Question

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 = 2.2. (Round your answers to two decimal places.)

(a) Compute a 95% CI for μ when n = 25 and x = 56.7.

(answer,answer) watts

(b) Compute a 95% CI for μ when n = 100 and x = 56.7.

(answer,answer) watts

(c) Compute a 99% CI for μ when n = 100 and x = 56.7.

(answer,answer) watts

(d) Compute an 82% CI for μ when n = 100 and x = 56.7.

(answer,answer) watts

(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.)

Answer 1

Solution :

Given that,

a) Z/2 = Z0.025 = 1.96

Margin of error = E = Z/2 * ( /n)

= 1.96 * ( 2.2 /  25 )

= 0.86

At 95% confidence interval estimate of the population mean is,

- E < < + E

56.7 - 0.86 <   < 56.7 + 0.86

( 55.84 <   < 57.56 )

b) Z/2 = Z0.025 = 1.96

Margin of error = E = Z/2 * ( /n)

= 1.96 * ( 2.2 /  100)

= 0.43

At 95% confidence interval estimate of the population mean is,

- E < < + E

56.7 - 0.43 <   < 56.7 + 0.43

( 56.27 <   < 57.13 )

c) Z/2 = Z0.005 = 2.576

Margin of error = E = Z/2 * ( /n)

= 2.576 * ( 2.2 /  100)

= 0.57

At 99% confidence interval estimate of the population mean is,

- E < < + E

56.7 - 0.57 <   < 56.7 + 0.57

( 56.13 <   < 57.27)

d) Z/2 = Z0.09 = 1.34

Margin of error = E = Z/2 * ( /n)

= 1.34 * ( 2.2 /  100)

= 0.29

At 82% confidence interval estimate of the population mean is,

- E < < + E

56.7 - 0.29 <   < 56.7 + 0.29

( 56.41 <   < 56.99)

e) E = 1.0

Z/2 = Z0.025 = 1.96  

sample size = n = [Z/2* / E] 2

n = [2.576 * 2.2 / 1.0 ]2

n = 32.11

Sample size = n = 33