91Ó°ÊÓ

Accidents or Illnesses The people who went to the emergency room at a local hospital were treated for an accident (A) or illness (I). Test the claim \(\alpha=0.10\) that the reason given occurred at random. $$ \begin{array}{llllllllll} \text { I } & \text { A } & \text { I } & \text { A } & \text { A } & \text { A } & \text { A } & \text { A } & \text { A } & \text { I } \\ \text { A } & \text { I } & \text { I } & \text { A } & \text { A } & \text { I } & \text { I } & \text { A } & \text { A } & \text { A } \\ \text { A } & \text { I } & \text { A } & \text { I } & \text { A } & \text { A } & \text { A } & \text { I } & \text { I } & \text { A } \\ \text { A } & \text { I } & \text { I } & \text { A } & \text { A } & \text { I } & \text { A } & \text { I } & \text { A } & \text { I } \\ \text { A } & \text { I } & \text { A } & \text { A } & \text { I } & \text { I } & \text { A } & \text { A } & \text { A } & \text { I } \\ \text { I } & \text { A } & \text { I } & \text { A } & \text { A } & \text { I } & \text { I } & \text { A } & \text { A } & \text { A } \end{array} $$

Step by step solution

01

Count the Occurrences

Count the number of occurrences of accidents (A) and illnesses (I) from the data provided. There are 60 entries in total.

02

Calculate Observed Frequencies

Determine the frequency of each category. Let's assume there are \( n_A \) accidents and \( n_I \) illnesses. Sum these to ensure total is 60.

03

Set Up the Hypotheses

The null hypothesis \( H_0 \) states that accidents and illnesses occur with equal probability, \( P(A) = P(I) = 0.5 \). The alternative hypothesis \( H_1 \) is that the probabilities are not equal, \( P(A) eq P(I) \).

04

Expected Frequencies Calculation

If occurrences are truly random with equal probability, each category should have an expected frequency of \( \frac{60}{2} = 30 \).

05

Use the Chi-Square Test

Set up the chi-square test. The formula is \( \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \), where \( O_i \) is the observed frequency and \( E_i \) is the expected frequency. Here, \( E_A = E_I = 30 \).

06

Calculate Chi-Square Statistic

Substitute observations into the formula. If \( n_A \) and \( n_I \) are found earlier: \( \chi^2 = \frac{(n_A - 30)^2}{30} + \frac{(n_I - 30)^2}{30} \).

07

Determine Critical Value

With \( \alpha = 0.10 \) and \( df = 1 \), find the critical chi-square value from tables or a calculator. It is approximately 2.71.

08

Conclusion on Hypothesis

Compare \( \chi^2 \) computed with the critical value. If \( \chi^2 \leq 2.71 \), fail to reject \( H_0 \). If \( \chi^2 > 2.71 \), reject \( H_0 \).

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Key Concepts

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

Null Hypothesis

In statistical hypothesis testing, the null hypothesis ( H_0 ) is a statement of no effect or no difference. It's essentially a default position that there is no association between two measured phenomena.

In our exercise, the null hypothesis is that there is no difference in the occurrence between accidents and illnesses among emergency room visits. In numerical terms, it states that the probability of an accident (A) is equal to the probability of an illness (I), i.e., P(A) = P(I) = 0.5 . We assume that both outcomes occur randomly, following this equal probability.

The null hypothesis sets a standard to be tested, and based on the test's outcome, we either reject it or fail to reject it. It's crucial to understand that failing to reject the null hypothesis does not prove it true; it simply suggests that there isn't strong enough evidence to disprove it.

Critical Value

The critical value is a threshold number that a calculated statistic must exceed in order to reject the null hypothesis. It's selected based on the significance level, symbolized by b1 (alpha), which represents the probability of rejecting a true null hypothesis.

In this exercise, the significance level is set at 0.10. This means that we are willing to accept a 10% chance of incorrectly rejecting a true null hypothesis. With 1 degree of freedom, the critical value for our chi-square distribution is approximately 2.71.

If the calculated chi-square statistic from our data is greater than this critical value, we reject the null hypothesis. Otherwise, we do not reject it, suggesting that any observed difference might be due to random chance.

Observed Frequencies

Observed frequencies are the actual counts of occurrences in each category based on the collected data. In this case, we need to count how many times 'accidents' and 'illnesses' were recorded from the hospital's emergency room data statements.

Counting carefully, we identify that if accidents (A) occurred, say, 38 times and illnesses (I), 22 times (just theoretical numbers here), these would serve as our observed frequencies. These values are significant because they provide the empirical data necessary for the chi-square test.

We use these figures to compare against what we would expect under the null hypothesis to see if there are significant deviations that could suggest a non-random pattern.

Expected Frequencies

Expected frequencies are what we would anticipate finding if the null hypothesis was true. These are calculated based on the presumption of equal probability for all categories.

In this scenario, assuming that accidents and illnesses occur with equal probability, the expected frequency for each would be half of the total observations. Given that the total number of observations is 60, each category - accidents and illnesses - would hence have an expected frequency of 30.

Expected frequencies are key in the chi-square test, as they help determine whether the observed frequencies deviate significantly from what would typically be expected in a random distribution scenario.