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 = 3.3. (Round your answers to two decimal places.)
(a) Compute a 95% CI for μ when n = 25 and x = 57.5.
,
watts
(b) Compute a 95% CI for μ when n = 100 and
x = 57.5.
,
watts
(c) Compute a 99% CI for μ when n = 100 and
x = 57.5.
,
watts
(d) Compute an 82% CI for μ when n = 100 and
x = 57.5.
,
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 =
Solution :
Given that,
a) Z/2 = Z0.025 = 1.96
Margin of error = E = Z/2
* (
/n)
= 1.96 * ( 3.3 / 25
)
= 1.29
At 95% confidence interval estimate of the population mean is,
± E
= 57.5 ± 1.29
= ( 56.21, 58.79 )
b) Z/2 = Z0.025 = 1.96
Margin of error = E = Z/2
* (
/n)
= 1.96 * ( 3.3 / 100
)
= 0.65
At 95% confidence interval estimate of the population mean is,
± E
= 57.5 ± 0.65
= ( 56.85, 58.15 )
c) Z/2 = Z0.005 = 2.576
Margin of error = E = Z/2
* (
/n)
= 2.576 * ( 3.3 / 100
)
= 0.85
At 99% confidence interval estimate of the population mean is,
± E
= 57.5 ± 0.85
= ( 56.65, 58.35 )
d) Z/2 = Z0.09 = 1.341
Margin of error = E = Z/2
* (
/n)
= 1.341 * ( 3.3 / 100
)
= 0.44
At 82% confidence interval estimate of the population mean is,
± E
= 57.5 ± 0.44
= ( 57.06, 57.94 )
e) margin of error = E = width / 2 = 1.0 / 2 = 0.5
sample size = n = [Z/2* / E] 2
n = [ 2.576 * 3.3 / 0.5]2
n = 289.05
Sample size = n = 290
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