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Let us denote by Fn(x) the cumulative distribution function for a chi-square variable with n degrees of freedom. Then what we want to prove is (5)
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P( u !p1Z1 Fn(x)
x and u!p1Z1 !p2Z2 c !pkZk u c !p2Z2 !pkZk u P)
for a suitable value of n. It is easy to establish (5) if we use our geometrical intuition. First of all, Theorem 4-3 shows that the unconditional distribution of Z 2 Z 2 c Z 2 is chi-square with k degrees of freedom. Hence, since the 1 2 k density function for each Zj is (2p) 1>2e z2>2,
CHAPTER 7 Tests of Hypotheses and Significance
Fk(x)
(2p)
z2 z2 1 2
2 2 (z1 z2
2 c zk )>2 dz1dz2
c dz k
c z2 k
Furthermore, we have for the numerator in (5): (7) Numerator (2p)
2 2 (z1 z2
2 c zk )> 2 dz1dz2
c dz k
z2 z2 c z2 x, 1 2 k u !p1z1 !p2z2 c !pkzk u
We recall from analytic geometry that in three-dimensional space, x2 x2 x2 a2 represents a spherical 1 2 3 solid of radius a centered at the origin, while a1x1 a2x2 a3x3 0 is a plane through the origin whose normal is the unit vector (a1, a2, a3). Figure 7-15 shows the intersection of the two bodies. It is obvious that when a function which depends only on distance from the origin, i.e., f(r) where r 2x2 1 x2 2 x2 3
is integrated over the circular area or throughout a thin slab lying on that area the value of the integral is completely independent of the direction-cosines a1, a2, a3. In other words, all cutting planes through the origin give the same integral.
Fig. 7-15
By analogy we conclude that in (7), where e r2 >2 is integrated over the intersection of a hypersphere about the origin and a hyperplane through the origin, the p s may be given any convenient values. We choose c p p p 0, p 1
1 2 k 1 k
and obtain (8) Numerator (2p) (2p)
z2 z2 1 2
2 2 (z1 z2 c z2 1)>2 dz dz k 1 2
c dz (2P) k 1
c z2 k
1>2F k 1(x)(2P)
using (6). The factor 2P is the thickness of the slab. To evaluate the denominator in (5), we note that the random variable c !p Z W !p Z !p Z
1 1 2 2
is normal (because it is a linear combination of the independent, normal Z s), and that E(W ) !p1(0) !p2(0) c !pk(0) 0 Var (W ) p (1) p (1) c p (1) 1
1 2 k
Therefore, the density function for W is f(w) (9) Denominator
(2p)
1>2e w2>2,
and (2p)
1>2(2P)
P( uW u
f(0)(2P) k 1.
Dividing (8) by (9), we obtain the desired result, where n
CHAPTER 7 Tests of Hypotheses and Significance
The above proof (which can be made rigorous) shows incidentally that every linear constraint placed on the Z s, and hence on the Y s or X s, reduces the number of degrees of freedom in x2 by 1. This provides the basis for the rules given on page 221.
SUPPLEMENTARY PROBLEMS
Tests of means and proportions using normal distributions
7.63. An urn contains marbles that are either red or blue. To test the hypothesis of equal proportions of these colors, we agree to sample 64 marbles with replacement, noting the colors drawn and adopt the following decision rule: (1) accept the hypothesis if between 28 and 36 red marbles are drawn; (2) reject the hypothesis otherwise. (a) Find the probability of rejecting the hypothesis when it is actually correct. (b) Interpret graphically the decision rule and the result obtained in (a). 7.64. (a) What decision rule would you adopt in Problem 7.63 if you require the probability of rejecting the hypothesis when it is actually correct to be at most 0.01, i.e., you want a 0.01 level of significance (b) At what level of confidence would you accept the hypothesis (c) What would be the decision rule if a 0.05 level of significance were adopted 7.65. Suppose that in Problem 7.63 you wish to test the hypothesis that there is a greater proportion of red than blue marbles. (a) What would you take as the null hypothesis and what would be the alternative (b) Should you use a one- or two-tailed test Why (c) What decision rule should you adopt if the level of significance is 0.05 (d) What is the decision rule if the level of significance is 0.01 7.66. A pair of dice is tossed 100 times, and it is observed that sevens appear 23 times. Test the hypothesis that the dice are fair, i.e., not loaded, using (a) a two-tailed test and (b) a one-tailed test, both with a significance level of 0.05. Discuss your reasons, if any, for preferring one of these tests over the other. 7.67. Work Problem 7.66 if the level of significance is 0.01. 7.68. A manufacturer claimed that at least 95% of the equipment which he supplied to a factory conformed to specifications. An examination of a sample of 200 pieces of equipment revealed that 18 were faulty. Test his claim at a significance level of (a) 0.01, (b) 0.05. 7.69. It has been found from experience that the mean breaking strength of a particular brand of thread is 9.72 oz with a standard deviation of 1.4 oz. Recently a sample of 36 pieces of thread showed a mean breaking strength of 8.93 oz. Can one conclude at a significance level of (a) 0.05, (b) 0.01 that the thread has become inferior 7.70. On an examination given to students at a large number of different schools, the mean grade was 74.5 and the standard deviation was 8.0. At one particular school where 200 students took the examination, the mean grade was 75.9. Discuss the significance of this result at a 0.05 level from the viewpoint of (a) a one-tailed test, (b) a two-tailed test, explaining carefully your conclusions on the basis of these tests. 7.71. Answer Problem 7.70 if the significance level is 0.01.
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