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where the last step uses the fact that the standard normal density function is even. Making the change of !t in the final integral, we obtain variable x P(Y y)
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dt 1 in (39), page 115, and
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But this is a chi-square distribution with 1 degree of freedom, as is seen by putting v using the fact that A 1 B !p. 2
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4.37. Prove Theorem 4-3, page 115, for v
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2 2 By Problem 4.36 we see that if X1 and X2 are normally distributed with mean 0 and variance 1, then X1 and X2 2 are chi square distributed with 1 degree of freedom each. Then, from Problem 4.35(b), we see that Z X1 X2 2 is chi square distributed with 1 1 2 degrees of freedom if X1 and X2 are independent. The general result for all positive integers v follows in the same manner.
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4.38. The graph of the chi-square distribution with 5 degrees of freedom is shown in Fig. 4-18. (See the remarks on notation on page 115.) Find the values x2,x2 for which 1 2
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(a) the shaded area on the right (b) the total shaded area 0.05, 0.10, 0.01. 0.05,
(c) the shaded area on the left (d) the shaded area on the right
Fig. 4-18
(a) If the shaded area on the right is 0.05, then the area to the left of x2 is (1 0.05) 0.95, and x2 represents 2 2 the 95th percentile, x2 . 0.95 Referring to the table in Appendix E, proceed downward under column headed v until entry 5 is reached. Then proceed right to the column headed x2 . The result, 11.1, is the required value of x2. 0.95 (b) Since the distribution is not symmetric, there are many values for which the total shaded area 0.05. For example, the right-hand shaded area could be 0.04 while the left-hand shaded area is 0.01. It is customary, however, unless otherwise specified, to choose the two areas equal. In this case, then, each area 0.025. If the shaded area on the right is 0.025, the area to the left of x2 is 1 0.025 0.975 and x2 represents 2 2 the 97.5th percentile, x2 , which from Appendix E is 12.8. 0.975 Similarly, if the shaded area on the left is 0.025, the area to the left of x2 is 0.025 and x2 represents the 1 1 2.5th percentile, x2 , which equals 0.831. 0.025 Therefore, the values are 0.831 and 12.8. (c) If the shaded area on the left is 0.10, x2 represents the 10th percentile, x2 , which equals 1.61. 0.10 1 (d) If the shaded area on the right is 0.01, the area to the left of x2 is 0.99, and x2 represents the 99th 2 2 percentile, x2 , which equals 15.1. 0.99
CHAPTER 4 Special Probability Distributions
4.39. Find the values of x2 for which the area of the right-hand tail of the x2 distribution is 0.05, if the number of degrees of freedom v is equal to (a) 15, (b) 21, (c) 50.
Using the table in Appendix E, we find in the column headed x2 the values: (a) 25.0 corresponding to 0.95 v 15; (b) 32.7 corresponding to v 21; (c) 67.5 corresponding to v 50.
4.40. Find the median value of x2 corresponding to (a) 9, (b) 28, (c) 40 degrees of freedom.
Using the table in Appendix E, we find in the column headed x2 (since the median is the 50th percentile) 0.50 the values: (a) 8.34 corresponding to v 9; (b) 27.3 corresponding to v 28; (c) 39.3 corresponding to v 40. It is of interest to note that the median values are very nearly equal to the number of degrees of freedom. In fact, for v 10 the median values are equal to v 0.7, as can be seen from the table.
4.41. Find x2 for (a) v 0.95
50, (b) v
100 degrees of freedom.
For v greater than 30, we can use the fact that ( !2x 2 !2v 1) is very closely normally distributed with mean zero and variance one. Then if zp is the (100p)th percentile of the standardized normal distribution, we can write, to a high degree of approximation, !2x2 p from which x2 p
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