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Chapter 19: Problem 758

Consider two independent multinomial distributions with parameters \(\mathrm{n}_{1}, \mathrm{P}_{11}, \mathrm{P}_{21}, \ldots \mathrm{P}_{\mathrm{k} 1}\) and \(\mathrm{n}_{2}, \mathrm{P}_{12}, \mathrm{P}_{22}, \ldots \mathrm{P}_{\mathrm{k} 2}\) Devise a test for the hypothesis that the parameters satisfy, \(\mathrm{P}_{11}=\mathrm{P}_{12}, \ldots \mathrm{P}_{\mathrm{k} 1}=\mathrm{P}_{\mathrm{k} 2}\) where all probabilities are unspecified.

Short Answer

Expert verified
To test the hypothesis that the parameters of two independent multinomial distributions are equal, perform the following steps: 1. Determine observed counts for each category. 2. Calculate the total counts for both distributions. 3. Calculate expected counts under the null hypothesis. 4. Perform the chi-square test using observed and expected counts. 5. Calculate degrees of freedom and critical value. 6. Compare the test statistic with the critical value. 7. Interpret the results in terms of the original problem.

Step by step solution

01

Determine Observed Counts

For each category, we need to determine the observed counts from both multinomial distributions. Let's denote these observed counts as: \(O_{i1}\) and \(O_{i2}\), where \(i = 1, 2, \ldots, k\).
02

Calculate the Total Counts

Compute the total counts for both multinomial distributions by summing the observed counts for all categories: $$ N_1 = \sum_{i=1}^k O_{i1} \quad \text{and} \quad N_2 = \sum_{i=1}^k O_{i2} $$
03

Calculate the Expected Counts

Compute the expected counts for both multinomial distributions under the null hypothesis that all probabilities are the same between the two distributions: $$ E_{i1} = \frac{O_{i1} + O_{i2}}{N_1 + N_2} \cdot N_1 \quad \text{and} \quad E_{i2} = \frac{O_{i1} + O_{i2}}{N_1 + N_2} \cdot N_2 $$
04

Perform the Chi-Square Test

Compute the chi-square test statistic using the observed and expected counts: $$ \chi^2 = \sum_{i=1}^k \frac{(O_{i1} - E_{i1})^2}{E_{i1}} + \sum_{i=1}^k \frac{(O_{i2} - E_{i2})^2}{E_{i2}} $$
05

Calculate Degrees of Freedom and Critical Value

Next, determine the degrees of freedom for the chi-square distribution, which is equal to the number of categories minus 1: $$ \text{df} = k - 1 $$ Then, find the critical value at a specific confidence level, say 95%, using the chi-square distribution table.
06

Compare Test Statistic with Critical Value

Compare the test statistic with the critical value. If the test statistic is greater than the critical value, we reject the null hypothesis. If the test statistic is less than or equal to the critical value, we fail to reject the null hypothesis.
07

Interpret the Results

Based on the comparison in Step 6, interpret the results in terms of the original problem. If we reject the null hypothesis, it means there is evidence to suggest that the probabilities are not the same between the two multinomial distributions. If we fail to reject the null hypothesis, it means we do not have enough evidence to suggest that the probabilities are different.

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Key Concepts

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

Chi-Square Test
The Chi-Square test is a statistical method used to determine if there is a significant difference between the expected frequencies and the observed frequencies in one or more categories of a contingency table. In the context of multinomial distributions, it helps us to test whether the observed outcomes fit a certain expected probability distribution.

When applying the Chi-Square test to multinomial distributions, we calculate the test statistic by summing the squared difference between observed counts and expected counts, divided by the expected counts for each category. If the calculated Chi-Square statistic is larger than the critical value from the Chi-Square distribution table, it suggests that the observed distribution deviates significantly from the expected distribution, leading us to reject the null hypothesis.
Null Hypothesis for Multinomial Distributions
In hypothesis testing for multinomial distributions, the null hypothesis typically states that there is no difference in the probability distribution across multiple categories or distributions. It asserts that any observed difference in sample counts is due to random chance.

Specifically, when testing the equality of two multinomial distributions with unknown probabilities, as in the provided exercise, the null hypothesis posits that the probabilities for corresponding categories are equal across the two distributions, such as \(P_{11} = P_{12}\), and so on for each category. Rejecting the null hypothesis would indicate that the differences in the observed counts are statistically significant and not due to random variation.
Degrees of Freedom
Degrees of freedom in statistics represent the number of values in a calculation that are free to vary. When performing a Chi-Square test for multinomial distributions, the degrees of freedom are crucial as they impact the critical value against which the test statistic is compared.

The degrees of freedom for this test are calculated as the number of categories minus one (\(df = k - 1\)). This accounts for the fact that once the counts for \(k - 1\) categories are known, the count for the last category is determined to ensure the total count equals the sample size. Therefore, in the context of the exercise, if there are \(k\) categories in each distribution, the degrees of freedom for our test would be \(k - 1\).
Independence in Probability
Independence in probability refers to a scenario where the occurrence of one event does not affect the probability of another event occurring. This concept is central to the idea of multinomial distributions, which assume that each trial is independent of others.

