Chapter 14.2, Problem 19E

Each headlight on an automobile undergoing an annual vehicle inspection can be focused either too high (H), too low (L), or properly (N). Checking the two headlights simultaneously (and not distinguishing between left and right) results in the six possible outcomes HH, LL, NN, HL, HN, and LN. If the probabilities (population proportions) for the single headlight focus direction are P(H) = θ₁, P(L) = θ₂, and P(N) = 1 − θ₁ − θ₂ and the two headlights are focused independently of one another, the probabilities of the six outcomes for a randomly selected car are the following:

$\begin{array}{l}{p}_1={\theta }_1^2{p}_2={\theta }_2^2{p}_3=\left(1-{\theta }_1-{\theta }_2\right){}^2\\ p_4=2{\theta }_1{\theta }_2{p}_5=2{\theta }_1\left(1-{\theta }_1-{\theta }_2\right)\\ p_6=2{\theta }_1\left(1-{\theta }_1-{\theta }_2\right)\end{array}$

Use the accompanying data to test the null hypothesis

$H_0:p_1={\pi }_1\left({\theta }_1,{\theta }_2\right),\dots ,p_6={\pi }_6\left({\theta }_1,{\theta }_2\right)$

where the π_i,( θ₁, θ₂)s are given previously.

Outcome	HH	LL	NN	HL	HN	LN
Frequency	49	26	14	20	53	38

 [Hint: Write the likelihood as a function of θ₁ and θ₂, take the natural log, then compute ∂/∂ θ₁ and ∂/∂θ₂, equate them to 0, and solve for ${\hat{\theta }}_1,{\stackrel{⌢}{\theta }}_2$.]

To determine

Test the null hypothesis that $H_0:p_1={\pi }_1\left({\theta }_1,{\theta }_2\right),\mathrm{...},p_1={\pi }_6\left({\theta }_1,{\theta }_2\right)$

Answer to Problem 19E

There is sufficient evidence to reject the null hypothesis that $H_0:p_1={\pi }_1\left({\theta }_1,{\theta }_2\right),\mathrm{...},p_1={\pi }_6\left({\theta }_1,{\theta }_2\right)$.

Explanation of Solution

Given info:

Assume that a vehicle inspection is conducted to check whether the headlamp is too high H or too low L and proper N. Samples of two vehicle headlamp was checked simultaneously.

Also, the single probability for each headlamp focus is given below:$P\left(H\right)={\theta }_1,P\left(L\right)={\theta }_2,P\left(N\right)=1-{\theta }_1-{\theta }_2$

The six possible outcomes are HH, LL, NN, HL, LN, HN and the six probabilities are:

$p_1={\theta }_1^2$,$p_2={\theta }_2^2$,$p_3={\left(1-{\theta }_1-{\theta }_2\right)}^2$,$p_4=2{\theta }_1{\theta }_2$,$p_5=2{\theta }_1\left(1-{\theta }_1-{\theta }_2\right)$ and $p_6=2{\theta }_2\left(1-{\theta }_1-{\theta }_2\right)$

Calculation:

There are three cell counts in the given data. Hence, there would be three cell proportions and it is given below:

$\begin{array}{c}{\pi }_1{\left(\theta \right)}^{n_1}+{\pi }_2{\left(\theta \right)}^{n_2}+{\pi }_3{\left(\theta \right)}^{n_3}={\pi }_1{\left(\theta \right)}^{2n_1+n_4+n_5}+{\pi }_2{\left(\theta \right)}^{2n_2+n_4+n_6}+{\pi }_3{\left(\theta \right)}^{2n_3+n_5+n_5}\\ ={\theta }_1^A{\theta }_2^B{\left(1-{\theta }_1-{\theta }_2\right)}^C\end{array}$

The likelihood estimate of $\theta$ is given below:

${\pi }_1{\left(\theta \right)}^{n_1}+{\pi }_2{\left(\theta \right)}^{n_2}+{\pi }_3{\left(\theta \right)}^{n_3}={\theta }_1^A{\theta }_2^B{\left(1-{\theta }_1-{\theta }_2\right)}^C$

