1 answer

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

Question:

A CI is desired for the true average stray-load loss u (watts) for a certain type of induction motor when the line current is

A CI is desired for the true average stray-load loss u (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 o = 2.2. (Round your answers to two decimal places.) (a) Compute a 95% CI for u when n = 25 and x = 55.3. watts (b) Compute a 95% CI for u when n = 100 and x = 55.3. D ) watts (c) Compute a 99% CI for u when n = 100 and x = 55.3. watts (d) Compute an 82% CI for u when n = 100 and x = 55.3. watts (e) How large must n be if the width of the 99% interval for u is to be 1.0? (Round your answer up to the nearest whole number.) n =

Answers

Solution

a )Given that,

\bar x = 55.3

\sigma = 2.2

n = 25

At 95% confidence level the z is ,

\alpha  = 1 - 95% = 1 - 0.95 = 0.05

\alpha / 2 = 0.05 / 2 = 0.025

Z\alpha/2 = Z0.025 = 1.960

Margin of error = E = Z\alpha/2* (\sigma\sqrt/n)

= 1.960 * (2.2/ \sqrt 25 )

= 0.86

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

\bar x - E < \mu < \bar x + E

55.3 - 0.86 < \mu < 55.3 + 0.86

54.44 < \mu < 56.16

(54.44 , 56.16)

b )Given that,

\bar x = 55.3

\sigma = 2.2

n = 100

At 95% confidence level the z is ,

\alpha  = 1 - 95% = 1 - 0.95 = 0.05

\alpha / 2 = 0.05 / 2 = 0.025

Z\alpha/2 = Z0.025 = 1.960

Margin of error = E = Z\alpha/2* (\sigma\sqrt/n)

= 1.960 * (2.2/ \sqrt 100 )

= 0.43

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

\bar x - E < \mu < \bar x + E

55.3 - 0.43< \mu < 55.3 + 0.43

54.87 < \mu < 55.73

(54.87 , 55.73)

c )Given that,

\bar x = 55.3

\sigma = 2.2

n = 100

At 99% confidence level the z is ,

\alpha = 1 - 99% = 1 - 0.99 = 0.01

\alpha / 2 = 0.01 / 2 = 0.005

Z\alpha/2 = Z0.005 = 2.576

Margin of error = E = Z\alpha/2* (\sigma\sqrt/n)

= 2.576 * (2.2/ \sqrt 100 )

= 0.57

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

\bar x - E < \mu < \bar x + E

55.3 - 0.57 < \mu < 55.3 + 0.57

54.73 < \mu < 56.87

(54.73 , 56.87)

d )Given that,

\bar x = 55.3

\sigma = 2.2

n = 100

At 82% confidence level the z is ,

\alpha = 1 - 82% = 1 - 0.82 = 0.18

\alpha / 2 = 0.18 / 2 = 0.09

Z\alpha/2 = Z 0.09 = 1.341

Margin of error = E = Z\alpha/2* (\sigma\sqrt/n)

= 1.341 * (2.2/ \sqrt 100 )

= 0.29

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

\bar x - E < \mu < \bar x + E

55.3 - 0.29 < \mu < 55.3 + 0.29

55.01 < \mu < 55.59

(55.01 , 55.59)

e ) Given that,

standard deviation = \sigma = 2.2

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

At 99% confidence level the z is ,

\alpha  = 1 - 99% = 1 - 0.99 = 0.01

\alpha / 2 = 0.01 / 2 = 0.005

Z\alpha/2 = Z0.005 = 2.576

Sample size = n = ((Z\alpha/2 * \sigma ) / E)2

= ((2.576 * 2.2) / 0.5)2

= 128

Sample size = 128

.

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