Understanding independence is key when determining whether the multinomial model is appropriate for a given situation. If the events are not independent, the probabilities could change after each trial, making a multinomial approach invalid. In our exercise, we assume the independence of the multinomial distributions when hypothesizing that the probabilities across the two distributions are theoretically equal.

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Most popular questions from this chapter

Let the result of a random experiment be classified by two attributes. Let these two attributes be eye color and hair color. One attribute of the outcome, eye color, can be divided into certain mutually exclusive and exhaustive events. Let these events be: \(\mathrm{A}_{1}:\) Blue eyes \(\mathrm{A}_{2}:\) Brown eyes \(\mathrm{A}_{3}:\) Grey eyes \(\mathrm{A}_{4}:\) Black eyes \(\quad \mathrm{A}_{5}:\) Green eyes The other attribute of the outcome can also be divided into certain mutually exclusive and exhaustive events. Let these events be: \(\mathrm{B}_{1}:\) Blond hair \(\mathrm{B}_{2}:\) Brown hair \(\quad \mathrm{B}_{3}:\) Black hair \(\mathrm{B}_{4}:\) Red hair. The experiment is performed by observing \(\mathrm{n}=500\) people and each of them are categorized according to eye color and hair color. Let \(\mathrm{A}_{\mathrm{i}} \cap \mathrm{B}_{\mathrm{j}}\) be the event that a person with eye color \(\mathrm{A}_{\mathrm{i}}\) and hair color \(\mathrm{B}_{\mathrm{i}}\) is observed, \(\mathrm{i}=1,2,3,4,5\) and \(j=1,2,3,4\). Let \(X_{i j}\) be the observed frequency of event \(\mathrm{A}_{\mathrm{i}} \cap \mathrm{B}_{\mathrm{j}}\). Test the hypothesis that \(\mathrm{A}_{\mathrm{i}}\) and \(\mathrm{B}_{\mathrm{j}}\) are independent attributes.

At the annual shareholders' meeting of the Syntho-Diamond Company, the managing director said he was pleased to report that they had captured a further \(6 \%\) of the gem market. 'How do you know?' asked a shareholder. The managing director replied, 'Last year we conducted a survey of 1,000 people who owned jewelry. They were selected quite haphazardly, so that we believe them to be a random sample. Of this group, 400, that is \(40 \%\), owned one or more of our artificial diamonds. Last week we conducted another such survey, and found 460 out of 1,000 owning our gemstones; this represents a rise to \(46 \% .\) ' This didn't impress the shareholder, who retaliated, 'It looks to me as though the difference might easily be due to chance.' The managing director was able to answer this doubt within a few minutes. Can you?

Monsieur Alphonse is director of Continental Mannequin Academy. He is desirous of a raise in salary, so is telling his Board of Management how his results are better than those of their competitor, Mrs. Batty's Establishment for Young Models. 'At the Institute's last examinations, 'he goes on, 'we got 10 honours, 45 passes, and 5 failures, whereas \(\mathrm{Mrs}\). Batty's Establishment only got 4 honours, 35 passes, and had 11 failures.' Do his figures prove his point, or might they be reasonably ascribed to chance? \begin{tabular}{|c|c|c|c|c|} \hline & \multicolumn{3}{|c|} { Exam results } & Row \\ \cline { 2 - 4 } & Honours & Passes & Failures & Totals \\ \hline Alphonse & 10 & 45 & 5 & 60 \\ Batty & 4 & 35 & 11 & 50 \\ \hline Column Totals & 14 & 80 & 16 & \(110=\mathrm{N}\) \\ \hline \end{tabular}

When you use a \(\mathrm{X}^{2}\) test to reject a null hypothesis, you must always look back at the data to make sure that they support your alternative. You have taken a census of the insect population of your rose garden. On the basis of several large samples you conclude that the insect population is distributed as follows: \(\begin{array}{llll}\text { Ladybugs } & 20 \% & \text { Inch worms } 20 \% & \text { Weevils }\end{array}\) \(30 \%\) Aphids \(\quad 25 \%\) Brown spiders \(5 \%\) Now you treat your garden with an insecticide that is supposed to control the undesirable weevils, aphids, and brown spiders without affecting ladybugs or inch worms. To check the effect of the insecticide you collect 150 insects at random. Your sample is composed as follows: \(\begin{array}{llllll}\text { Ladybugs } & 25 & \text { Ladybugs } & 25 & \text { Inch worms } 45\end{array}\) \(\begin{array}{lllll}\text { Weevils } & 45 & \text { Aphids } & 25 & \text { Brown spiders } 10\end{array}\) Use the census to determine the predicted frequencies. Compute \(\mathrm{X}^{2}\) and answer these two questions: (a) Has the distribution of insects changed significantly (at the \(5 \%\) level 1 ? (b) Has the insecticide had the intended effect?

Find the measure of association, \(\varphi^{2}\) for the following data relating reaction to a drug and sex. \begin{tabular}{|c|c|c|c|} \hline & Male & Female & Row Totals \\ \hline Severe reaction & 26 & 24 & 50 \\ \hline Mild reaction & 24 & 26 & 50 \\ \hline Column totals & 50 & 50 & 100 \\ \hline \end{tabular}
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