Taking ln yields,

$\ln \left[{\theta }_1^A{\theta }_2^B{\left(1-{\theta }_1-{\theta }_2\right)}^C\right]=A\ln \left({\theta }_1\right)+B\ln \left({\theta }_2\right)+C\ln \left(1-{\theta }_1-{\theta }_2\right)$

Maximizing the equation with respect to ${\theta }_1$,

$\begin{array}{c}\frac{\partial }{\partial {\theta }_1}\left[A\ln \left({\theta }_1\right)+B\ln \left({\theta }_2\right)+C\ln \left(1-{\theta }_1-{\theta }_2\right)\right]=\frac{A}{{\theta }_1}+\frac{C}{1-{\theta }_1-{\theta }_2}\left(-1\right)\\ =\frac{A}{{\theta }_1}-\frac{C}{1-{\theta }_1-{\theta }_2}\end{array}$

Equating to zero yields,

$\begin{array}{c}\frac{A}{{\theta }_1}-\frac{C}{1-{\theta }_1-{\theta }_2}=0\\ \frac{A}{{\theta }_1}=\frac{C}{1-{\theta }_1-{\theta }_2}\end{array}$

Also,

Maximizing the equation with respect to ${\theta }_2$ and equating to zero yields,

$\begin{array}{c}\frac{\partial }{\partial {\theta }_2}\left[A\ln \left({\theta }_1\right)+B\ln \left({\theta }_2\right)+C\ln \left(1-{\theta }_1-{\theta }_2\right)\right]=\frac{B}{{\theta }_2}+\frac{C}{1-{\theta }_1-{\theta }_2}\left(-1\right)\\ =\frac{B}{{\theta }_2}-\frac{C}{1-{\theta }_1-{\theta }_2}\\ \frac{\partial }{\partial {\theta }_2}=0\\ \frac{B}{{\theta }_2}-\frac{C}{1-{\theta }_1-{\theta }_2}=0\end{array}$

$\frac{B}{{\theta }_2}=\frac{C}{1-{\theta }_1-{\theta }_2}$

The value of ${\theta }_2$ is calculated by using the $\frac{A}{{\theta }_1}=\frac{C}{1-{\theta }_1-{\theta }_2}$ and $\frac{B}{{\theta }_2}=\frac{C}{1-{\theta }_1-{\theta }_2}$

Subtracting the two equations,

$\begin{array}{l}\underset{\_}{\begin{array}{l}\frac{A}{{\theta }_1}=\cancel{\frac{C}{1-{\theta }_1-{\theta }_2}}\\ \frac{B}{{\theta }_2}=\cancel{\frac{C}{1-{\theta }_1-{\theta }_2}}\end{array}}\\ \left(\frac{A}{{\theta }_1}-\frac{B}{{\theta }_2}\right)=0\\ \frac{A}{{\theta }_1}=\frac{B}{{\theta }_2}\end{array}$

${\theta }_2=\frac{B{\theta }_1}{A}$

Substitute ${\theta }_2=\frac{B{\theta }_1}{A}$ in the equation $\frac{A}{{\theta }_1}=\frac{C}{1-{\theta }_1-{\theta }_2}$ to get ${\theta }_1$

$\begin{array}{c}\frac{A}{{\theta }_1}=\frac{C}{1-{\theta }_1-\left(\frac{B{\theta }_1}{A}\right)}\\ \frac{A}{{\theta }_1}=\frac{AC}{A-{\theta }_{1}A-B{\theta }_1}\\ AC{\theta }_1=A^2-{\theta }_1{A}^2-AB{\theta }_1\\ AC{\theta }_1+AB{\theta }_1+{\theta }_1{A}^2=A^2\end{array}$

$\begin{array}{c}A{\theta }_1\left(C+B+A\right)=A^2\\ {\theta }_1=\frac{A}{A+B+C}\end{array}$

Similarly,

${\theta }_1=\frac{A{\theta }_2}{B}$ in the equation $\frac{B}{{\theta }_2}=\frac{C}{1-{\theta }_1-{\theta }_2}$ to get ${\theta }_2$.

