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.
A CI is desired for the true average stray-load loss u (watts) for a certain type...
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A Cl is desired for the true average stray-load loss p (watts) for a certain type induction motor when the line current is held at 10 amps for a speed of 1500 rpm. Assume that stray-load loss is normally distributed with o = 2.3. (Round your answers to two decimal places.) (a) Compute a 95% CI for u when n = 25 and x = 56.6. watts (b) Compute a 95% CI for u when n = 100 and x...