#1 Please write legibly. Use this as the coversheet for your work Read Haldar, C

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#1

Please write legibly. Use this as the coversheet for your work Read Haldar, Chapters 5 and 6. Complete the following problems: 1. Haldar, 5.18 5.18 The biological oxygen demand (BOD) level is measured for 10 days for a river at a station and is found to be 3.6, 4.2, 2.8, 4.5, 3.0, 2.9, 2.8, 5.0, 3.1, and 3.3 mg/L. Assume that the daily BOD level is a normal random variable. (a) Calculate the mean and standard deviation of the daily BOD level. (b) Determine the 99% confidence interval for the mean BOD. (c) Determine the 99% upper confidence limit for the mean BOD. (d) Determine the 99% confidence interval for the variance. 2. The link lengths of a robot manipulator (shown) are given by L1 L,-10±0.1 in L2 = 15 ± 0.3 in L3 = 12 ± 0.2 in Assume norm al distributions with mean±standard deviation. Find the probability that the manipulator end P can reach a point located at a distance of 37.5 inches from the base point O.

Explanation / Answer

A.
sample mean, x =3.52
standard deviation, s =0.7843
sample size, n =10

B.
TRADITIONAL METHOD
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.7843/ sqrt ( 10) )
= 0.248
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 9 d.f is 3.25
margin of error = 3.25 * 0.248
= 0.806
III.
CI = x ± margin of error
confidence interval = [ 3.52 ± 0.806 ]
= [ 2.714 , 4.326 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =3.52
standard deviation, s =0.7843
sample size, n =10
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 9 d.f is 3.25
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 3.52 ± Z a/2 ( 0.7843/ Sqrt ( 10) ]
= [ 3.52-(3.25 * 0.248) , 3.52+(3.25 * 0.248) ]
= [ 2.714 , 4.326 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 99% sure that the interval [ 2.714 , 4.326 ] 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

C.
UPPER INTERVAL
DIRECT METHOD
given that,
sample mean, x =3.52
standard deviation, s =0.7843
sample size, n =10
level of significance, = 0.01
from standard normal table,right tailed value of |t /2| with n-1 = 9 d.f is 2.821
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 3.52 ± Z a/2 ( 0.7843/ Sqrt ( 10) ]
= [ 3.52-(2.821 * 0.248) , 3.52+(2.821 * 0.248) ]
= [ 2.82 , 4.22 ]

D.
CONFIDENCE INTERVAL FOR STANDARD DEVIATION
ci = (n-1) s^2 / ^2 right < ^2 < (n-1) s^2 / ^2 left
where,
s = standard deviation
^2 right = (1 - confidence level)/2
^2 left = 1 - ^2 right
n = sample size
since alpha =0.01
^2 right = (1 - confidence level)/2 = (1 - 0.99)/2 = 0.01/2 = 0.005
^2 left = 1 - ^2 right = 1 - 0.005 = 0.995
the two critical values ^2 left, ^2 right at 9 df are 23.5894 , 1.735
s.d( s^2 )=0.7843
sample size(n)=10
confidence interval for ^2= [ 9 * 0.6151/23.5894 < ^2 < 9 * 0.6151/1.735 ]
= [ 5.5361/23.5894 < ^2 < 5.5361/1.7349 ]
[ 0.2347 < ^2 < 3.191 ]

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