$\begin{array}{c}\frac{B}{{\theta }_2}=\frac{C}{1-\frac{A{\theta }_2}{B}-{\theta }_2}\\ \frac{B}{{\theta }_2}=\frac{BC}{B-{\theta }_{2}B-A{\theta }_2}\\ BC{\theta }_2=B^2-{\theta }_2{B}^2-AB{\theta }_2\\ BC{\theta }_2+AB{\theta }_2+{\theta }_2{B}^2=B^2\end{array}$

$\begin{array}{c}B{\theta }_2\left(C+A+B\right)=B^2\\ {\theta }_2=\frac{B}{A+B+C}\end{array}$

The estimates ${\widehat{\theta }}_1,{\widehat{\theta }}_2,{\widehat{\theta }}_3$ are given below:

$\begin{array}{c}{\widehat{\theta }}_1=\frac{2n_1+n_4+n_5}{2n_1+n_4+n_5+2n_2+n_4+n_6+2n_3+n_5+n_6}\\ =\frac{2n_1+n_4+n_5}{2\left(n_1+n_4+n_5+n_2+n_4+n_6+n_3+n_5+n_6\right)}\\ =\frac{2n_1+n_4+n_5}{2n}\end{array}$

$\begin{array}{c}{\widehat{\theta }}_2=\frac{2n_2+n_4+n_6}{2n_1+n_4+n_5+2n_2+n_4+n_6+2n_3+n_5+n_6}\\ =\frac{2n_2+n_4+n_6}{2n}\end{array}$

$\begin{array}{c}{\widehat{\theta }}_3=\frac{2n_2+n_5+n_6}{2n_1+n_4+n_5+2n_2+n_4+n_6+2n_3+n_5+n_6}\\ =\frac{2n_2+n_5+n_6}{2n}\end{array}$

Substitute the observed values in each of the estimates ${\widehat{\theta }}_1,{\widehat{\theta }}_2,{\widehat{\theta }}_3$ to find the values of ${\widehat{\theta }}_1,{\widehat{\theta }}_2,{\widehat{\theta }}_3$.

$\begin{array}{c}{\widehat{\theta }}_1=\frac{2n_1+n_4+n_5}{2n}\\ =\frac{2\left(49\right)+20+53}{2\left(200\right)}\\ =\frac{98+20+53}{400}\\ =\frac{171}{400}\end{array}$

$=0.4275$

$\begin{array}{c}{\widehat{\theta }}_2=\frac{2n_2+n_4+n_6}{2n}\\ =\frac{2\left(26\right)+20+38}{2\left(200\right)}\\ =\frac{52+20+38}{400}\\ =\frac{110}{400}\end{array}$

$=0.2750$

$\begin{array}{c}{\widehat{\theta }}_3=\frac{2n_3+n_5+n_6}{2n}\\ =\frac{2\left(14\right)+53+38}{2\left(200\right)}\\ =\frac{28+53+38}{400}\\ =\frac{119}{400}\end{array}$

$=0.2975$

From these values the expected proportions are calculated as follows:

$p_1={\theta }_1^2$,$p_2={\theta }_2^2$,$p_3={\left(1-{\theta }_1-{\theta }_2\right)}^2$,$p_4=2{\theta }_1{\theta }_2$,$p_5=2{\theta }_1\left(1-{\theta }_1-{\theta }_2\right)$ and $p_6=2{\theta }_2\left(1-{\theta }_1-{\theta }_2\right)$

Substitute ${\widehat{\theta }}_1$ as 0.4275.

$\begin{array}{c}{\widehat{p}}_1={\widehat{\theta }}_1^2\\ ={\left(0.4275\right)}^2\\ =0.183\end{array}$

Substitute ${\widehat{\theta }}_2$ as 0.2750.

