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 = 2.3. (Round your answers to two decimal places.)

(a) Compute a 95% CI for μ when n = 25 and x = 59.6.

  ,

watts

(b) Compute a 95% CI for μ when n = 100 and x = 59.6.

  ,

watts

(c) Compute a 99% CI for μ when n = 100 and x = 59.6.

  ,

watts

(d) Compute an 82% CI for μ when n = 100 and x = 59.6.

  ,

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.)
n =

Answer 1

Solution :

Given that,

a) Z/2 = Z0.025 = 1.96

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

= 1.96 * ( 2.3 /  25 )

= 0.90

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

  ± E

= 59.6  ± 0.90

= ( 58.70, 60.50 )

b) Z/2 = Z0.025 = 1.96

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

= 1.96 * ( 2.3 /  100 )

= 0.45

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

  ± E

= 59.6  ± 0.45

= ( 59.15, 60.05 )

c) Z/2 = Z0.005 = 2.576

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

= 2.576 * ( 2.3 /  100 )

= 0.59

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

  ± E

= 59.6  ± 0.59

= ( 59.01, 60.19 )

d) Z/2 = Z0.09 = 1.34

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

= 1.34 * ( 2.3 /  100 )

= 0.31

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

  ± E

= 59.6  ± 0.31

= ( 59.29, 59.91 )

e) margin of error = E = 1.0 / 2 = 0.5

Z/2 = Z0.005 = 2.576

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

n = [2.576 * 2.3 /0.5 ]2

n = 140.41

Sample size = n = 141