Question

Conduct the appropriate test of specified probabilities using the information given. Write the null and alternative hypotheses, give the rejection region with \(\alpha=.05\) and calculate the test statistic. Find the approximate \(p\) -value for the test. Conduct the test and state your conclusions. The five categories are equally likely to occur, and the category counts are shown in the table: $$ \begin{array}{l|ccccc} \text { Category } & 1 & 2 & 3 & 4 & 5 \\ \hline \text { Observed Count } & 47 & 63 & 74 & 51 & 65 \end{array} $$

Short answer

Answer: There is not enough evidence to reject the assumption that the categories are equally likely to occur.

Step-by-step solution

State the null and alternative hypotheses

The null hypothesis (\(H_0\)) asserts that all categories are equally likely to occur. The alternative hypothesis (\(H_1\)) states that at least one category has a different probability. $$H_0: p_1 = p_2 = p_3 = p_4 = p_5$$ $$H_1: \text{At least one } p_i \neq p_j$$

Determine the critical value and rejection region

To find the critical value for the chi-squared test, we need the degrees of freedom, which in this case is equal to the number of categories minus one: \(df = 5 - 1 = 4\). For a significance level of \(\alpha = 0.05\), and \(df=4\), the critical value can be found in a chi-squared distribution table or by using a calculator or statistical software. The critical value is \(\approx 9.488\). Thus, the rejection region is: $$\chi^2 > 9.488$$

Calculate the test statistic

First, we need to calculate the expected counts for each category, which are computed as the product of the total count and the probability for each category. Since the five categories are equally likely, the probability for each category is \(1/5 = 0.2\). The total count is the sum of observed counts: $$n = 47 + 63 + 74 + 51 + 65 = 300$$ Now, we can calculate the expected counts: $$E_i = n \cdot p_i = 300 \cdot 0.2 = 60$$ Next, we will calculate the test statistic using the formula \(\chi^2 = \sum_{i=1}^{5} \frac{(O_i - E_i)^2}{E_i}\), where \(O_i\) represents the observed count and \(E_i\) represents the expected count: $$\chi^2 = \frac{(47-60)^2}{60}+\frac{(63-60)^2}{60}+\frac{(74-60)^2}{60}+\frac{(51-60)^2}{60}+\frac{(65-60)^2}{60} = 5.6867$$

Find the approximate \(p\)-value for the test

To find the \(p\)-value, we need the cumulative distribution function (CDF) for the chi-squared distribution given our test statistic (\(\chi^2 = 5.6867\)) and the degrees of freedom (\(df=4\)). Using a calculator or statistical software, the \(p\)-value can be found to be approximately 0.224.

Conduct the test and state your conclusions

Now, we will compare the test statistic to the critical value and the \(p\)-value to the significance level: - The test statistic (\(\chi^2 = 5.6867\)) is less than the critical value (\(9.488\)), so it does not fall within the rejection region. - The \(p\)-value (0.224) is greater than the significance level (\(\alpha = 0.05\)), so we do not reject the null hypothesis. Both results above indicate that we do not have enough evidence to reject the null hypothesis. Therefore, our conclusion is that there is not enough evidence to support the claim that the probabilities of the five categories are different. We cannot reject the assumption that the categories are equally likely to occur.

Key concepts

Null and Alternative Hypotheses

Understanding the null and alternative hypotheses is crucial for any statistical test. The null hypothesis (\( H_0 \) is a statement of no effect or no difference and serves as the basis for testing. In the context of a chi-squared test, it often asserts that all categories have the same probability of occurring.

For instance, in the example provided, the null hypothesis claims that the five categories are equally likely to occur, as stated mathematically by \( H_0: p_1 = p_2 = p_3 = p_4 = p_5 \).

On the other hand, the alternative hypothesis (\( H_1 \) represents the outcome that the study is aiming to detect. It is the hypothesis that there is an effect or a difference. In our problem, the alternative hypothesis suggests that at least one category has a different probability of occurring than the others, noted as \( H_1: \text{At least one } p_i eq p_j \).The role of a hypothesis test, like the chi-squared test, is to determine whether there is enough evidence to reject the null hypothesis in favor of the alternative hypothesis.

Test Statistic Calculation

Calculating the test statistic is a crucial part of hypothesis testing. For the chi-squared test, the test statistic measures how far the observed data deviate from what would be expected under the null hypothesis. It is calculated using an equation that quantifies the difference between observed and expected frequencies.

In the educational exercise, the test statistic is computed with the formula \( \chi^2 = \sum_{i=1}^{5} \frac{(O_i - E_i)^2}{E_i} \), where \( O_i \) represents the observed counts and \( E_i \) represents the expected counts under the null hypothesis. Given that the categories are presumed equally likely, the expected count for each is \( 300 \times 0.2 = 60 \).

By applying the formula, the test statistic was calculated as \( 5.6867 \). This value will then be used to determine whether it falls into the rejection region of the chi-squared distribution given the degrees of freedom.

P-value

A p-value is a measure of the probability that the observed data would occur if the null hypothesis were true. It is a crucial concept in hypothesis testing used to make decisions about the hypotheses.

In the given exercise, the p-value is found by looking at the chi-squared distribution cumulative distribution function (CDF) with the calculated test statistic \( \chi^2 = 5.6867 \) and degrees of freedom \( df = 4 \). The resulting p-value was approximately 0.224.

Interpreting this p-value is straightforward: it tells us that there is a 22.4% chance of observing data as extreme as or more extreme than our observed data if the null hypothesis is true. To decide if this provides enough evidence to reject the null hypothesis, we compare the p-value to our significance level, typically \( \alpha = 0.05 \). Since \( 0.224 > 0.05 \) in our example, we do not reject \( H_0 \), suggesting the observed data do not significantly deviate from that expected under the null hypothesis.