Question

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

A CI is desired for the true average stray-load loss μ (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 σ = 3.3. (Round your answers to two decimal places.)

(a) Compute a 95% CI for μ when n = 25 and x = 56.0. . watts

(b) Compute a 95% CI for μ when n = 100 and x = 56.0.

(c) Compute a 99% CI for μ when n = 100 and x = 56.0.

(d) Compute an 82% CI for μ when n = 100 and x = 56.0.

(e) How large must n be if the width of the 99% interval for μ is to be 1.0? (Round your answer up to the nearest whole number.) n = 256 Incorrect:

a)

z value at 95% = 1.96

CI = mean +/ z *(s/sqrt(n))
= 56 +/- 1.96 *(3.3/sqrt(25)))
= (54.71 , 57.29)

The 95% cI is (54.71 , 57.29)

b)

z value at 95% = 1.96

CI = mean +/ z *(s/sqrt(n))
= 56 +/- 1.96 *(3.3/sqrt(100))
= (55.35 , 56.65)

The 95% cI is (55.35 , 56.65 )

c)

z value at 99% = 2.576

CI = mean +/ z *(s/sqrt(n))
= 56 +/- 2.576 *(3.3/sqrt(100))
= (55.15 , 56.85 )

The 99% cI is (55.15 , 56.85 )

d)

z value at 82% = 1.341

CI = mean +/ z *(s/sqrt(n))
= 56 +/- 1.341*(3.3/sqrt(100))
= (55.56 , 56.44)

The 82% cI is (55.56 , 56.44 )

e)

z value at 99% = 2.576
s = 3.3 , ME = 1/2 = 0.5

ME = z *(s/sqrt(n))
0.5= 2.576 *(3.3/sqrt(n))

n = ( 2.576 *3.3/0.5)^2
n = 289