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Draw QR Code 2d barcode in Software (a) If v 50, x2 !2(50) 0.95 2 (z0.95 value 67.5 given in Appendix E. 1 2 (zp

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Student s t distribution 4.42. Prove Theorem 4-6, page 116.
Since Y is normally distributed with mean 0 and variance 1, its density function is 1 e y2>2 !2p Since Z is chi-square distributed with v degrees of freedom, its density function is (1) (2) 1 z(v>2) 1e 2v>2 (v>2)
Because Y and Z are independent, their joint density function is the product of (1) and (2), i.e., 1 z(v>2) !2p 2v>2 (v>2) for ` y `, z 0. Y> !Z>v is P(T x) P(Y x!Z>v)
( y2 z)>2 1 e (y2 z)>2
The distribution function of T F(x)
1 z(v>2) 1 e !2p2v>2 (v>2) 6
dy dz
where the integral is taken over the region 5 of the yz plane for which y x !z>v. We first fix z and integrate with respect to y from ` to x !z>v. Then we integrate with respect to z from 0 to ` . We therefore have F(x) !2p2v>2 (v>2) 3z 1
z(v>2) 1 e
z>2 c
x1z>v
y2>2
dy d dz
CHAPTER 4 Special Probability Distributions
u 2z> v in the bracketed integral, we find F(x) 1 !2p 2v>2
Letting y
(v>2) 3z
3 0 u
z(v>2) 1e
z>2 !z>v
u2z>2v du dz
x 1 v>2 (v>2) 3 !2pv 2 u
1)>2
(z>2)[1 (u2>v)] dz d
Letting w
z a1 2
u2 v b , this can then be written 1 2(v !2pv 2v>2 (v>2) a v 2 1 b
x x 1)>2
F(x)
w(v 1)>2e w dw d du u2 >v)(v 1)>2 0 (1
v 2pv a b 2 as required.
` (1
du u2 >v)(v
1)>2
4.43. The graph of Student s t distribution with 9 degrees of freedom is shown in Fig. 4-19. Find the value of t1 for which (a) the shaded area on the right (b) the total shaded area (c) the total unshaded area 0.05, 0.99, 0.01, 0.05,
(d) the shaded area on the left
(e) the area to the left of t1 is 0.90.
Fig. 4-19
(a) If the shaded area on the right is 0.05, then the area to the left of t1 is (1 0.05) 0.95, and t1 represents the 95th percentile, t0.95. Referring to the table in Appendix D, proceed downward under the column headed v until entry 9 is reached. Then proceed right to the column headed t0.95. The result 1.83 is the required value of t. (b) If the total shaded area is 0.05, then the shaded area on the right is 0.025 by symmetry. Therefore, the area to the left of t1 is (1 0.025) 0.975, and t1 represents the 97.5th percentile, t0.975. From Appendix D, we find 2.26 as the required value of t. (c) If the total unshaded area is 0.99, then the total shaded area is (1 right is 0.01> 2 0.005. From the table we find t0.995 3.25. 0.99) 0.01, and the shaded area to the
(d) If the shaded area on the left is 0.01, then by symmetry the shaded area on the right is 0.01. From the table, t0.99 2.82. Therefore, the value of t for which the shaded area on the left is 0.01 is 2.82. (e) If the area to the left of t1 is 0.90, then t1 corresponds to the 90th percentile, t0.90, which from the table equals 1.38.
CHAPTER 4 Special Probability Distributions
4.44. Find the values of t for which the area of the right-hand tail of the t distribution is 0.05 if the number of degrees of freedom v is equal to (a) 16, (b) 27, (c) 200.
Referring to Appendix D, we find in the column headed t0.95 the values: (a) 1.75 corresponding to v 16; (b) 1.70 corresponding to v 27; (c) 1.645 corresponding to v 200. (The latter is the value that would be obtained by using the normal curve. In Appendix D this value corresponds to the entry in the last row marked .)
The F distribution 4.45. Prove Theorem 4-7.
The joint density function of V1 and V2 is given by f (v1, v2) a 1 v (n1>2) 1e 2v1>2 (n1 >2) 1
n2)>2 v1>2
1 n(n2>2) 1e 2n2>2 (n2 >2) 2
(v1 v2)>2
v2>2
2(n1 if v1 0, v2
1 (n v(n1>2) 1v2 2>2) 1e (n1 >2) (n2 >2) 1
0 and 0 otherwise. Make the transformation u v1 >v1 v2 >v2 v2v1 v1v2 , w v2 or v1 v1uw v2 v2 w
Then the Jacobian is '(v1, v2) '(u, w) 2
Denoting the density as a function of u and w by g(u, w), we thus have g(u, w) v1uw (v1>2) 1 a v b 2 (v1 >2) (v2 >2)
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