Question

Q test for outliers and calculation of mean for the Spectronic 20 data

Table 1. %Transmittance of Samples Using Different Sp ectrophotom Spectrophotometer Spectronic 20 Genesys 10UV %Transmittance Measurements 2 30.o 2s.9 *Mean, X 30 0 2 25.98 *Median *Do not include data that has been rejected by the Q-test. Q-test for outliers at the 95% confidence level. Show work for each data set in the space provided in the table below. If there are no questionable values, state that the Q-test is not needed. If the Q-test is applied, clearly state the conclusion (retain or reject the value). Instrument Spectronic 20 Genesys 10UV Qeritical (from Table C) Suspect data point, X Calculation of Qexp Compare (circle one) Qexp Qcrit or exp2 Qcrit conclusion Calculation of mean for the Spectronic 20 data. Show all work. Do not include data that have been rejected by the Q-test

Answer 1

Let us first do Q-test for data set 1

arrange the data into ascending order

30,30.1

there are repeating values so not writing it all values

there has to be questionable value for which we have to conduct the test whether it is outlier or not

Let us say we will conduct the test for value 30

Formula for Q test is given as

$Q=\frac{x_{2}-x_{1}}{x_{n}-x_{1}}$

Where

x₁ is smallest (Suspect value)

x₂ is second smallest value

x_n is largest value

Here Suspect value X_q=30

Q Exp =(30.1-30)/30.10-30)=0.1/0.1=1

Q critical value at 95% Confidence interval is 0.73

Q_critical<Q_calculated So the point 30 is an outlier

b)

Let us first do Q-test for data set 2

arrange the data into ascending order

25.9,26,26.1

there are repeating values so not writing it all values

there has to be questionable value for which we have to conduct the test whether it is outlier or not

Let us say we will conduct the test for value 25.9

Formula for Q test is given as

$Q=\frac{x_{2}-x_{1}}{x_{n}-x_{1}}$

Where

x₁ is smallest (Suspect value)

x₂ is second smallest value

x_n is largest value

Here Suspect value X_q=25.9

Q Exp =(26.0-25.9)/26.10-25.9)=0.1/0.2=0.5

Q critical value at 95% Confidence interval is 0.73

Q_critical>Q_calculated So the point 25.9  is not an outlier

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