Mathematical Statistics with Applications
Mathematical Statistics with Applications
7th Edition
ISBN: 9780495110811
Author: Dennis Wackerly, William Mendenhall, Richard L. Scheaffer
Publisher: Cengage Learning
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Chapter 14, Problem 41SE
To determine

State whether the model fits the data.

Expert Solution & Answer
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Answer to Problem 41SE

The data provide evidence that the model fits the data.

Explanation of Solution

In this context, there are three classes n1,n2,n3, and n4 with corresponding probabilities p2,(p22)+pq,q2, and q22, respectively. It is known that the observed values are n1=880,n2=1,032,n3=80, and n4=8, respectively.

The likelihood function is obtained below:

L(p)=n!n1!n2!n3!n4!(p2)n1×[(p22)+pq]n2×(q2)n3×(q22)n4=n!i=14ni!(p2)n1×[(p22)+p(1p)]n2×(1p2)n3×((1p)22)n4=Dpn1+n2(2p)n2(1p)n3+2n4

In this context, D=n!i=14ni!.

Take natural logarithm on both the sides.

lnL=ln[Dpn1+n2(2p)n2(1p)n3+2n4]=ln(D)+(n2)ln(2p)+n2ln2+(n3+2n4)ln(1p)

Here, differentiate lnL with respect to pj.

ddplnL=n1+n2pn2(2p)n3+2n41p.

Equate ddplnL=0,

n1+n2pn2(2p)n3+2n41p=0(n1+n2)(2p)(1p)n2p(1p)p(2p)(n3+2n4)=0(n1+2n2+n3+2n4)p2(3n1+4n2+2n3+4n4)p+2(n1+n2)=0{(880+2(1,032)+80+2(8))p2(3(880)+4(1,032)+2(80)+4(8))p+2(880+1,032)}=03,040p26,960p+3,824=0p=6,960±17,411,7606,080=(0.9155,1.3739)

In general, the probability value should not exceed 1. The value of p is 0.9155.

The test hypothesis is given below:

Null hypothesis: H0: The model fits the data.

Alternative hypothesis: Ha: The model does not fit the data.

Therefore, the expected cell frequency is obtained below:

E(n1)=np2=2,000×(0.91552)=915.5

E(n2)=2,000×(p22)+p(1p)=2,000×(0.915522)+0.9155(10.9155)=992.86

E(n3)=2,000(1p2)=2,000(10.91552)=84.5

E(n4)=2,000([1p]22)=2,000([10.9155]22)=7.14

Thus, the observed and expected frequencies are given below:

 1234
Observed8801,032808
Expected915.5992.8684.57.14

Here, the test statistic follows a chi-square distribution denoted by χ2.

Test statistic:

χ2=i=1n[(niE(n^i))2E(n^i)]where,k1degrees of freedom.kis the number of categories.niis an observed frequency.E(n^i)is an expected frequency.

The value of test statistic is obtained below:

χ2={(880915.5)2915.5+(1,032992.86)2992.86+(8084.5)284.5+(87.14)27.14}=1.3766+1.5430+0.2396+0.1036=3.26

The degrees of freedom is as follows:

k2=42=2

Decision rule:

  • If χ2>χ0.052, reject the null hypothesis.
  • Otherwise, fail to reject the null hypothesis.

In Appendix 3, Table 6 “Percentage Points of the χ2 Distributions”, the critical value at 2 df is 5.99. The test statistic is less than the critical value. That is, χ2(=3.26)<χ0.052(=5.99). The null hypothesis is not rejected. Therefore, it can be concluded that the model fits the data.

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