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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:,T3=0.870,T4=0.499,Σxi2=0.456919)

Short Answer

Expert verified

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

P>0.05⇒FailtoRejectH0.

Step by step solution

01

Given information

The given data is

T1=0.638

T2=1.206

T3=0.870

T4=0.499

Σxi2=0.456919.

02

Explanation

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

SST=∑x2-∑x2n

SST=0.4569-(3.2130)225

=0.0440

SSTR=∑xi2ni-∑x2n

=0.090125+0.1558210+0.143926+0.067224-(3.2130)225

SSTR=0.0023

SSE=SST-SSTR

=0.0417

Then,

dfT=k-1

=4-1

=3

dfE=n-k

=25-4

=21

MSTR=SSTRdfT

=0.00233

=0.00077

MSE=SSEdfE

=0.041721

=0.001986

F=MSTRMSE

=0.000770.001986≈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⇒FailtoRejectH0.

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Most popular questions from this chapter

Fill in the missing entries in the partially completed one-way ANOVA tables.

Staph Infections. In the article "Using EDE, ANOVA and Regression to Optimize Some Microbiology Data" (Journal of Statistics Education, Vol. 12, No. 2, online), N. Binnie analyzed bacteria culture data collected by G. Cooper at the Auckland University of Technology. Five strains of cultured Staphylococcus aureus bacteria that cause staph infections were observed for 24hours at 27oC. The following table reports bacteria counts, in millions, for different cases from each of the five strains.

At the 5%significance level, do the data provide sufficient evidence to conclude that a difference exists in mean bacteria counts among the five strains of Staphylococcus aureus? (Note: T1=104,T2=129, T3=185,T4=98,T5=194,Σx2i=25,424.)

An F-curve has df=(12,5). In each case, find the F-value having the specified area to its right.

a.0.01

b.0.05

c.0.005

We have provided data from independent simple random samples from several populations. In each case, determine the following items.

a. SSTR

b. MSTR

c. SSE

d. MSE

e. F

Sample 1 Sample 2 Sample 3
5 10 4
9 4 16

8 10

6

2

In section\(13.2\) we considered two hypothetical examples to explain the logic behind one-way ANOVA. Now you are to further examine those examples.

a. Refer to Table \(13.1\) on page \(528\). Perform a one-way ANOVA on the data and compare your conclusion to that stated in the corresponding "what does it mean"? box. Use \(\alpha =0.05\).

b. Repeat part (a) for the data in Table \(13.2\) on page \(528\).

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