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 specified probabilities are \(p_{1}=.4, p_{2}=.3\), \(p_{3}=.3\) and the category counts are shown in the table: $$ \begin{array}{l|ccc} \text { Category } & 1 & 2 & 3 \\ \hline \text { Observed Count } & 130 & 98 & 72 \end{array} $$

Short answer

Answer: Yes, we fail to reject the null hypothesis, meaning there is not enough evidence to claim that the observed counts do not match the specified probability distribution with a significance level of 0.05. The observed data is consistent with the specified probabilities of \(p_1=0.4, p_2=0.3,\) and \(p_3=0.3\).

Step-by-step solution

1. Null and Alternative Hypotheses

Define the null and alternative hypotheses as follows: Null Hypothesis (\(H_0\)): The observed counts match the specified probability distribution, i.e., \(p_1=0.4, p_2=0.3,\) and \(p_3=0.3\). Alternative Hypothesis (\(H_a\)): The observed counts do not match the specified probability distribution.

2. Find the Expected Counts

To find the expected counts for each category, first compute the total number of counts: $$N = 130 + 98 + 72 = 300$$ Now calculate the expected count for each category using the specified probabilities: Category 1: \((0.4)(300) = 120\) Category 2: \((0.3)(300) = 90\) Category 3: \((0.3)(300) = 90\)

3. Compute the Test Statistic

To compute the test statistic, we use the Chi-Square Goodness-of-Fit test statistic formula: $$\chi^2 = \sum_{i=1}^{k} \frac{(O_i - E_i)^2}{E_i}$$ Where \(O_i\) are the observed counts, \(E_i\) are the expected counts, and \(k\) is the number of categories. Plugging in the observed and expected counts, we get: $$\chi^2 = \frac{(130-120)^2}{120} + \frac{(98-90)^2}{90} + \frac{(72-90)^2}{90} = 3.4167$$

4. Rejection Region

The problem asks us to find the rejection region with a significance level \(\alpha = 0.05\). Since we have 3 categories, the degrees of freedom (df) for the Chi-Square distribution is \((3-1)=2\). Using a Chi-Square table or calculator, we find the critical value \(\chi^2_{0.05, 2} \approx 5.9915\). The rejection region is defined as: $$\chi^2 > \chi^2_{0.05, 2}$$

5. Calculate the p-value and Conduct the Test

Now, we will use the Chi-Square test statistic computed earlier and the degrees of freedom to find the approximate p-value for the test. Using a Chi-Square table or calculator, we find the p-value \(\approx 0.1815\). Since the p-value \(0.1815 > \alpha=0.05\), we fail to reject the null hypothesis.

6. State the Conclusions

As we fail to reject the null hypothesis, we conclude that there is not enough evidence to claim that the observed counts do not match the specified probability distribution. The observed data is consistent with the specified probabilities of \(p_1=0.4, p_2=0.3,\) and \(p_3=0.3\).

Key concepts

Null and Alternative Hypotheses

In the context of the Chi-Square Goodness-of-Fit test, constructing the null and alternative hypotheses is a critical first step. The null hypothesis (\(H_0\)) posits that the observed data aligns with a specific probability distribution. In simpler terms, we assume that there's no significant difference between the observed counts and what we would expect if the specified probabilities are correct.

Conversely, the alternative hypothesis (\(H_a\) or \(H_1\(H_1\)) suggests that the observed counts deviate from the expected distribution, indicating that there might be some sort of effect or difference present. It's essential to precisely state these hypotheses because they guide the entire analysis process and determine the criteria for assessing whether or not the observed frequencies significantly differ from the expected frequencies.

Expected Count Calculation

The next step in the Chi-Square Goodness-of-Fit test is to calculate the expected counts for each category. These are the counts we'd expect if our null hypothesis was true. To determine the expected count for a category, you multiply the total sample size by the probability specified for that category.

For example, if the total number of observations is 300 and the expected probability for category 1 is 0.4, the expected count would be \(300 \times 0.4 = 120\). It's essential to compute these figures accurately as they are used to compare against the actual observed counts. Any discrepancies between observed and expected counts serve as the basis for our Chi-Square test statistic calculation.

Test Statistic Calculation

Once we have the expected counts, we move on to calculating the test statistic. The Chi-Square test statistic measures how much the observed counts differ from the expected counts. It's calculated by summing the squared differences between observed (\(O_i\)) and expected counts (\(E_i\)), divided by the expected counts, for all categories.

The formula \[\chi^2 = \sum_{i=1}^{k} \frac{(O_i - E_i)^2}{E_i}\] is used, where \(O_i\) is the observed count for each category, \(E_i\) is the expected count, and \(k\) is the number of categories. If the calculated test statistic is large enough, it may lead us to reject the null hypothesis, indicating that the observed counts significantly differ from what we expected.

Significance Level and Rejection Region

The significance level (\(\alpha\)), often set at 0.05 or 5%, is the probability of rejecting the null hypothesis when it's actually true — this is our threshold for deciding if a result is statistically significant. The rejection region is the range of values for which we reject \(H_0\(H_0\)); it depends on \(\alpha\) and the degrees of freedom (df), which is the number of categories minus one.

In the Chi-Square distribution, we look up the critical value corresponding to our \(\alpha\) level and df to define our rejection region. If our calculated test statistic exceeds this critical value, then it falls within the rejection region, and we reject the null hypothesis, inferring that our observed counts are significantly different from the expected.

P-Value Interpretation

The p-value is the probability of obtaining test results as extreme as the observed results, assuming that the null hypothesis is true. It provides a measure of the evidence against the null hypothesis: a small p-value (typically \( \leq \alpha \) suggests strong evidence against \(H_0\), while a large p-value suggests weak evidence.

If the p-value is smaller than or equal to the significance level, we reject the null hypothesis. In our case, the p-value is greater than \(\alpha = 0.05\(\alpha = 0.05\), leading us to fail to reject \(H_0\). This means that there isn't enough statistical evidence to declare that the observed counts are inconsistent with the proposed probability distribution. It's vital for students to understand the meaning of the p-value to correctly interpret the results of hypothesis testing.