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

Question

A CI is desired for the true average stray-load loss (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 = 2.8. (Round your answers to two decimal places.)

a)Compute a 95% CI for when n = 25 and x = 51.0.

b)Compute a 95% CI for when n = 100 and x = 51.0.

c)Compute a 99% CI for when n = 100 and x = 51.0v

d)Compute an 82% CI for when n = 100 and x = 51.0.

e)How large must n be if the width of the 99% interval for is to be 1.0? (Round your answer up to the nearest whole number.)

Explanation / Answer

as we know that std error =std deviation/(n)1/2

margin of error E=z*std error

and confidence interval =sample mean -/+ E

lower confidence interval =sample mean -margin of error

upper confidence inteval =sample mean +margin of error

hence from above below is table for CI:

e)

margin of error E =1

for 99% CI ;z =2.5758

required sample size n =(z*std deviation/E)2 =~53

sample mean sample size for CI lower Upper x n std error z E confidence interval confidence interval a 51 25 0.56 1.9600 1.10 49.90 52.10 b 51 100 0.28 1.9600 0.55 50.45 51.55 c 51 100 0.28 2.5758 0.72 50.28 51.72 d 51 100 0.28 1.3408 0.38 50.62 51.38

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