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 held at 10 amps for a speed of 1500 rpm. Assume that stray-load loss is normally distributed with o = 3.3. (Round your answers to two decimal places.) (a) Compute a 95% CI for u when n = 25 and x = 59.1. watts C (b) Compute a 95% CI for u when n = 100 and x = 59.1. ( C ) watts (c) Compute a 99% CI for u when n = 100 and x = 59.1. watts (d) Compute an 82% CI for u when n = 100 and x = 59.1. 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 =

Answer 1

Solution :

Given that,

$\bar x$ = 59.1

$\sigma$ = 3.3

a ) 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 = Z_0.025 = 1.960

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

= 1.960 * (3.3 / $\sqrt$ 25 )

= 1.29

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

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

59.1 - 1.29 < $\mu$ < 59.1 + 1.29

57.81 < $\mu$ < 60.39

(59.8 , 60.39)

b ) 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 = Z_0.025 = 1.960

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

= 1.960 * (3.3 / $\sqrt$ 100)

= 0.65

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

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

59.1 - 0.65 < $\mu$ < 59.1 + 0.65

58.45 < $\mu$ < 59.75

(58.45 , 59.75)

c ) 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 = Z_0.005 = 2.576

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

= 2.576 * (3.3 / $\sqrt$ 100)

= 0.85

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

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

59.1 - 0.85 < $\mu$ < 59.1 + 0.85

58.25 < $\mu$ < 59.95

(58.25 , 59.95)

d ) 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* (3.3 / $\sqrt$ 100)

= 0.44

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

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

59.1 - 0.44 < $\mu$ < 59.1 + 0.44

58.66 < $\mu$ < 59.55

(58.66 , 59.55)

e ) margin of error = E = 1.0 = 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 = Z_0.005 = 2.576

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

= ((2.576 * 3.3 ) / 0.5 )²

= 289.0272

= 289

Sample size = 289