$\begin{array}{c}{\widehat{p}}_2={\widehat{\theta }}_2^2\\ ={\left(0.2750\right)}^2\\ =0.076\end{array}$

$\begin{array}{c}{\widehat{p}}_3={\left(1-{\widehat{\theta }}_1-{\widehat{\theta }}_2\right)}^2\\ ={\left(1-0.4275-0.2750\right)}^2\\ ={0.2975}^2\\ =0.089\end{array}$

$\begin{array}{c}{\widehat{p}}_4=2{\widehat{\theta }}_1{\widehat{\theta }}_2\\ =2\left(0.4275\right)\left(0.2750\right)\\ =0.235\end{array}$

$\begin{array}{c}{\widehat{p}}_5=2{\widehat{\theta }}_1\left(1-{\widehat{\theta }}_1-{\widehat{\theta }}_2\right)\\ =2\left(0.4275\right)\left(0.2975\right)\\ =0.254\end{array}$

$\begin{array}{c}{\widehat{p}}_6=2{\widehat{\theta }}_2\left(1-{\widehat{\theta }}_1-{\widehat{\theta }}_2\right)\\ =2\left(0.2750\right)\left(0.2975\right)\\ =0.164\end{array}$

Testing the hypothesis:

Null hypothesis:

$H_0:$ All the six observed proportions are equal to the expected proportion.

Alternative hypothesis:

$H_a:$ At least one of the six observed proportions is not equal to the expected proportion.

Expected frequency:

The expected frequency for each group is calculated as follows,

$\text{Expected frequency}=n{p}_i$

Where,

n is the total number of observed frequency.

$p_i$ is the proportion corresponding to a particular group.

The expected frequency is calculated as follows:

Category	Observed frequency	Expected frequency
HH	49	$200\left(0.183\right)=36.6$
LL	26	$200\left(0.076\right)=15.2$
NN	14	$200\left(0.089\right)=17.8$
HL	20	$200\left(0.235\right)=47$
HN	53	$200\left(0.254\right)=50.8$
LN	38	$200\left(0.164\right)=32.8$
Total	200

Test statistic:

${\chi }^2={\displaystyle \sum _i\frac{{\left(n_i-n{p}_{i0}\right)}^2}{n{p}_{i0}}}$

Where,

$n_i$ represents the observed frequency.

$n{p}_i$ represents the expected frequency.

The table shows the calculation for chi-square test statistic:

Category	Observed $n_i$	Expected $n{p}_i$	${\chi }^2={\displaystyle \sum _i\frac{{\left(n_i-n{p}_{i0}\right)}^2}{n{p}_{i0}}}$
HH	49	36.6	4.201
LL	26	15.2	7.674
NN	14	17.8	0.811
HL	20	47.0	15.511
HN	53	50.8	0.095
LN	38	32.8	0.824
Total			29.1

Thus, the test statistic is 29.1.

Degrees of freedom:

If there are k categories given, then the degrees of freedom would be $k-1$

Here, there are six categories, the degrees of freedom is,

$\begin{array}{c}k-1=6-1\\ =5\end{array}$

Critical value:

Use the Table A.7 to find the chi-square critical values.

Locate 5 under the column of v.

In the column of $\alpha$ locate the least value 0.005.

The value intersecting these two numbers will give the critical value corresponding to ${\chi }_{0.005,5}^2$

Thus, the critical value is ${\chi }_{0.005,5}^2$ is16.748.

Decision rule:

The null hypothesis would be rejected if the P-value is lesser than or equal to the level of significance $\alpha$ and this would occur if the test statistic value of chi-square is greater than or equal to the critical value.

Conclusion:

The test statistic value is 29.1 and the critical value is 16.748.

The test statistic value is greater than the critical value.

Hence, the P-value would be lesser than the level of significance.

That is, $P\text{-value}<0.005\left(=\alpha \right)$

Hence, the null hypothesis is rejected.

Thus, there is no sufficient evidence to conclude that the six observed proportions are equal to the expected proportion.