Question

1. A paper in the Journal of the Association of Asphalt Paving Technologies (1990) described an experiment to determine the effects of air voids on percentage retained strength of asphalt. For purposes of the experiment, air voids are controlled at three levels: low (2-4%), medium (4-6%), and high (6-8%). The data are shown in the following table:

Air Voids	Retained Strength (%)
Low	106	90	103	90	79	88	92	95
Medium	80	69	94	91	70	83	87	83
High	78	80	62	69	76	85	69	85

1. Write down the Null and Alternative hypotheses to test if the different levels of air voids significantly affect mean retained strength. Define all the terms used.

2. Find the calculated Ftest statistic then answer what are the critical F value and the rejection region of H0 at α=5% ?

1. c. Test the hypotheses using the test statistic / critical value method (use your answers to (b), and (c)). Clearly interpret the test result in the context of the problem.

d. What is the P value for the test? Test the hypotheses using the P value. Do you get the same answer as (d)? If your answer is different, clearly interpret it.

Answer 1

a)

Ho: µ1=µ2=µ3
H1: not all means are equal
.........

b)

treatment	G1	G2	G3	G4
count, ni =	8	8	8
mean , x̅ i =	92.875	82.13	75.50
std. dev., si =	8.6	9.0	8.2
sample variances, si^2 =	73.268	81.268	67.714
total sum	743	657	604				2004	(grand sum)
grand mean , x̅̅ =	Σni*x̅i/Σni =	83.50
( x̅ - x̅̅ )²	87.891	1.891	64.000
							TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	703.125	15.125	512.000				1230.25
SS(within ) = SSW = Σ(n-1)s² =	512.875	568.875	474.000				1555.7500

no. of treatment , k =   3  
df between = k-1 =    2  
N = Σn =   24  
df within = N-k =   21  
      
mean square between groups , MSB = SSB/k-1 =    1230.25/2=   615.1250
mean square within groups , MSW = SSW/N-k =    1555.75/21=   74.0833
      
F-stat = MSB/MSW =    615.125/74.0833=   8.30
      

anova table
	SS	df	MS	F		F-critical
Between:	1230.3	2	615.1	8.30		3.467
Within:	1555.8	21	74.1
Total:	2786.0	23

F stat = 8.30

F-critical = 3.467

rejection region :

f stat > 3.467

................

f stat > critical value , reject Ho

so, there is enough evidence to say that the different levels of air voids significantly affect mean retained strength

........

P value =   0.0022
p value < 0.05, reject Ho

same result as critical value

..................

Please let me know in case of any doubt.

Thanks in advance!

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