Question

Permeation Sampling. Permeation sampling is a method o sampling air in buildings for pollutants. It can be used over a long period of time and is not affected by humidity, air currents, or temperature. In the paper "Calibration of Permeation Passive Samplers With Silicone Membranes Based on Physicochemical Properties of the Analytes" (Analytical Chemistry, Vol. 75, No, 13, pp. 3182-3192). B. Zabiegata et al. obtained calibration constants experimentally for samples of compounds in each of four compound groups. The following data summarize their results.

At the $5\%$significance level, do the data provide sufficient evidence to conclude that a difference exists in the mean calibration constant among the four compound groups? (Note:,$T_3=0.870,T_4=0.499$,$\Sigma x_i^2=0.456919$)

Short answer

Data do not provide sufficient evidence at the $5\%$significance level since the $p-$value fails to reject the null hypothesis.

$P>0.05\Rightarrow FailtoReject{H}_0$.

Step-by-step solution

Given information

The given data is

$T_1=0.638$

$T_2=1.206$

$T_3=0.870$

$T_4=0.499$

$\Sigma x_i^2=0.456919$.

Explanation

Using the given data, find SST, SSRT, SSE with the relation

$SST=\sum x^2-\frac{{\left(\sum x\right)}^2}{n}$

$SST=0.4569-\frac{(3.2130{)}^2}{25}$

$=0.0440$

$SSTR=\frac{\sum {\left(x_i\right)}^2}{n_i}-\frac{{\left(\sum x\right)}^2}{n}$

$=\frac{0.{0901}^2}{5}+\frac{0.{1558}^2}{10}+\frac{0.{1439}^2}{6}+\frac{0.{0672}^2}{4}-\frac{(3.2130{)}^2}{25}$

$SSTR=0.0023$

$SSE=SST-SSTR$

$=0.0417$

Then,

$d{f}_T=k-1$

$=4-1$

$=3$

$d{f}_E=n-k$

$=25-4$

$=21$

$MSTR=\frac{SSTR}{d{f}_T}$

$=\frac{0.0023}{3}$

$=0.00077$

$MSE=\frac{SSE}{d{f}_E}$

$=\frac{0.0417}{21}$

$=0.001986$

$F=\frac{MSTR}{MSE}$

$=\frac{0.00077}{0.001986}\approx 0.3877$

Let's create the ANOVA table

Data do not provide sufficient evidence at the $5\%$significance level since the $p-$value fails to reject the null hypothesis.

$P>0.05\Rightarrow FailtoReject{H}_0$.