A CI is desired for the true average stray-load loss mu (watts) for a certain ty

Question

A CI is desired for the true average stray-load loss mu (watts) for a certain type of 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 sigma = 2.8. (Round your answers to two decimal places.) (a) Compute a 95% CI for mu when n = 25 and x = 52.0 (b) Compute a 95% CI for mu when n = 100 and x = 52.0 (c) Compute a 99% CI for mu when n = 100 and x = 52.0 (d) Compute an 82% CI for mu when n = 100 and x = 52.0. (e) How large must n be if the width of the 99% interval for mu is to be 1.0? (Round your answer up to the nearest whole number.) You may need to use the appropriate table in the Appendix of Tables to answer this question.

Explanation / Answer

a.
Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=52
Standard deviation( sd )=2.8
Sample Size(n)=25
Confidence Interval = [ 52 ± Z a/2 ( 2.8/ Sqrt ( 25) ) ]
= [ 52 - 1.96 * (0.56) , 52 + 1.96 * (0.56) ]
= [ 50.9,53.1 ]
b.
Sample Size(n)=100
Confidence Interval = [ 52 ± Z a/2 ( 2.8/ Sqrt ( 100) ) ]
= [ 52 - 1.96 * (0.28) , 52 + 1.96 * (0.28) ]
= [ 51.45,52.55 ]
c.
AT 0.01
Confidence Interval = [ 52 ± Z a/2 ( 2.8/ Sqrt ( 100) ) ]
= [ 52 - 2.58 * (0.28) , 52 + 2.58 * (0.28) ]
= [ 51.28,52.72 ]
d.
AT 82% CI
Confidence Interval = [ 52 ± Z a/2 ( 2.8/ Sqrt ( 100) ) ]
= [ 52 - 1.34 * (0.28) , 52 + 1.34 * (0.28) ]
= [ 51.62,52.38 ]
e.
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.01% LOS is = 2.58 ( From Standard Normal Table )
Standard Deviation ( S.D) = 2.8
ME =1
n = ( 2.58*2.8/1) ^2
= (7.22/1 ) ^2
= 52.19 ~ 53      

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