CHAPTER 5 Sampling Theory

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The result indicates why a corrected variance for samples is often defined as S 2 then follows that mS 2 s2 (see also remarks on page 158).

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n n 1

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S2, since it

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(b) The variance of the sampling distribution of variances s22 is obtained by subtracting the mean 5.40 from S each of the 25 numbers in the above table, squaring these numbers, adding them, and then dividing the result by 25. Therefore, s22 575.75>25 23.03 or sS2 4.80. S

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5.19. Work the previous problem if sampling is without replacement.

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(a) There are 10 samples whose variances are given by the numbers above (or below) the diagonal of zeros in the table of Problem 5.18(a). Then mS2 0.25 4.00 9.00 20.25 2.25 10 a 6.25 16.00 1.00 6.25 2.25 6.75

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This is a special case of the general result mS2 verified by putting N 5, n 2, and s2

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N n 1 b a n bs2 [equation (19), page 158] as is N 1 5 1 10.8 on the right-hand side to obtain mS2 A 4 B A 2 B (10.8) 6.75.

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(b) Subtracting 6.75 from each of the 10 numbers above the diagonal of zeros in the table of Problem 5.18(a), squaring these numbers, adding them, and dividing by 10, we find s22 39.675, or sS2 6.30. S

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5.20. Prove that

E(S 2) n n 1 s2

where S2 is the sample variance for a random sample of size n, as defined on pages 157 158, and s2 is the variance of the population.

Method 1 We have X1 # X X1 1 n [(n Then (X1 # X)2 1 [(n n2 1)2(X1 m)2 (X2 m)2 c (Xn m)2 cross-product terms] 1 n (X1 1)(X1 c m) Xn) (X2 1 n [(n m) 1)X1 c X2 (Xn c m)] Xn]

Since the X s are independent, the expectation of each cross-product term is zero, and we have E[(X1 # X)2] 1 5(n n2 1 5(n n2 1 5(n n2 (n E(S2) 1)2E[(X1 1)2s2 1)2s2 1)s2 >n for k 1 n E[(X1 # X)2 s2 (n m)2] c 1)s26 E[(X2 s26 n n 1 s2 m)2] c E[(Xn m)2]6

Similarly, E[(Xk

# X)2]

2, c, n. Therefore, c c n (Xn 1 # X)2] s2 d n 1 s2

1 n 1 2 nc n s

CHAPTER 5 Sampling Theory

Method 2

We have Xj

m) (Xj

# (X # X)2

m). Then (Xj m)2 2(Xj # 2(X # m)(X m) # (X m)2

and (1) a (Xj # X)2 a (Xj m)2 m) a (Xj m) # a (X m)2

where the sum is from j (2)

1 to n. This can be written a (Xj # X)2 a (Xj a (Xj m)2 m)2 # 2n(X # n(X m)2 m)2 # n(X m)2

since g(Xj we find

nm E S a (Xj

# n(X # X)2 T

m). Taking the expectation of both sides of (2) and using Problem 5.7, E S a (Xj ns2 m)2 T # nE [(X (n 1) s2 m)2]

s2 na n b

from which

5.21. Prove Theorem 5-4, page 156.

If Xj, j 1, 2, c, n, is normally distributed with mean (see Table 4-2, page 110) fj(v) The characteristic function of X1 f(v) X2 eimv and variance

(s2v2)>2

, then its characteristic function is

cX is then, by Theorem 3-12, n einmv

(ns2v2)>2

f1(v)f2(v) c fn(v)

since the Xj are independent. Then, by Theorem 3-11, the characteristic function of # X is fX(v) X1 X2 n eimv

[(s2>n)v2]>2

v fa n b

But this is the characteristic function for a normal distribution with mean result follows from Theorem 3-13.

and variance s2 >n, and the desired

5.22. Prove Theorem 5-6, page 158.

By definition, (n 1) S2 V V1 V2,where V

g n 1(Xj j

# X)2. It then follows from (2) of Method 2 in Problem 5.20 that

(Xj s2

m)2 , V1

1)S2 s2

# (X m)2 s2 >n

Now by Theorem 4-3, page 115, V is chi-square distributed with n degrees of freedom [as is seen on replacing # Xj by (Xj m)>s]. Also, by Problem 5.21, X is normally distributed with mean m and variance 2 >n. Therefore, from Theorem 4-3 with n # s 1 and X1 replaced by (X m)> 2s2 >n, we see that V2 is chisquare distributed with 1 degree of freedom. It follows from Theorem 4-5, page 115, that if V1 and V2 are independent, then Vl, is chi-square distributed with n 1 degrees of freedom. Since it can be shown that V1 and V2 are indeed independent, the required result follows.

5.23. (a) Use Theorem 5-6 to determine the expected number of samples in Problem 5.1 for which sample variances are greater than 7.2. (b) Check the result in (a) with the actual result.