91Ó°ÊÓ

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 \(H H, L L, N N, H L, H N\), and \(L N\). If the probabilities (population proportions) for the single headlight focus direction are \(P(H)=\theta_{1}\), \(P(L)=\theta_{2}\), and \(P(N)=1-\theta_{1}-\theta_{2}\) and the two headlights are focused independently of each other, the probabilities of the six outcomes for a randomly selected car are the following: $$ \begin{aligned} &p_{1}=\theta_{1}^{2} \quad p_{2}=\theta_{2}^{2} \quad p_{3}=\left(1-\theta_{1}-\theta_{2}\right)^{2} \\ &p_{4}=2 \theta_{1} \theta_{2} \quad p_{5}=2 \theta_{1}\left(1-\theta_{1}-\theta_{2}\right) \\ &p_{6}=2 \theta_{2}\left(1-\theta_{1}-\theta_{2}\right) \end{aligned} $$ Use the accompanying data to test the null hypothesis $$ H_{0}: p_{1}=\pi_{1}\left(\theta_{1}, \theta_{2}\right), \ldots, p_{6}=\pi_{6}\left(\theta_{1}, \theta_{2}\right) $$ where the \(\pi_{i}\left(\theta_{1}, \theta_{2}\right)\) 's are given previously. \(\begin{array}{lllllll}\text { Outcome } & H H & L L & N N & H L & H N & L N \\\ \text { Frequency } & 49 & 26 & 14 & 20 & 53 & 38\end{array}\)

Step by step solution

01

Define the Hypotheses

We are testing the null hypothesis \(H_0\), which states that the probabilities of the outcomes (\(p_1\) to \(p_6\)) match the provided models based on \(\theta_1\) and \(\theta_2\).

02

Set up the Expected Probabilities

The expressions for the probabilities of each outcome according to the model are: - \(p_1 = \theta_1^2\)- \(p_2 = \theta_2^2\)- \(p_3 = (1-\theta_1-\theta_2)^2\)- \(p_4 = 2\theta_1\theta_2\)- \(p_5 = 2\theta_1(1-\theta_1-\theta_2)\)- \(p_6 = 2\theta_2(1-\theta_1-\theta_2)\).

03

Calculate Sample Proportions

To test \(H_0\), compute the sample proportions from the data: - Number of trials: \(n = 49 + 26 + 14 + 20 + 53 + 38 = 200\) - \(\hat{p}_1 = \frac{49}{200}, \hat{p}_2 = \frac{26}{200}, \hat{p}_3 = \frac{14}{200}, \hat{p}_4 = \frac{20}{200}, \hat{p}_5 = \frac{53}{200}, \hat{p}_6 = \frac{38}{200}\).

04

Solve the Parameter Equations

We need to estimate \(\theta_1\) and \(\theta_2\) by solving the system of equations from the equations of expected probabilities and the sample proportions. Solving: - \(\theta_1^2 \approx \frac{49}{200}\)- \(\theta_2^2 \approx \frac{26}{200}\)- \((1-\theta_1-\theta_2)^2 \approx \frac{14}{200}\).

05

Estimate Parameters

Solve for \(\theta_1\) and \(\theta_2\) using numerical methods or substitutions. Suppose solutions yield choices for \(\theta_1\) and \(\theta_2\), calculate their approximate values. Full analytic solutions may require numerical solutions.

06

Compute Expected Frequencies

Using the estimated \(\theta_1\) and \(\theta_2\), calculate the expected frequencies:- \(E_1 = 200 \times \theta_1^2\)- \(E_2 = 200 \times \theta_2^2\)- \(E_3 = 200 \times (1-\theta_1-\theta_2)^2\)- \(E_4 = 200 \times 2\theta_1\theta_2\)- \(E_5 = 200 \times 2\theta_1(1-\theta_1-\theta_2)\)- \(E_6 = 200 \times 2\theta_2(1-\theta_1-\theta_2)\).

07

Conduct Chi-Square Test

Use the chi-square test with the formula \(\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\), where \(O_i\) are observed frequencies and \(E_i\) are expected frequencies.Calculate \(\chi^2\) and compare it with the critical value of \(\chi^2\) distribution at the suitable significance level (e.g., 0.05) and degrees of freedom \((df = 3)\).

08

Conclusion

If the calculated \(\chi^2\) value is greater than the critical \(\chi^2\) value, reject the null hypothesis \(H_0\). Otherwise, do not reject \(H_0\), indicating a good fit between the model and the data.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chi-Square Test

Imagine you are trying to check whether your data matches a certain combination of expected outcomes. This is when the Chi-Square Test becomes your buddy. It is a statistical method used to see if there are significant differences between the expected and observed data.

The Chi-Square Test formula is quite friendly:

• It runs as \( \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \), where \( O_i \) is the observed frequency (what you actually see happening)
• and \( E_i \) is the expected frequency (what you expect should happen).

For example, if you counted how often car headlights are focused in pairs (either too high, too low, or properly), you can use this test to see if your observed results match the probabilities based on hypothetical focus direction parameters.

In practice, once you've calculated the \( \chi^2 \) value, you compare it to a critical value from the Chi-Square distribution table. This basically tells you whether to accept or reject your null hypothesis (\( H_0 \)). If your calculated value is larger, you're looking at a mismatch that needs attention.

Remember, the degrees of freedom (number of categories minus parameters estimated) determine which critical value you should use. This provides a solid backbone for hypothesis testing in statistics.

Probability Distributions

Probability distributions offer a secure way to forecast the potential outcomes of an event. Think of it like a recipe that tells you the likelihood of different ingredients in your dish (in this case, outcomes of car headlight focus direction):

• A probability distribution assigns a probability to each possible outcome of a random event.

If you have three possibilities for a single headlight (too high, too low, or just right), the probability values are labeled as \( P(H) = \theta_1 \), \( P(L) = \theta_2 \), and \( P(N) = 1 - \theta_1 - \theta_2 \). With two headlights, these probabilities lead to six possible outcomes like \( HH, LL, NN, HL, HN, \) and \( LN \).

Each outcome's probability can be described more mathematically using equations derived from these base probabilities. For example, the probability of both headlights being 'too high' is \( p_1 = \theta_1^2 \).

• This helps statisticians not only in organizing data, but it also sets foundation points for parameter estimation and further statistical tests, like the Chi-Square Test.

Probability distributions are essential in determining the likelihood of diverse real-world outcomes and preparing for statistical analysis.

Parameter Estimation

Parameter estimation is like tinkering with your model until it fits your data just right. When you're given observed frequencies (like how often car headlights are improperly aligned), you estimate parameters like \( \theta_1 \) and \( \theta_2 \) to align your model's theoretical probabilities with these observations.

The process involves:

• Solving equations crafted from theoretical probabilities.
• Adjusting parameters to match observed sample proportions.

In our headlight example, you'd solve equations such as \( \theta_1^2 \approx \frac{49}{200} \) to initially estimate \( \theta_1 \) and \( \theta_2 \). However, these initial guesses often need adjustments, employing numerical methods or optimization techniques. Easy, right? Well, finding the correct parameters requires patience and often systematic approaches.

Once accurate parameter values are in hand, they become the key input for calculating expected frequencies based on these parameters. This enables a clear comparison against observed data and is a cornerstone of hypothesis testing methods such as the Chi-Square Test.

A well-tuned parameter estimate ensures the model truly reflects real-world phenomena, striking a balance between mathematical elegance and practical application.