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 [tex]\mathbf{ H_o : \mu = 10 V \ against \ H_1 : \mu \neq 10V}[/tex], using n = 10 units. Statistical Tables and Charts
(a) The critical region is [tex]\mathbf{\overline X < 9.83}[/tex] or [tex]\mathbf{\overline X < 10.17}[/tex] . Find the value of [tex]\mathbf{\alpha }[/tex]
Answer:
∝ = 0.032 (to 3 decimal place)
Step-by-step explanation:
From the given information:
[tex]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 )[/tex]
[tex]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 )[/tex]
[tex]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 )[/tex]
[tex]P(\overline X < 9.83 ) + P( X> 10.17) = \bigg (1- P \bigg (-2.15 \bigg )< Z< \bigg ( 2.15 \bigg ) \bigg )[/tex]
From the z - tables;
[tex]P(\overline X < 9.83 ) + P( X> 10.17) = \bigg (1- (0.9842 -0.0158) \bigg )[/tex]
[tex]\alpha = \mathbf{P(\overline X < 9.83 ) + P( X> 10.17) = 0.032}[/tex]
