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CHAPTER 7 Tests of Hypotheses and Significance
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Let p probability of heads in a single toss of the coin. Under the hypothesis (H0: p 0.5) (i.e., the coin is fair), f (x) P(x heads in 6 tosses)
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1 6 6 1 Then the probabilities of 0, 1, 2, 3, 4, 5, and 6 heads are given, respectively, by 64, 64, 15, 20, 15, 64, and 64. 64 64 64
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One-tailed test Here we wish to decide between the hypotheses (H0: p 0.5) and (H1: p 0.5). Since P(6 heads) 1 6 1 0.01562 and P(5 or 6 heads) 64 0.1094, we can reject H0 at a 0.05 but not a 0.01 level 64 64 (i.e., the result observed is significant at a 0.05 but not a 0.01 level). Two-tailed test Here we wish to decide between the hypotheses (H0: p 0.5) and (H1: p 2 0.5). Since P(0 or 6 heads) 1 1 0.03125, we can reject H0 at a 0.05 but not a 0.01 level. 64 64
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7.59. Work Problem 7.58 if the coin comes up heads 5 times.
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One-tailed test Since P(5 or 6 heads)
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0.1094, we cannot reject H0 at a level of 0.05 or 0.01. 0.2188, we cannot reject H0 at a level of 0.05 or 0.01.
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Two-tailed test Since P(0 or 1 or 5 or 6 heads)
2 A 64 B
7.60. Show that a chi-square test involving only two categories is equivalent to the significance test for proportions (page 216).
If P is the sample proportion for category I, p is the population proportion, and n is the total frequency, we can describe the situation by means of Table 7-23. Then by definition, x2 (nP np n2(P np p)2 n2(P nq p)2 np)2 [n(1 P) nq n(P n(1 p)]2 1 p)2 p 1 q n(P pq p)2 (P p)2 pq>n
which is the square of the Z statistic (5) on page 216. Table 7-23 I Observed Frequency Expected Frequency nP np
II n(l n(1 p) P) nq
TOTAL n n
7.61. Suppose X1, X2, c, Xk have a multinomial distribution, with expected frequencies np1, np2, c, npk, respectively. Let Y1, Y2, c, Yk be mutually independent, Poisson-distributed variables, with parameters l1 np1, l2 np2, c, lk npk, respectively. Prove that the conditional distribution of the Y s given that
Y1 Y2 c Yk n
is precisely the multinomial distribution of the X s.
For the joint probability function of the Y s, we have (1) P(Y1 y1, Y2 y2, c, Yk yk) B RB R c B R
(np1)y1e y1!
(np2)y2e y2!
(npk)yke yk!
y2 c ykpy1py2 1
c pyk k e cy! y1!y2! k
CHAPTER 7 Tests of Hypotheses and Significance
where we have used the fact that p1 given by (2) P(Y1 y1, Y2 p2 c pk
1. The conditional distribution we are looking for is c
y2, c, Yk P(Y1 y1, Y2
yk uY1
n) Y2 n) c Yk n)
y2, c, Yk P(Y1 Y2
yk and Y1 c Y
Now, the numerator in (2) has, from (1), the value nnpy1py2 c pyk 1 2 k e y !y ! c y !
1 2 k n
As for the denominator, we know from Problem 4.94, page 146, that Y1 Y2 c Yk is itself a Poisson variable with parameter np1 np2 c npk n. Hence, the denominator has the value nne n! Therefore, (2) becomes P(Y1 y1, Y2 y2, c,Yk yk uY1 Y2 c Yk n) n! py1py2 c pyk k y1!y2! c yk! 1 2
which is just the multinomial distribution of the X s [compare (16), page 112].
7.62. Use the result of Problem 7.61 to show that x2, as defined by (21), page 220, is approximately chi-square distributed.
As it stands, (21) is difficult to deal with because the multinomially distributed X s are dependent, in view of the restriction (22). However, Problem 7.61 shows that we can replace the X s by the independent, Poissondistributed Y s if it is given that Y1 Y2 c Yk n..Therefore, we rewrite (21) as (1) x2 Y1 l1 !l1
As n S ` , all the l s approach ` , and the central limit theorem for the Poisson distribution [(14), page 112] gives (2) x2 < Z 2 1 Z2 2 c Z2 k
where the Z s are independent normal variables having mean 0 and variance 1 whose distribution is conditional upon the event (3) !l1Z1 !l2Z2 c !lkZk 0 or !p1Z1 !p2Z2 c !pkZk 0
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