Question

Problem 2

An article in the American Industrial Hygiene Association Journal (1976, Vol. 37, pp. 418-422) described a field test for detecting the presence of arsenic in urine samples. The test has been proposed for use among forestry workers because of the increasing use of organic arsenics in that industry. The experiment compared the test as performed by both a trainee and an experienced trainer to an analysis at a remote laboratory. Four subjects were selected for testing and are considered as blocks. The response variable is arsenic content (in ppm) in the subject's urine. The data are as follows. Use Excel to do an ANOVA to determine if there is any difference between the Trainee, Trainer and Lab in measuring arsenic.

Subject	Trainee	Trainer	Lab
A	.05	.05	.04
B	.05	.05	.04
C	.04	.04	.03
D	.15	.17	.10

Part A.) Find the data for Problem 2a in the Ch. 13 homework problem data set file posted to Bb in the Ch. 13 folder. Use Excel to perform the ANOVA. In looking at the data, what is the definition of an entry that reads ".05"?

multiple choice:

A. It is the fraction of samples that were inspected by a trainer or trainee
B. It is the percent accuracy of a trainer or a trainee
C. It is the measured arsenic content in a subject's urine
D.It is the measured arsenic content in the urine of a trainee or trainer

E.None of the above

Part B.) What is the p-value for the hypothesis test associated with Problem 2?

Part C.) What will be the conclusion of the hypothesis test associated with Problem 2?

A. There is no difference in mean arsenic levels in the urine of the four subjects

B. There is no difference in mean arsenic levels in the urine of the trainer, trainee and lab

C. One of the subjects has a higher concentration of arsenic in his/her urine

D. The trainer, trainee and lab, when measuring arsenic in urine, are all yielding the same mean measured arsenic level.

E. none of the above

Answer 1

using excel we get following output-

Anova: Two-Factor Without Replication
SUMMARY	Count	Sum	Average	Variance
A	3	0.14	0.046666667	3.33333E-05
B	3	0.14	0.046666667	3.33333E-05
C	3	0.11	0.036666667	3.33333E-05
D	3	0.42	0.14	0.0013
Trainee	4	0.29	0.0725	0.002691667
Trainer	4	0.31	0.0775	0.003825
Lab	4	0.21	0.0525	0.001025
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Rows	0.021225	3	0.007075	30.32142857	0.000506103	4.757062663
Columns	0.0014	2	0.0007	3	0.125	5.14325285
Error	0.0014	6	0.000233333
Total	0.024025	11

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1)

It is the measured arsenic content in a subject's urine

2)

p-value=0.125

3)

since p-value=0.125>alpha=0.05,null hypothesis is rejected

The trainer, trainee and lab, when measuring arsenic in urine, are all yielding the same mean measured arsenic level