The National Institute of Standards and Technology (NIST) supplies \"standard ma

Question

The National Institute of Standards and Technology (NIST) supplies "standard materials" whose physical properties are supposed to be known. For example, you can buy from NIST an iron rod whose electrical conductivity is supposed to be 10.1 at 293 kelvin. (The units for conductivity are microsiemens per centimeter. Distilled water has conductivity 0.5.) Of course, no measurement is exactly correct. NIST knows the variability of its measurements very well, so it is quite realistic to assume that the population of all measurements of the same rod has the Normal distribution with mean equal to the true conductivity and standard deviation = 0.1. Here are six measurements on the same standard iron rod, which is supposed to have conductivity 10.1.

10.06    9.89    10.02    10.13    10.21    10.11

NIST wants to give the buyer of this iron rod a 90% confidence interval for its true conductivity. What is this interval? (Round your answers to three decimal places.)

______ to ______ microsiemens per centimeter.

Explanation / Answer

TRADITIONAL METHOD
given that,
sample mean, x =10.07
standard deviation, s =0.1094
sample size, n =6
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.1094/ sqrt ( 6) )
= 0.045
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 5 d.f is 2.015
margin of error = 2.015 * 0.045
= 0.09
III.
CI = x ± margin of error
confidence interval = [ 10.07 ± 0.09 ]
= [ 9.98 , 10.16 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =10.07
standard deviation, s =0.1094
sample size, n =6
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 5 d.f is 2.015
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 = [ 10.07 ± t a/2 ( 0.1094/ Sqrt ( 6) ]
= [ 10.07-(2.015 * 0.045) , 10.07+(2.015 * 0.045) ]
= [ 9.980 , 10.160]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ 9.98 , 10.16 ] 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

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