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Mathematics

1 answer:

Naya [18.7K]3 years ago

4 0

Answer:

<em>Hope that helps!</em>

Step-by-step explanation:

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A manufacturer is interested in the output voltage of a power supply used in a PC. Output voltage is assumed to be normally dist

Luda [366]

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:

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

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

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