Springer Texts in Business and Economics
The t-distribution with r Degrees of Freedom
a. IfX1, ..,Xn are IIN(p, o2),thenX~ N(p, o2/n) andz = (X-^. isN(0,1).
b. (n — 1)s2/o2 ~ x2—1. Dividing our N(0, 1) random variables z in part (a), by the square-root of our x^-1 random variable in part (b), divided by its degrees of freedom, we get
(X— p) _ (X— p)
Using the fact that X is independent of s2, this has a t-distribution with (n — 1) degrees of freedom.
c. The 95% confidence interval for p would be based on the t-distribution derived in part (b) with (n — 1) = 15 degrees of freedom.
X — p 20 — p 20 — p
s/Vn 2/рТб 1/2
Pr[—ta/2 < t < ta/2] = 1 — a = 0.95
From the t-tables with 15 degrees of freedom, t0 025 = 2.131. Hence Pr[—2.131 <40 — 2p < 2.131] = 0.95. rearranging terms, one gets
Pr[37.869/2 < p < 42.131/2] = 0.95 Pr[18.9345 < p < 21.0655] = 0.95.
2.12 The F-distribution.
(n1 - 1) ^/CTj2 = 24(15.6)/ct12 - /^4 also
(n2 - 1)sI/of = 30 (18.9) /of - xP
Therefore, under Ho; ol = ol 2 2 18.9
F = s|/sf = = 1.2115 - F3o,24.
15.6
Using the F-tables with 30 and 24 degrees of freedom, we find F.05 30 24 = 1.94. Since the observed F-statistic 1.2115 is less than 1.94, we do not reject Ho that the variance of the two shifts is the same.
