Question

In a Laboratory work, values of DC voltage circuit load were recorded drop as on 9.0. 8,8,912, 910, 9.1, 8.9, 9.0, 8.8, 9.2,9.0, AC signal noise are The mean and variance of =0 and 2 = 4, respectively. Calculate the AC

Answer 1

TRADITIONAL METHOD
given that,
standard deviation, σ =2
sample mean, x =9
population size (n)=10
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 2/ sqrt ( 10) )
= 0.632
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
since our test is two-tailed
value of z table is 1.645
margin of error = 1.645 * 0.632
= 1.04
III.
CI = x ± margin of error
confidence interval = [ 9 ± 1.04 ]
= [ 7.96,10.04 ]
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DIRECT METHOD
given that,
standard deviation, σ =2
sample mean, x =9
population size (n)=10
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
since our test is two-tailed
value of z table is 1.645
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 = [ 9 ± Z a/2 ( 2/ Sqrt ( 10) ) ]
= [ 9 - 1.645 * (0.632) , 9 + 1.645 * (0.632) ]
= [ 7.96,10.04 ]
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interpretations:
1. we are 90% sure that the interval [7.96 , 10.04 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean
Answer:
Upper confidence interval for the mean value of DC voltage values is 10.04