In: Statistics and Probability
A sample of 10 longleaf pine trees each are taken from the western and eastern halves of the Wade Tract in Thomas County, Georgia. The diameters for the trees are given:
Western | 17.2 | 44.6 | 44.1 | 35.5 | 51 | 21.6 | 44.1 | 11.2 | 36 | 42.1 |
Eastern | 23.5 | 43.5 | 6.6 | 11.5 | 17.2 | 38.7 | 2.3 | 31.5 | 10.5 | 23.7 |
(a) Is there a difference between the mean diameters of the
trees in the two halves? Perform a test of significance at level α
= .05 based on Satterthwaite’s approximation (without assuming
equality of popu- lation standard deviations). Carefully show all
steps of your test.
(b) Now, assume that the population standard deviations are equal.
Perform a test of significance at levelα = .05. Carefully show all
steps of your test.
For Western
= 34.74, s1 = 13.45, n1 = 10
For Eastern
= 20.9, s2 = 13.81, n2 = 10
a) H0:
H1:
The test statistic t = ()/sqrt(s1^2/n1 + s2^2/n2)
= (34.74 - 20.9)/sqrt((13.45)^2/10 + (13.81)^2/10)
= 2.27
DF = (s1^2/n1 + s2^2/n2)^2/((s1^2/n1)^2/(n1 - 1) + (s2^2/n2)^2/(n2 - 1))
= ((13.45)^2/10 + (13.81)^2/10)^2/(((13.45)^2/10)^2/9 + ((13.81)^2/10)^2/9)
= 18
At = 0.05, the critical vallue is t0.025, 18 = +/- 2.101
Since the test statistic value is greater than the upper critical value (2.27 > 2.101), we should reject the null hypothesis.
So at = 0.05, we can conclude that there is a difference between the mean diameters of the trees in the two halves.
b) H0:
H1:
sp2 = ((n1 - 1)s1^2 + (n2 - 1)s2^2)/(n1 + n2 -2)
= (9 * (13.45)^2 + 9 * (13.81)^2)/(10 + 10 - 2)
= 185.8093
The test statistic t = ()/sqrt(sp2/n1 + sp2/n2)
= (34.74 - 20.9)/sqrt(185.8093/10 + 185.8093/10)
= 2.27
DF = 10 + 10 - 2 = 18
At = 0.05, the critical value is t0.025, 18 = +/- 2.101
Since the test statistic value is greater than the upper critical value (2.27 > 2.101), so we should reject the null hypothesis.
So at = 0.05, we can conclude that there is a difference between the mean diameters of the trees in the two halves.