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 = 2.2. (Round your answers to two decimal places.)

(a) Compute a 95% CI for μ when n = 25 and x = 50.8.

,

watts

(b) Compute a 95% CI for μ when n = 100 and x = 50.8.

,

watts

(c) Compute a 99% CI for μ when n = 100 and x = 50.8.

,

watts

(d) Compute an 82% CI for μ when n = 100 and x = 50.8.

,

watts

(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

a)
sample mean, xbar = 50.8
sample standard deviation, σ = 2.2
sample size, n = 25

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

ME = zc * σ/sqrt(n)
ME = 1.96 * 2.2/sqrt(25)
ME = 0.86

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (50.8 - 1.96 * 2.2/sqrt(25) , 50.8 + 1.96 * 2.2/sqrt(25))
CI = (49.94 , 51.66)

b)

sample mean, xbar = 50.8
sample standard deviation, σ = 2.2
sample size, n = 100

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

ME = zc * σ/sqrt(n)
ME = 1.96 * 2.2/sqrt(100)
ME = 0.43

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (50.8 - 1.96 * 2.2/sqrt(100) , 50.8 + 1.96 * 2.2/sqrt(100))
CI = (50.37 , 51.23)

c)

sample mean, xbar = 50.8
sample standard deviation, σ = 2.2
sample size, n = 100

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.5758

ME = zc * σ/sqrt(n)
ME = 2.5758 * 2.2/sqrt(100)
ME = 0.57

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (50.8 - 2.5758 * 2.2/sqrt(100) , 50.8 + 2.5758 * 2.2/sqrt(100))
CI = (50.23 , 51.37)

d)

sample mean, xbar = 50.8
sample standard deviation, σ = 2.2
sample size, n = 100

Given CI level is 82%, hence α = 1 - 0.82 = 0.18
α/2 = 0.18/2 = 0.09, Zc = Z(α/2) = 1.3408

ME = zc * σ/sqrt(n)
ME = 1.3408 * 2.2/sqrt(100)
ME = 0.29

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (50.8 - 1.3408 * 2.2/sqrt(100) , 50.8 + 1.3408 * 2.2/sqrt(100))
CI = (50.51 , 51.09)

e)
The following information is provided,
Significance Level, α = 0.01, Margin or Error, E = 0.5, σ = 2.2

The critical value for significance level, α = 0.01 is 2.58.

The following formula is used to compute the minimum sample size required to estimate the population mean μ within the required margin of error:
n >= (zc *σ/E)^2
n = (2.58 * 2.2/0.5)^2
n = 128.87

Therefore, the sample size needed to satisfy the condition n >= 128.87 and it must be an integer number, we conclude that the minimum required sample size is n = 129
Ans : Sample size, n = 129