Question

can someone help me do the t-test for this data?

White fir Blue spruce Conifer Needle Table Conifer Needle White pine ID# U - 5 4.5 3.25 10 9.8 2.7 2.4 2.0 LI 29 8.0 9.2 9. 1.9 2.2 1.9 2.1 2.2 4.25 92 3.3 3.5 10 Mean Standard Deviation Standard Error of Mean how your calculation and t-test work here. Use the backside of this page and additional sheets needed.

Answer 1

	White Pine	Blue Spruce	White fir
	4	2.7	10
	5	2.4	9.8
	4.5	2	10.1
	3.25	2.1	10.2
	3.5	2.3	7.9
	4.75	1.9	8
	4.25	2.2	9.2
	3.8	1.9	9.1
	3.3	2.1	9.2
	3.5	2.2	8.9
Sample size (n)=	10	10	10
Mean=	3.985	2.18	9.24
Standard deviation=	0.6205	0.244	0.8181
Standard error of mean=	0.1962	0.0772	0.2587

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
White Pine	10	39.85	3.985	0.385028
Blue Spruce	10	21.8	2.18	0.059556
White fir	10	92.4	9.24	0.669333
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	269.0555	2	134.5278	362.3101	3.14E-20	3.354131
Within Groups	10.02525	27	0.371306
Total	279.0808	29

Hence all populations means are not equal.

Formula = Standard error of mean = = where, s = standard deviation, n = sample size. Let pi = true population mean for White Pine 12 = true population mean for Blue Spruce U3 = true population mean for White fir Null hypothesis, Holi,i'): Hi= Hius. Alternative hypothesis, H(i,i'): Mit Mir We use Fisher's t test (assume that three populations are independent and normally distributed with equal variance) Test statistic = ti,i') = - MSE (1) where, MSE = mean sum of square due to error = 0.3713, n= 10 We reject Holi,i') at 5% level of significance if Ti-Til > to.025,27 VMSE/5 = 0.5591 = LSD (= least significant difference) where, t0.025,27 = 2.0518. T1 - 72] = 13.985 – 2.180 = 1.805 > LSD, 71 – 73 = 3.985 – 9.24 = 5.255 > LSD, -73 = 2.18 – 9.24 = 7.06 > LSD.