Question

LURESTA19 7.E.004 NOTES 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 mes for some of 1500 eur load loss is normally distributed with a = 2.3. (Round your answers to two decimal places) (a) Compute a 95% CI for when - 25 and x = 53.7. watts (b) Compute a 95% CI for when n - 100 and X - 53.7. 1) watts (c) Compute a 99% CI for w when n 100 and X - 53.7. watts (d) Compute an 82% CI for a when n - 100 and x = 53.7. Watts (e) How large must be if the width of the 99% interval for is to be 1.07 (Round your answer up to the nearest whole number.)

Answer 1

a.
TRADITIONAL METHOD
given that,
standard deviation, σ =2.3
sample mean, x =53.7
population size (n)=25
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 2.3/ sqrt ( 25) )
= 0.46
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 0.46
= 0.9
III.
CI = x ± margin of error
confidence interval = [ 53.7 ± 0.9 ]
= [ 52.8,54.6 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =2.3
sample mean, x =53.7
population size (n)=25
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 53.7 ± Z a/2 ( 2.3/ Sqrt ( 25) ) ]
= [ 53.7 - 1.96 * (0.46) , 53.7 + 1.96 * (0.46) ]
= [ 52.8,54.6 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [52.8 , 54.6 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean
b.
TRADITIONAL METHOD
given that,
standard deviation, σ =2.3
sample mean, x =53.7
population size (n)=100
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 2.3/ sqrt ( 100) )
= 0.23
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 0.23
= 0.45
III.
CI = x ± margin of error
confidence interval = [ 53.7 ± 0.45 ]
= [ 53.25,54.15 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =2.3
sample mean, x =53.7
population size (n)=100
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 53.7 ± Z a/2 ( 2.3/ Sqrt ( 100) ) ]
= [ 53.7 - 1.96 * (0.23) , 53.7 + 1.96 * (0.23) ]
= [ 53.25,54.15 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [53.25 , 54.15 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean
c.
TRADITIONAL METHOD
given that,
standard deviation, σ =2.3
sample mean, x =53.7
population size (n)=100
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 2.3/ sqrt ( 100) )
= 0.23
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
value of z table is 2.576
margin of error = 2.576 * 0.23
= 0.59
III.
CI = x ± margin of error
confidence interval = [ 53.7 ± 0.59 ]
= [ 53.11,54.29 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =2.3
sample mean, x =53.7
population size (n)=100
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
value of z table is 2.576
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 53.7 ± Z a/2 ( 2.3/ Sqrt ( 100) ) ]
= [ 53.7 - 2.576 * (0.23) , 53.7 + 2.576 * (0.23) ]
= [ 53.11,54.29 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [53.11 , 54.29 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population mean
d.
TRADITIONAL METHOD
given that,
standard deviation, σ =2.3
sample mean, x =53.7
population size (n)=100
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 2.3/ sqrt ( 100) )
= 0.23
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.18
from standard normal table, two tailed z α/2 =1.341
since our test is two-tailed
value of z table is 1.341
margin of error = 1.341 * 0.23
= 0.31
III.
CI = x ± margin of error
confidence interval = [ 53.7 ± 0.31 ]
= [ 53.39,54.01 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =2.3
sample mean, x =53.7
population size (n)=100
level of significance, α = 0.18
from standard normal table, two tailed z α/2 =1.341
since our test is two-tailed
value of z table is 1.341
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 53.7 ± Z a/2 ( 2.3/ Sqrt ( 100) ) ]
= [ 53.7 - 1.341 * (0.23) , 53.7 + 1.341 * (0.23) ]
= [ 53.39,54.01 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 82% sure that the interval [53.39 , 54.01 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 82% of these intervals will contains the true population mean
e.
given data,
population standard deviation =2.3
margin of error = 1
confidence level is 99%
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Zα/2 at 0.01% LOS is = 2.576 ( From Standard Normal Table )
Standard Deviation ( S.D) = 2.3
ME =1
n = ( 2.576*2.3/1) ^2
= (5.92/1 ) ^2
= 35.1 ~ 36