Question

Suppose you wish to test the null hypothesis that three binomial parameters \(p_{A}, p_{B},\) and \(p_{c}\) are equal versus the alternative hypothesis that at least two of the parameters differ. Independent random samples of 100 observations were selected from each of the populations. Use the information in the table to answer the questions in Exercises \(5-7 .\) $$ \begin{array}{lrrrr} \hline & {\text { Population }} & \\ & \text { A } & \text { B } & \text { C } & \text { Total } \\ \hline \text { Successes } & 24 & 19 & 33 & 76 \\ \text { Failures } & 76 & 81 & 67 & 224 \\ \hline \text { Total } & 100 & 100 & 100 & 300 \end{array} $$ Calculate the test statistic and find the approximate \(p\) -value for the test in Exercise 5.

Short answer

Answer: __________

Step-by-step solution

Calculate the overall proportions of successes and failures

To begin, we need to find the proportion of successes and failures across all three populations combined. We are given that there is a total of 76 successes and 224 failures out of 300 total observations. Therefore, the overall proportions of successes and failures are $$p_s = \frac{76}{300}$$ and $$p_f = \frac{224}{300}$$.

Calculate the expected counts for all six cells

Now, we need to determine the expected counts for each cell, assuming that the null hypothesis is true. The expected count for the successes in Population A would be the product of the total number of observations in Population A and the overall proportion of successes, and so on for the rest of the cells. The expected counts are as follows: $$E_{A_s} = 100 \cdot p_s = 100 \cdot \frac{76}{300}$$ $$E_{A_f} = 100 \cdot p_f = 100 \cdot \frac{224}{300}$$ $$E_{B_s} = 100 \cdot p_s = 100 \cdot \frac{76}{300}$$ $$E_{B_f} = 100 \cdot p_f = 100 \cdot \frac{224}{300}$$ $$E_{C_s} = 100 \cdot p_s = 100 \cdot \frac{76}{300}$$ $$E_{C_f} = 100 \cdot p_f = 100 \cdot \frac{224}{300}$$

Calculate the chi-square test statistic

The chi-square test statistic is calculated as $$\chi^2 = \sum \frac{(O - E)^2}{E}$$, where O refers to the observed counts (the counts given in the table) and E refers to the expected counts calculated in Step 2. Here, we will calculate the chi-square test statistic: $$\chi^2 = \frac{(24 - E_{A_s})^2}{E_{A_s}} + \frac{(76 - E_{A_f})^2}{E_{A_f}} + \frac{(19 - E_{B_s})^2}{E_{B_s}} + \frac{(81 - E_{B_f})^2}{E_{B_f}} + \frac{(33 - E_{C_s})^2}{E_{C_s}} + \frac{(67 - E_{C_f})^2}{E_{C_f}}$$

Calculate the p-value for the test statistic

Now that we have calculated the chi-square test statistic, we need to find the p-value associated with it. Since this is a chi-square test for independence, our degrees of freedom are \(df = (3 - 1) \times (2 - 1) = 2\). Using a chi-square distribution table or a calculator, find the p-value associated with the test statistic and degrees of freedom. Finally, compare the p-value to the desired significance level (usually 0.05) to determine if there is enough evidence to reject the null hypothesis. If the p-value is less than the significance level, then there is evidence to support that at least two of the population parameters differ.

Key concepts

Binomial Parameters

In the context of the chi-square test for independence, understanding binomial parameters is essential. Binomial parameters refer to the probability of success (or failure) in a binary outcome scenario within a specific population. For a given population, let's denote this probability of success by the parameter \( p \.

In the exercise provided, we are working with three separate populations, each with its own success probability denoted as \( p_{A} \), \( p_{B} \), and \( p_{C} \). To understand whether these probabilities are statistically the same or different across the populations, we would compare the observed successes and failures against what we would expect if the probabilities were equal (as asserted by the null hypothesis).

A crucial aspect of this comparison is the expectation that these proportions will remain consistent across each population if there is true independence. Therefore, when we collect data in the form of successes and failures for each population, we can then analyze any significant deviations from what we'd expect under a uniform probability of success \( p \). This deviation could suggest that not all binomial parameters are equal.

Null Hypothesis

The null hypothesis in the context of a chi-square test for independence serves as a statement of 'no effect' or 'no difference.' In simpler terms, it implies that any observed differences in the data are due to random chance rather than actual disparities between the groups or conditions being tested.

For the exercise at hand, the null hypothesis claims that the binomial parameters for each of the three populations (\( p_{A} \), \( p_{B} \), and \( p_{C} \)) are equal. Formally, we state this null hypothesis as \( H_0: p_{A} = p_{B} = p_{C} \). We essentially say that there is no significant difference in the success rates of the three populations being studied, and whatever variations exist are random and not statistically significant.

When we perform the chi-square test, we are testing this null hypothesis against the alternative hypothesis, which posits that at least two of the binomial parameters are not equal. The aim is to determine whether we have sufficient evidence to reject the null hypothesis and conclude that there is a statistically significant difference in the success rates of at least two populations.

P-Value

The p-value is a crucial concept in hypothesis testing, representing the probability of observing a test statistic as extreme as, or more extreme than, the one calculated from our sample data, assuming that the null hypothesis is true. It quantifies the evidence against the null hypothesis.

In the provided step-by-step solution, once we've calculated the chi-square test statistic, we then seek to find the p-value. The smaller the p-value, the stronger the evidence against the null hypothesis. Typically, researchers use a significance level (commonly denoted by \( \alpha \)), such as 0.05, as a threshold for determining statistical significance. If the p-value is less than \( \alpha \), one may conclude there is sufficient evidence to reject the null hypothesis in favor of the alternative.

Therefore, in the provided exercise, if our calculated p-value from the chi-square test statistic is below the chosen significance level, it would imply that there is a statistically significant difference in the success rates among the three populations. This could potentially lead to the conclusion that not all the binomial parameters (\( p_{A} \), \( p_{B} \), and \( p_{C} \)) are equal.