Aconfidence interval (ci) is desired for the true average stray-load loss u (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 strayload loss is normally distributed with o = 3.0. (1). construct a 95% ci for u when n = 25 and x = 58.3. (10 points) (2). construct a 95% ci for u when n = 100 and 1 = 58.3. (10 points) (3). construct a 99% ci for u when n = 100 and x = 58.3. (10 points) (4). how large must n be if the half-width of the 99% interval for u is to be 0.5? (10 points)

A) CI = (57.12 , 59.48)

B) CI = (57.71 , 58.89)

C) CI = (57.53 , 59.07)

D) n = 239.63

Step-by-step explanation:

a)

given data:

mean,

standard deviation, σ = 3

sample size, n = 25Given CI level is 95%, hence α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025,

Zc = Z(α/2) = 1.96

ME = 1.18

CI = (57.12 , 59.48)

b)

Given data:

mean,

standard deviation, σ = 3

sample size, n = 100

Given CI level is 95%, hence α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

ME = 0.59

CI = (57.71 , 58.89)

c)

sample mean,

sample standard deviation, σ = 3

sample size, n = 100

Given CI level is 99%, hence α = 1 - 0.99 = 0.01

α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.58

ME = 0.77

CI = (57.53 , 59.07)

D)

Given data:

Significance Level, α = 0.01,

Margin or Error, E = 0.5,

σ = 3

The critical value for α = 0.01 is 2.58.

for calculating population mean we used

n = 239.63

only 10/60 isn't equivalent

hope this : )

i cant answer this, theres no table. sorry man

step-by-step explanation: