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A CI is desired for the true average stray-load loss u (watts) for a certain type of induction motor when the line current is

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Answer #1

Part a)

Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.05 /2) = 1.96
57.5 ± Z (0.05/2 ) * 2.2/√(25)
Lower Limit = 57.5 - Z(0.05/2) 2.2/√(25)
Lower Limit = 56.64
Upper Limit = 57.5 + Z(0.05/2) 2.2/√(25)
Upper Limit = 58.36
95% Confidence interval is ( 56.64 , 58.36 )

Part b)

Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.05 /2) = 1.96
57.5 ± Z (0.05/2 ) * 2.2/√(100)
Lower Limit = 57.5 - Z(0.05/2) 2.2/√(100)
Lower Limit = 57.07
Upper Limit = 57.5 + Z(0.05/2) 2.2/√(100)
Upper Limit = 57.93
95% Confidence interval is ( 57.07 , 57.93 )

Part c)

Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.01 /2) = 2.576
57.5 ± Z (0.01/2 ) * 2.2/√(100)
Lower Limit = 57.5 - Z(0.01/2) 2.2/√(100)
Lower Limit = 56.93
Upper Limit = 57.5 + Z(0.01/2) 2.2/√(100)
Upper Limit = 58.07
99% Confidence interval is ( 56.93 , 58.07 )

Part d)

Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.18 /2) = 1.341
57.5 ± Z (0.18/2 ) * 2.2/√(100)
Lower Limit = 57.5 - Z(0.18/2) 2.2/√(100)
Lower Limit = 57.20
Upper Limit = 57.5 + Z(0.18/2) 2.2/√(100)
Upper Limit = 57.80
82% Confidence interval is ( 57.20 , 57.80 )

Part e)

Sample size can be calculated by below formula
n = (( Z(α/2) * σ) / e )2
n = (( Z(0.01/2) * 2.2 ) / 1 )2
Critical value Z(α/2) = Z(0.01/2) = 2.5758
n = (( 2.5758 * 2.2 ) / 1 )2
n = 33
Required sample size at 99% confidence is 33.




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