A rum producer monitors the position of its label on the bottle by sampling two

Question

A rum producer monitors the position of its label on the bottle by sampling two bottles from each batch. One quantity measured is the distance from the bottom of the bottle neck to the top of the label. The process mean should be 1.8 inches. Past experience indicates that the distance varies with = 0.15 inch. (a) The mean distance x for each batch sample is plotted on an xcontrol chart. Calculate the center line and control limits for this chart. (Round your answers to three decimal places.) CL: LCL = UCL = in in (b) The sample standard deviation s for each batch's sample is plotted on an s control chart. What are the center line and control limits for this chart? (Round your answers to four decimal places.) CL LCL UCL in in in

Explanation / Answer

a.
TRADITIONAL METHOD
given that,
standard deviation, =0.15
sample mean, x =1.8
population size (n)=2
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 0.15/ sqrt ( 2) )
= 0.106
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 0.106
= 0.208
III.
CI = x ± margin of error
confidence interval = [ 1.8 ± 0.208 ]
= [ 1.592,2.008 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =0.15
sample mean, x =1.8
population size (n)=2
level of significance, = 0.05
from standard normal table, two tailed z /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 = [ 1.8 ± Z a/2 ( 0.15/ Sqrt ( 2) ) ]
= [ 1.8 - 1.96 * (0.106) , 1.8 + 1.96 * (0.106) ]
= [ 1.592,2.008 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [1.592 , 2.008 ] 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
[ANSWERS]
best point of estimate = mean = 1.8
standard error =0.106
z table value = 1.96
margin of error = 0.208
confidence interval = [ 1.592 , 2.008 ]

b.
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 =0.05
^2 right = (1 - confidence level)/2 = (1 - 0.95)/2 = 0.05/2 = 0.025
^2 left = 1 - ^2 right = 1 - 0.025 = 0.975
the two critical values ^2 left, ^2 right at 1 df are 5.0239 , 0.001
s.d( s^2 )=0.15
sample size(n)=2
confidence interval for ^2= [ 1 * 0.0225/5.0239 < ^2 < 1 * 0.0225/0.001 ]
= [ 0.0225/5.0239 < ^2 < 0.0225/0.001 ]
[ 0.0045 < ^2 < 22.5 ]
and confidence interval for = sqrt(lower) < < sqrt(upper)
= [ sqrt (0.0045) < < sqrt(22.5), ]
= [ 0.0669 < < 4.7434 ]

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