132k views
2 votes
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

1 Answer

5 votes

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 \\eq 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)

Explanation:

From the given information:


P(\overline X < 9.83 ) + P( X> 10.17) = \bigg (1- P \bigg ( (X - \mu)/((\sigma)/(√(n))) \bigg )< Z< \bigg ( (X - \mu)/((\sigma)/(√(n))) \bigg ) \bigg )


P(\overline X < 9.83 ) + P( X> 10.17) = \bigg (1- P \bigg ( (9.83 - 10)/((0.25)/(√(10))) \bigg )< Z< \bigg ( (10.17- 10)/((0.25)/(√(10))) \bigg ) \bigg )


P(\overline X < 9.83 ) + P( X> 10.17) = \bigg (1- P \bigg ( (-0.17)/((0.25)/(√(10))) \bigg )< Z< \bigg ( (0.17)/((0.25)/(√(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}

User Ashish Balchandani
by
5.5k points