Concentrations of atmospheric pollutants such as carbon monoxide (CO) can be mea
ID: 3361517 • Letter: C
Question
Concentrations of atmospheric pollutants such as carbon monoxide (CO) can be measured with a spectrophotometer. In a calibration test, 50 measurements were taken of a laboratory gas sample that is known to have a CO concentration of 70 parts per million (ppm). A measurement is considered to be satisfactory if it is within 5 ppm of the true concentration. Of the 50 measurements, 37 were satisfactory. (a) [5 Points] What proportion of the sample measurements was satisfactory? (b) [15 Points] Find a 95% confidence interval for the proportion of measurements made by this instrument that will be satisfactory?
Explanation / Answer
a.
possibile chances (x)=37
sample size(n)=50
success rate ( p )= x/n = 0.74
sample proportion of measurements = 0.74
b.
TRADITIONAL METHOD
given that,
possibile chances (x)=37
sample size(n)=50
success rate ( p )= x/n = 0.74
I.
sample proportion = 0.74
standard error = Sqrt ( (0.74*0.26) /50) )
= 0.062
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
margin of error = 1.96 * 0.062
= 0.1216
III.
CI = [ p ± margin of error ]
confidence interval = [0.74 ± 0.1216]
= [ 0.6184 , 0.8616]
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DIRECT METHOD
given that,
possibile chances (x)=37
sample size(n)=50
success rate ( p )= x/n = 0.74
CI = confidence interval
confidence interval = [ 0.74 ± 1.96 * Sqrt ( (0.74*0.26) /50) ) ]
= [0.74 - 1.96 * Sqrt ( (0.74*0.26) /50) , 0.74 + 1.96 * Sqrt ( (0.74*0.26) /50) ]
= [0.6184 , 0.8616]
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interpretations:
1. We are 95% sure that the interval [ 0.6184 , 0.8616] contains the true population proportion
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 proportion
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