Pollution Control Two types of instruments for measuring the amount of sulfur mo
ID: 2930269 • Letter: P
Question
Pollution Control
Two types of instruments for measuring the amount of sulfur monoxide in the atmosphere are being compared in an air-pollution experiment. The data set below contains the readings of sulfur monoxide that were recorded by each of the instruments each day for a period of two weeks. The variables are day, instrA and instrB which contain respectively the day the reading was made and the level of sulfur monoxide recorded by instrument A and instrument B respectively.
a.) Use the sign test to test whether the two instrument readings are the same.
b.) Use the rank-sign test to test whether the two instrument readings are the same.
c.) Are there any differences in the conclusions of the two tests?
d.) When might we get two different conclusions from the sign test and the rank-sign test if we are using the same data set?
Explanation / Answer
Solution :
Answer a)
Let
H0 :Amount of sulfur monoxide in air measuring by instrument A and Instrument B ate same.
VS
H1:Amount of sulfur monoxide in air measuring by instrument A and Instrument B are not same.
Here Total 12 positive and 2 Negative
So here Number of trial is 14 and number of success is 12
Now find the P value using Binomial distribution table where the 0.5 is the probability.
The null hypothesis is that there are an equal number of signs (i.e. 50/50).
Therefore the 0.5 of chance being negative sign and .5 for positive assuming null hypothesis is true
Here n=14, X=12, p=0.5, q=0.5
P(x=12) = nCx *P^x * q^n-x
=0.9999
P(x=12) = 0.9999
P value is 0.9999 which is greater than alpha level of significance so we can accept the test
We can conclude that measuring sulfur monoxide by instrument A and Instrument B are same.
Day InstrA InstrB InstrA-InstrB Sign 1 0.96 0.87 0.09 ` 2 0.82 0.74 0.08 + 3 0.75 0.63 0.12 + 4 0.61 0.55 0.06 + 5 0.89 0.78 0.11 + 6 0.64 0.7 -0.06 - 7 0.81 0.69 0.12 + 8 0.68 0.57 0.11 + 9 0.65 0.53 0.12 + 10 0.84 0.88 -0.04 - 11 0.59 0.51 0.08 + 12 0.94 0.79 0.15 + 13 0.91 0.84 0.07 + 14 0.77 0.63 0.14 +Let
H0 :Amount of sulfur monoxide in air measuring by instrument A and Instrument B ate same.
VS
H1:Amount of sulfur monoxide in air measuring by instrument A and Instrument B are not same.
Here Total 12 positive and 2 Negative
So here Number of trial is 14 and number of success is 12
Now find the P value using Binomial distribution table where the 0.5 is the probability.
The null hypothesis is that there are an equal number of signs (i.e. 50/50).
Therefore the 0.5 of chance being negative sign and .5 for positive assuming null hypothesis is true
Here n=14, X=12, p=0.5, q=0.5
P(x=12) = nCx *P^x * q^n-x
=0.9999
P(x=12) = 0.9999
P value is 0.9999 which is greater than alpha level of significance so we can accept the test
We can conclude that measuring sulfur monoxide by instrument A and Instrument B are same.