286 Some Elementary Statistical Inferences

ofQ 3 =

∑ 4

1 (Xi−npi^0 )

(^2) /(np
i 0 )is
(6−5)^2
5

• (18−15)^2
  15


• 
  

  (20−25)^2
  25

  
  


• 
  

  (36−35)^2
  35

  
  

  64
  35
  =1. 83.
  The following R segment calculates the test andp-value:
  x=c(6,18,20,36); ps=c(1,3,5,7)/16; chisq.test(x,p=ps)
  X-squared = 1.8286, df = 3, p-value = 0.6087
  Hence, we fail to rejectH 0 at level 0.0250.
  Thus far we have used the chi-square test when the hypothesisH 0 is a simple
  hypothesis. More often we encounter hypothesesH 0 in which the multinomial prob-
  abilitiesp 1 ,p 2 ,...,pkare not completely specified by the hypothesisH 0 .Thatis,
  underH 0 , these probabilities are functions of unknown parameters. For an illustra-
  tion, suppose that a certain random variableYcan take on any real value. Let us
  partition the space{y:−∞<y<∞}intokmutually disjoint setsA 1 ,A 2 ,...,Ak
  so that the eventsA 1 ,A 2 ,...,Akare mutually exclusive and exhaustive. LetH 0 be
  the hypothesis thatY isN(μ, σ^2 )withμandσ^2 unspecified. Then each
  pi=
  ∫
  Ai
  1
  √
  2 πσ
  exp[−(y−μ)^2 / 2 σ^2 ]dy, i=1, 2 ,...,k,
  is a function of the unknown parametersμandσ^2. Suppose that we take a random
  sampleY 1 ,...,Ynof sizenfrom this distribution. If we letXidenote the frequency
  ofAi,i=1, 2 ,...,k,sothatX 1 +X 2 +···+Xk=n, the random variable
  Qk− 1 =
  ∑k
  i=1
  (Xi−npi)^2
  npi
  cannot be computed onceX 1 ,...,Xkhave been observed, since eachpi, and hence
  Qk− 1 , is a function ofμandσ^2. Accordingly, choose the values ofμandσ^2 that
  minimizeQk− 1. These values depend upon the observedX 1 =x 1 ,...,Xk=xkand
  are calledminimum chi-square estimatesofμandσ^2. These point estimates of
  μandσ^2 enable us to compute numerically the estimates of eachpi. Accordingly,
  if these values are used, Qk− 1 can be computed onceY 1 ,Y 2 ,...,Yn, and hence
  X 1 ,X 2 ,...,Xk, are observed. However, a very important aspect of the fact, which
  we accept without proof, is that nowQk− 1 is approximatelyχ^2 (k−3). That is, the
  number of degrees of freedom of the approximate chi-square distribution ofQk− 1 is
  reduced by one for each parameter estimated by the observed data. This statement
  applies not only to the problem at hand but also to more general situations. Two
  examples are now be given. The first of these examples deals with the test of the
  hypothesis that two multinomial distributions are the same.
  Remark 4.7.1.In many cases, such as that involving the meanμand the variance
  σ^2 of a normal distribution, minimum chi-square estimates are difficult to com-
  pute. Other estimates, such as the maximum likelihood estimates of Example 4.1.3,
  μˆ=Y andσˆ^2 =(n−1)S^2 /n, are used to evaluatepiandQk− 1. In general,Qk− 1
  is not minimized by maximum likelihood estimates, and thus its computed value