Chapter 14, Problem 41SE

To determine

State whether the model fits the data.

Answer to Problem 41SE

The data provide evidence that the model fits the data.

Explanation of Solution

In this context, there are three classes $n_1,n_2,n_3\text{, and }{n}_4$ with corresponding probabilities $\frac{p}{2},\left(\frac{p^2}{2}\right)+pq,\frac{q}{2}\text{, and }\frac{q^2}{2}$, respectively. It is known that the observed values are $n_1=880,n_2=1,032,n_3\text{=80, and }{n}_4=8$, respectively.

The likelihood function is obtained below:

$\begin{array}{c}L\left(p\right)=\frac{n!}{n_1!n_2!n_3!n_4!}{\left(\frac{p}{2}\right)}^{n_1}\times {\left[\left(\frac{p^2}{2}\right)+pq\right]}^{n_2}\times {\left(\frac{q}{2}\right)}^{n_3}\times {\left(\frac{q^2}{2}\right)}^{n_4}\\ =\frac{n!}{{\displaystyle \prod _{i=1}^4{n}_i!}}{\left(\frac{p}{2}\right)}^{n_1}\times {\left[\left(\frac{p^2}{2}\right)+p\left(1-p\right)\right]}^{n_2}\times {\left(\frac{1-p}{2}\right)}^{n_3}\times {\left(\frac{{\left(1-p\right)}^2}{2}\right)}^{n_4}\\ =D{p}^{n_1+n_2}{\left(2-p\right)}^{n_2}{\left(1-p\right)}^{n_3+2n_4}\end{array}$

In this context, $D=\frac{n!}{{\displaystyle \prod _{i=1}^4{n}_i!}}$.

Take natural logarithm on both the sides.

$\begin{array}{c}\ln L=\ln \left[D{p}^{n_1+n_2}{\left(2-p\right)}^{n_2}{\left(1-p\right)}^{n_3+2n_4}\right]\\ =\ln \left(D\right)+\left(n_2\right)\ln \left(2-p\right)+n_2\ln 2+\left(n_3+2n_4\right)\ln \left(1-p\right)\end{array}$

Here, differentiate $\ln L$ with respect to $p_j$.

$\frac{d}{dp}\ln L=\frac{n_1+n_2}{p}-\frac{n_2}{\left(2-p\right)}-\frac{n_3+2n_4}{1-p}$.

Equate $\frac{d}{dp}\ln L=0$,

$\begin{array}{c}\frac{n_1+n_2}{p}-\frac{n_2}{\left(2-p\right)}-\frac{n_3+2n_4}{1-p}=0\\ \left(n_1+n_2\right)\left(2-p\right)\left(1-p\right)-n_{2}p\left(1-p\right)-p\left(2-p\right)\left(n_3+2n_4\right)=0\\ \left(n_1+2n_2+n_3+2n_4\right)p^2-\left(3n_1+4n_2+2n_3+4n_4\right)p+2\left(n_1+n_2\right)=0\\ \left\{\begin{array}{l}\left(880+2\left(1,032\right)+80+2\left(8\right)\right)p^2-\\ \left(3\left(880\right)+4\left(1,032\right)+2\left(80\right)+4\left(8\right)\right)p\\ +2\left(880+1,032\right)\end{array}\right\}=0\\ 3,040p^2-6,960p+3,824=0\\ p=\frac{6,960\pm \sqrt{17,411,760}}{6,080}\\ =\left(0.9155,1.3739\right)\end{array}$

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

The test hypothesis is given below:

Null hypothesis: $H_0:$ The model fits the data.

Alternative hypothesis: $H_a:$ The model does not fit the data.

Therefore, the expected cell frequency is obtained below:

$\begin{array}{c}E\left(n_1\right)=n\frac{p}{2}\\ =2,000\times \left(\frac{0.9155}{2}\right)\\ =915.5\end{array}$

$\begin{array}{c}E\left(n_2\right)=2,000\times \left(\frac{p^2}{2}\right)+p\left(1-p\right)\\ =2,000\times \left(\frac{{0.9155}^2}{2}\right)+0.9155\left(1-0.9155\right)\\ =992.86\end{array}$

$\begin{array}{c}E\left(n_3\right)=2,000\left(\frac{1-p}{2}\right)\\ =2,000\left(\frac{1-0.9155}{2}\right)\\ =84.5\end{array}$

$\begin{array}{c}E\left(n_4\right)=2,000\left(\frac{{\left[1-p\right]}^2}{2}\right)\\ =2,000\left(\frac{{\left[1-0.9155\right]}^2}{2}\right)\\ =7.14\end{array}$

Thus, the observed and expected frequencies are given below:

	1	2	3	4
Observed	880	1,032	80	8
Expected	915.5	992.86	84.5	7.14

Here, the test statistic follows a chi-square distribution denoted by ${\chi }^2$.

Test statistic:

$\begin{array}{l}{\chi }^2={\displaystyle \sum _{i=1}^n\left[\frac{{\left(n_i-E\left({\widehat{n}}_i\right)\right)}^2}{E\left({\widehat{n}}_i\right)}\right]}\\ where,\\ k-1\to \text{degrees of freedom}\text{.}\\ k\text{is the number of categories}\text{.}\\ n_i\text{is an observed frequency}\text{.}\\ E\left({\widehat{n}}_i\right)\text{is an expected frequency}\text{.}\end{array}$

The value of test statistic is obtained below:

$\begin{array}{c}{\chi }^2=\left\{\frac{{\left(880-915.5\right)}^2}{915.5}+\frac{{\left(1,032-992.86\right)}^2}{992.86}+\frac{{\left(80-84.5\right)}^2}{84.5}+\frac{{\left(8-7.14\right)}^2}{7.14}\right\}\\ =1.3766+1.5430+0.2396+0.1036\\ =3.26\end{array}$

The degrees of freedom is as follows:

$\begin{array}{c}k-2=4-2\\ =2\end{array}$

Decision rule:

• If ${\chi }^2>{\chi }_{0.05}^2$, reject the null hypothesis.
• Otherwise, fail to reject the null hypothesis.

In Appendix 3, Table 6 “Percentage Points of the ${\chi }^2$ Distributions”, the critical value at 2 df is 5.99. The test statistic is less than the critical value. That is, ${\chi }^2\left(=3.26\right)<{\chi }_{0.05}^2\left(=5.99\right)$. The null hypothesis is not rejected. Therefore, it can be concluded that the model fits the data.