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 against , using units. Statistical Tables and Charts (a) The critical region is or . Find the value of

Answer 1

The missing values in the question are shown in bold forms below.

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 of 0.25 V, and the manufacturer wished to test $\mathbf{ H_o : \mu = 10 V \ against \ H_1 : \mu \neq 10V}$, using  n = 10 units. Statistical Tables and Charts

(a) The critical region is $\mathbf{\overline X < 9.83}$ or $\mathbf{\overline X < 10.17}$ . Find the value of  $\mathbf{\alpha }$

Answer:

∝ = 0.032   (to 3 decimal place)

Step-by-step explanation:

From the given information:

$P(\overline X < 9.83 ) + P( X> 10.17) = \bigg (1- P \bigg ( \dfrac{X - \mu}{\dfrac{\sigma}{\sqrt{n}}} \bigg )< Z< \bigg ( \dfrac{X - \mu}{\dfrac{\sigma}{\sqrt{n}}} \bigg ) \bigg )$

$P(\overline X < 9.83 ) + P( X> 10.17) = \bigg (1- P \bigg ( \dfrac{9.83 - 10}{\dfrac{0.25}{\sqrt{10}}} \bigg )< Z< \bigg ( \dfrac{10.17- 10}{\dfrac{0.25}{\sqrt{10}}} \bigg ) \bigg )$

$P(\overline X < 9.83 ) + P( X> 10.17) = \bigg (1- P \bigg ( \dfrac{-0.17}{\dfrac{0.25}{\sqrt{10}}} \bigg )< Z< \bigg ( \dfrac{0.17}{\dfrac{0.25}{\sqrt{10}}} \bigg ) \bigg )$

$P(\overline X < 9.83 ) + P( X> 10.17) = \bigg (1- P \bigg (-2.15 \bigg )< Z< \bigg ( 2.15 \bigg ) \bigg )$

From the z - tables;

$P(\overline X < 9.83 ) + P( X> 10.17) = \bigg (1- (0.9842 -0.0158) \bigg )$

$\alpha = \mathbf{P(\overline X < 9.83 ) + P( X> 10.17) = 0.032}$