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 = 2.0. (Round your answers to two decimal places.)

(a) Compute a 95% CI for μ when n = 25 and x =59.4

(b) Compute a 95% CI for μ when n = 100 and x = 59.4.

(c) Compute a 99% CI for μ when n = 100 and x = 59.4.

(d) Compute an 82% CI for μ when n = 100 and x = 59.4.

(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,

= 59.4

= 2.0

n = 25

A ) At 95% confidence level the z is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.960

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

= 1.960 * (2.0 / 25 ) = 0.784

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

- E < < + E

59.4 - 0.784 < < 59.4 + 0.784

58.62 < < 60.18

(58.62 , 60.18)

b)n = 100

At 95% confidence level the z is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.960

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

= 1.960 * (2.0 / 100 ) = 0.392

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

- E < < + E

59.4 - 0.392< < 59.4 + 0.392

59.01< < 59.79

(59.01 ,59.79)

c) n = 100

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 2.576 * (2.0 / 100 ) = 0.515

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

- E < < + E

59.4 - 0.515< < 59.4 + 0.515

58.89 < < 59.92

(58.89  ,59.92)

d) n = 100

At 82% confidence level the z is ,

  = 1 - 82% = 1 - 0.82 = 0.18

/ 2 = 0.18 / 2 = 0.09

Z_/2 = Z_0.09 = 1.34

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

= 1.34* (2.0 / 100 ) = 0.268

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

- E < < + E

59.4 - 0.268< < 59.4 + 0.268

59.13 < < 59.67

(59.13 ,59.67)

e )standard deviation = = 2.0

margin of error = E = 1

A ) At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Sample size = n = ((Z_/2 * ) / E)²

= ((2.576 * 2.0) / 1)²

= 26.54 = 27

Sample size = 27