A CI is desired for the true average stray-load loss (watts) for a certain type
ID: 3153416 • Letter: A
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 = 2.1. (Round your answers to two decimal places.)
(a) Compute a 95% CI for when n = 25 and x = 57.3.
,
watts
(b) Compute a 95% CI for when n = 100 and x = 57.3.
,
watts
(c) Compute a 99% CI for when n = 100 and x = 57.3.
,
watts
(d) Compute an 82% CI for when n = 100 and x = 57.3.
,
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
A)
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.025
X = sample mean = 57.3
z(alpha/2) = critical z for the confidence interval = 1.959963985
s = sample standard deviation = 2.1
n = sample size = 25
Thus,
Margin of Error E = 0.823184874
Lower bound = 56.47681513
Upper bound = 58.12318487
Thus, the confidence interval is
( 56.47681513 , 58.12318487 ) [ANSWER]
*************************
b)
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.025
X = sample mean = 57.3
z(alpha/2) = critical z for the confidence interval = 1.959963985
s = sample standard deviation = 2.1
n = sample size = 100
Thus,
Margin of Error E = 0.411592437
Lower bound = 56.88840756
Upper bound = 57.71159244
Thus, the confidence interval is
( 56.88840756 , 57.71159244 ) [ANSWER]
***************************
c)
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.005
X = sample mean = 57.3
z(alpha/2) = critical z for the confidence interval = 2.575829304
s = sample standard deviation = 2.1
n = sample size = 100
Thus,
Margin of Error E = 0.540924154
Lower bound = 56.75907585
Upper bound = 57.84092415
Thus, the confidence interval is
( 56.75907585 , 57.84092415 ) [ANSWER]
****************************
d)
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.09
X = sample mean = 57.3
z(alpha/2) = critical z for the confidence interval = 1.340755034
s = sample standard deviation = 2.1
n = sample size = 100
Thus,
Margin of Error E = 0.281558557
Lower bound = 57.01844144
Upper bound = 57.58155856
Thus, the confidence interval is
( 57.01844144 , 57.58155856 ) [ANSWER]
***************************
e)
Note that
n = z(alpha/2)^2 s^2 / E^2
where
alpha/2 = (1 - confidence level)/2 = 0.005
Using a table/technology,
z(alpha/2) = 2.575829304
Also,
s = sample standard deviation = 2.1
E = margin of error = 1.0/2 = 0.5
Thus,
n = 117.039576
Rounding up,
n = 118 [ANSWER]
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