Question

XVI. (Hint: See 8.1 and recall the work we did in class for this section.)

a. Test the fairness of the die with α = 0.10

Numbers: 1 2 3 4 5 6

Obs: 43 58 42 39 57 61
b. Can you conclude that the die is fair? Why or why not.

XVII. (Hint: See 8.1 and recall the work we did in class for this section.) From your text, page 420, do #4. Can you conclude that the number of speeding tickets received is independent of the type of motorist? Why or why not

XVIII. (Hint: See 8.2 and recall the work we did in class for this section.)

A farmer conducts a test on 3 types of superfeed. The weight gain (in lbs.) are as follows:

Type A Type B Type C

12 11   14
17 12 21
19 10   17

13   18

Test the claim that the mean weight gain is the same for all three types of superfeed at the 99% level.

XVI. (Hint: See 8.1 and recall the work we did in class for this section.) a. Test the fairness of the die with a = 0.10 Numbers: 2 زرا 4. 5 6 Obs: 43 58 42 39 57 61 b. Can you conclude that the die is fair? Why or why not. XVII. (Hint: See 8.1 and recall the work we did in class for this section.) From your text, page 420, do #4. Can you conclude that the number of speeding tickets received is independent of the type of motorist? Why or why not XVIII. (Hint: See 8.2 and recall the work we did in class for this section.) A farmer conducts a test on 3 types of superfeed. The weight gain (in lbs.) are as follows: Type A Type B Type C 12 11 14 17 12 21 19 10 17 13 18 Test the claim that the mean weight gain is the same for all three types of superfeed at the 99% level.

Answer 1

XVI)

observed frequencey, O	expected proportion	expected frequency,E	(O-E)		(O-E)²/E
43	0.167	50.00	-7.00		0.980
58	0.167	50.00	8.00		1.280
42	0.167	50.00	-8.00		1.280
39	0.167	50.00	-11.00		2.420
57	0.167	50.00	7.00		0.980
61	0.167	50.000	11.00		2.420

chi square test statistic,X² = Σ(O-E)²/E =   9.360              
                  
level of significance, α=   0.1              
Degree of freedom=k-1=   6   -   1   =   5
                  
P value =   0.0955   [ excel function: =chisq.dist.rt(test-stat,df) ]          
Decision: P value < α, Reject Ho                  

DIE IS NOT FAIR

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

XVII) DATA NOT GIVEN

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XVIII)

treatment	G1	G2	G3	G4
count, ni =	4	3	4
mean , x̅ i =	15.250	11.00	17.50
std. dev., si =	3.3	1.0	2.9
sample variances, si^2 =	10.917	1.000	8.333
total sum	61	33	70				164	(grand sum)
grand mean , x̅̅ =	Σni*x̅i/Σni =	14.91
( x̅ - x̅̅ )²	0.116	15.281	6.713
							TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	0.465	45.843	26.851				73.15909091
SS(within ) = SSW = Σ(n-1)s² =	32.750	2.000	25.000				59.7500

no. of treatment , k =   3  
df between = k-1 =    2  
N = Σn =   11  
df within = N-k =   8  
      
mean square between groups , MSB = SSB/k-1 =    73.1591/2=   36.5795
mean square within groups , MSW = SSW/N-k =    59.75/8=   7.4688
      
F-stat = MSB/MSW =    36.5795/7.4688=   4.90

anova table
	SS	df	MS	F	p-value	F-critical
Between:	73.16	2	36.58	4.9	0.041	8.65
Within:	59.75	8	7.47
Total:	132.91	10

α =    0.01  
Decision:   p-value>α , do not reject null hypothesis    
there is enough evidence of significant mean difference among three treatments

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

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