Question

Please answer only question 4

QUESTION 3 [20] A manufacturer of car shock absorbers was interested in comparing the durability of its shocks with that of the shocks produced by its biggest competitor. To make the comparison, one of the manufacturer's and one of the competitor's shocks were randomly selected and installed on the rear wheels of each of six cars. After the cars had been driven 20000 km, the strength of each test shock was measured and recorded. The data are in the table below Shocks Car 2 4 Manufacturers Competitor's (Manufacturer's - Competitor's) 8.8 10.5 12.5 9.7 9.6 13.2 8.4 10.1 12.0 9.3 9.0 13.0 0.4 0.5 0.4 0.6 0.2 Sample mean Sample variance 3.0697 3.3040 0.0177 QUESTION 4 [33] Refer to QUESTION 3 The experimental design used by the manufacturer was a randomized complete block de- sign (i) What is the response variable in the experiment? (ii) What is the treatment factor to be studied in the experiment? (ii) What are the experimental units? (iv) What is the blocking factor in the experiment. (b) Complete the two-way ANOV A table (of the data in QUESTION 3) below (15) Source df SS MS F Shocks?? ?? ???? Car Error ? ?? otal (c) Do the data provide sufficient evidence to indicate a difference between the mean strength of (5) the manufacturer's and the competitor's shocks? Use the 0.05 level of significance. (d) Verify that: i) F- in QUESTION 3 (b); and (ii) t2025 (with 5 degrees of freedom) Fo.os (with 1 and 5 numerator and denominator degrees of freedom, respectively) (2+3)

Answer 1

Answer:

a).

i). strength of shock

ii). Maker of the shock

iii). shock

iv).car

b).

ANOVA
Source of Variation	SS	df	MS	F
shocks	0.5208	1	0.5208	58.9623
car	31.8242	5	6.3648	720.5472
Error	0.0442	5	0.0088
Total	32.3892	11

c).

calculated F= 58.96 which is > critical F(1,5) =5.05 at 0.05 level. Ho is rejected. There is sufficient evidence to indicate a difference between the mean strength of the manufacturers and the competitors shocks.

d).

F_0.05( 1,5) =6.608

Critical t with 5 df at 0.025 level ( one side)= 2.57

t²=2.57*2.57=6.605 which is approximately same as the above F value.

Excel Addon PHStat used

Randomized Block Design (ANOVA: Two-Factor Without Replication)
SUMMARY	Count	Sum	Average	Variance
c1	2	17.2	8.6000	0.0800
c2	2	20.6	10.3000	0.0800
c3	2	24.5	12.2500	0.1250
c4	2	19	9.5000	0.0800
c5	2	18.6	9.3000	0.1800
c6	2	26.2	13.1000	0.0200
manufacturer	6	64.3	10.7167	3.0697
competitor	6	61.8	10.3000	3.3040
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Rows	31.8242	5	6.3648	720.5472	0.0000	5.0503
Columns	0.5208	1	0.5208	58.9623	0.0006	6.6079
Error	0.0442	5	0.0088
Total	32.3892	11
				Level of significance		0.05