Question

In a collaborative trial, four laboratories were sent samples from the same batch of a pharmaceutical product and requested to perform ten assays and report the results based on percentage of a labeled amount of the drug

Lab	Score
Lab A	100
Lab A	99.8
Lab A	99.5
Lab A	100.1
Lab A	99.7
Lab A	99.9
Lab A	100.4
Lab A	100
Lab A	99.7
Lab A	99.9
Lab B	99.5
Lab B	100
Lab B	99.3
Lab B	99.9
Lab B	100.3
Lab B	99.5
Lab B	99.6
Lab B	98.9
Lab B	99.8
Lab B	100.1
Lab C	99.6
Lab C	99.3
Lab C	99.1
Lab C	99.7
Lab C	99.6
Lab C	99.4
Lab C	99.5
Lab C	99.5
Lab C	99.5
Lab C	99.9
Lab D	99.8
Lab D	100.5
Lab D	100
Lab D	100.1
Lab D	99.4
Lab D	99.6
Lab D	100.2
Lab D	99.9
Lab D	100.4
Lab D	100.1

1-Which labs have means that are significantly different? (mark all that apply)

		a. Lab A and Lab B
		b. Lab A and Lab C
		c. Lab A and Lab D
		d. Lab B and Lab C
		e. Lab B and Lab D
		f. Lab C and Lab D

2- What is the conclusion of your hypothesis test?

		a. All the means are different between the labs
		b. At least one of the labs has a mean that is significantly different from the others
		c. None of the above

Answer 1

To answer the given questions, first we will have to conduct a one-way ANOVA and obtain the ANOVA table.

We have to test

H₀: μ_A = μ_B = μ_C = μ_D

H₁: at least one μ_i is different.

	Original observation			squared observation
	LAB A	LAB B	LAB C	LAB D		LAB A	LAB B	LAB C	LAB D
	100	99.5	99.6	99.8		10000	9900.25	9920.16	9960.04
	99.8	100	99.3	100.5		9960.04	10000	9860.49	10100.25
	99.5	99.3	99.1	100		9900.25	9860.49	9820.81	10000
	100.1	99.9	99.7	100.1		10020.01	9980.01	9940.09	10020.01
	99.7	100.3	99.6	99.4		9940.09	10060.09	9920.16	9880.36
	99.9	99.5	99.4	99.6		9980.01	9900.25	9880.36	9920.16
	100.4	99.6	99.5	100.2		10080.16	9920.16	9900.25	10040.04
	100	98.9	99.5	99.9		10000	9781.21	9900.25	9980.01
	99.7	99.8	99.5	100.4		9940.09	9960.04	9900.25	10080.16
	99.9	100.1	99.9	100.1		9980.01	10020.01	9980.01	10020.01
Total	999	996.9	995.1	1000	GT=3991	99800.66	99382.51	99022.83	100001	Σx2=398207.04
Mean	99.9	99.69	99.51	100

We have n= 40

            t=4

            r=10

ANOVA Table

Source of Variation	SS	df	MS	F	p value
Between Groups	1.437	3	0.48	4.82	0.006
Within groups (Error)	3.578	36	0.10
Total	5.015	39

In the above ANOVA table we have p-value= 0.006, which is less than 0.05 and it suggests that we have enough evidence against null hypothesis to reject the null and conclude that at least one of the labs has a mean that is significantly different from the others.

Answer(a): To know Which labs have means that are significantly different we need to do pairwise comparison of means using Tukey’s HSD as below:

The critical value for Tukey’s HSD can be obtained as below:

We have q=q_(0.95,4,36)= 3.808798

MSE =0.1

r=10

Hence the critical value is

We have pairwise mean difference for all treatment groups which is presented below. If the magnitude of pairwise difference is greater than critical value, it is considered to be significantly different.

pair	Pair wise difference of means	S/NS
Lab A - Lab B	0.21	NS
Lab A - Lab C	0.39	S
Lab A - Lab D	0.1	NS
Lab B - Lab C	0.18	NS
Lab B - Lab D	0.31	NS
Lab C - Lab D	0.49	S

We can see that the pair Lab A - Lab C and Lab C – Lab D are significantly different from each other while other pairs are not significantly different.

Hence the correct option is b and f

Answer(2):

We can obtain the conclusion of hypothesis test from the ANOVA table which we have already stated that at least one of the labs has a mean that is significantly different from the others.

Hence the correct option is b.