\sigma / \sqrt{n} } > \dfrac{4.85-5}{0.25\sqrt{8} })\\\alpha = 2P ( z <-1.697)= 0.0897$

Where, $z = \dfrac{\bar{x} - \mu}{\sigma / \sqrt{n} }$

b.  type II error

$\beta = P(4.85 \leq \bar{x} \leq 5.15\ when\ \mu = 5.1)\\\beta = (\dfrac{4.85 - \mu}{0.25/\sqrt{8} }\leq \dfrac{\bar{x} - \mu}{0.25/\sqrt{8} }\leq \dfrac{5.15 - \mu}{0.25/\sqrt{8} }\ when\ \mu=5.1 )\\\beta = P(-2.8284\leq z\leq 0.5657)\\\beta = P(z\leq 0.5657)- P(z\leq -2.8284)$

therefore, the power of the test for detecting a true mean output voltage of 5.1 volt is

$\begin{aligned} 1 - \beta &= 0.28281 \\\beta &= 0.71186\\\end{aligned}$

Thus, the value of $\alpha$ and $\beta$ are 0.0897 and 0.71186 respectively.

More about the probability link is given below.

https://brainly.com/question/795909