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

Question

A CI is desired for the true average stray-load loss (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 = 2.3. (Round your answers to two decimal places.) (a) Compute a 95% CI for when n = 25 and x = 53.0. watts (b) Compute a 95% CI for when n = 100 and x = 53.0. watts (c) Compute a 99% CI for when n = 100 and x = 53.0. watts (d) Compute an 82% CI for when n = 100 and x = 53.0. watts (e) How large must n be if the width of the 99% interval for is to be 1.0? (Round your answer up to the nearest whole number.) n-

Explanation / Answer

Here population standard deviation = 2.3

(a) Here x = 53.0 ; n = 25

95% CI for = x +- Z95% ( /n) = 53.0 +- 1.96 * (2.3 / 25) = 53.0 +- 1.96 * 0.46 = (52.10, 53.90)

(b) Here x = 53.0 ; n = 100

95% CI for = x +- Z95% ( /n) = 53.0 +- 1.96 * (2.3 / 100) = 53.0 +- 1.96 * 0.23 = (52.55, 53.45)

(c) Here x = 53.0 ; n = 100

99% CI for = x +- Z99% ( /n) = 53.0 +- 2.575 * (2.3 / 100) = 53.0 +- 2.575 * 0.23 = (52.41, 53.59)

(d) Here x = 53.0 ; n = 100

82% CI for = x +- Z82% ( /n) = 53.0 +- 2.575 * (2.3 / 100) = 53.0 + 1.3408 * 0.23 = (52.69, 53.31)

(e) Here width of population intervl is 1.0

so Margin of error = 1 = critical test statistic * standar error of the sample mean

1.0 = 2.575 * 2.3/ n

n = 2.575 * 2.3/1

n = 35

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