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

Solution

a )Given that,

= 55.3

= 2.2

n = 25

At 95% confidence level the z is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.960

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

= 1.960 * (2.2/ 25 )

= 0.86

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

- E < < + E

55.3 - 0.86 < < 55.3 + 0.86

54.44 < < 56.16

(54.44 , 56.16)

b )Given that,

= 55.3

= 2.2

n = 100

At 95% confidence level the z is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.960

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

= 1.960 * (2.2/ 100 )

= 0.43

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

- E < < + E

55.3 - 0.43< < 55.3 + 0.43

54.87 < < 55.73

(54.87 , 55.73)

c )Given that,

= 55.3

= 2.2

n = 100

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

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

55.3 - 0.57 < < 55.3 + 0.57

54.73 < < 56.87

(54.73 , 56.87)

d )Given that,

= 55.3

= 2.2

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 = 1.341

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

= 1.341 * (2.2/ 100 )

= 0.29

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

- E < < + E

55.3 - 0.29 < < 55.3 + 0.29

55.01 < < 55.59

(55.01 , 55.59)

e ) Given that,

standard deviation = = 2.2

margin of error = E = ( width / 2 = 1 / 2 ) =0.5

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576

Sample size = n = ((Z/2 * ) / E)2

= ((2.576 * 2.2) / 0.5)2

= 128

Sample size = 128