Question

(a) Test the hypothesis that each sample comes from a population with the same mean at the a = 0.05 level of significance. That is, test Ho: My = H2 = 43. P-value = (Round to three decimal places as needed.) Choose the correct conclusion below. Do not reject the null hypothesis. Reject the null hypothesis. 0 - (b) If you rejected the null hypothesis in part (a), use Tukey's test to determine which pairwise means differ using a familywise error rate of a = 0.05. Find the Tukey 95% simultaneous confidence intervals for all pairwise comparisons. Select the correct choice below and, if necessary, fill in any answer boxes to complete your choice. comparison A. The interval for the pairwise comparison 12 - Hy is (OD. The interval for the pairwise comparison H3 – 47 is (OD. Data Table The interval for the pairwise comparison 43 - H2 is (Round to two decimal places as needed.) O B. The null hypothesis was not rejected. Sample 1 Sample 2 Sample 3 Choose the correct conclusion below. 35.3 41.7 43.8 35.1 39.1 48.2 39.4 33.5 42.3 O A. H2 = 137 11 44.4 34.8 53.7 36.6 32.8 47.2 B. My = H2+H3 28.5 21.6 51.8 0 C. H = 13 + H2 u OD. The null hypothesis was not rejected. Print Done

Answer 1

a)

treatment	G1	G2	G3	L4
count, ni =	6	6	6
mean , x̅ i =	36.550	33.92	47.83
std. dev., si =	5.3	6.9	4.4
sample variances, si^2 =	27.643	48.270	19.515
total sum	219.3	203.5	287				709.8	(grand sum)
grand mean , x̅̅ =	Σni*x̅i/Σni =	39.43
( x̅ - x̅̅ )²	8.314	30.434	70.560
							TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	49.882	182.602	423.360				655.8433333
SS(within ) = SSW = Σ(n-1)s² =	138.215	241.348	97.573				477.1367

no. of treatment , k =   3  
df between = k-1 =    2  
N = Σn =   18  
df within = N-k =   15  
      
mean square between groups , MSB = SSB/k-1 =    655.8433/2=   327.9217
mean square within groups , MSW = SSW/N-k =    477.1367/15=   31.8091
      
F-stat = MSB/MSW =    327.9217/31.8091=   10.31
      
P value =   0.002

anova table
	SS	df	MS	F	p-value
Between:	655.8	2	327.9	10.3	0.002
Within:	477.1	15	31.8
Total:	1133.0	17

Reject the null hypothesis.

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b)

TUKEY HSD/KRAMER
Level of significance	0.05
no. of treatments,k	3
DF error =N-k=	15
MSE =	31.8091
q-critical value(0.05,3,15)=	3.6734

			confidence interval		result
	point estimate	critical value	lower limit	upper limit
µ2-µ1	-2.63	8.46	-11.09	5.82	not sig
µ3-µ1	11.28	8.46	2.83	19.74	significant
µ3-µ2	13.92	8.46	5.46	22.37	significant

B. p1 = p + да

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