Question

Researchers from Germany have concluded that the risk of a heart attack for a working person may be as much as \(50 \%\) greater on Monday than on any other day. \({ }^{1}\) The researchers kept track of heart attacks and coronary arrests over a period of 5 years among 330,000 people who lived near Augsburg, Germany. In an attempt to verify their claim, you survey 200 working people who had recently had heart attacks and recorded the day on which their heart attacks occurred: Do the data present sufficient evidence to indicate that there is a difference in the incidence of heart attacks depending on the day of the week? Test using \(\alpha=.05 .\)

Short answer

In conclusion, after performing a chi-squared test and comparing the p-value to the significance level of 0.05, we can determine whether there is enough evidence to reject the null hypothesis and support the claim that there is a difference in the incidence of heart attacks depending on the day of the week. If the p-value is less than 0.05, we have enough evidence to reject the null hypothesis. If the p-value is greater than or equal to 0.05, we cannot reject the null hypothesis and cannot support the claim. The results of this hypothesis test will help us understand the relationship between heart attack risks and specific days of the week, specifically if Mondays have a higher risk for working individuals.

Step-by-step solution

State the null and alternative hypotheses

The null hypothesis (\(H_0\)) is that there is no difference in the incidence of heart attacks depending on the day of the week. The alternative hypothesis (\(H_a\)) is that there is a difference in the incidence of heart attacks depending on the day of the week.

Calculate the expected frequencies for each day of the week

Assuming the null hypothesis is true and there is no difference in the incidence of heart attacks, we would expect an equal number of heart attacks on each day of the week. With 200 heart attacks recorded, we would expect: - \(\frac{200}{7}\approx28.57\) heart attacks on each day of the week.

Perform a chi-squared test to calculate the test statistic and p-value

Using the data from the survey and the expected frequencies calculated above, we can perform a chi-squared test. The test statistic is calculated as \(\chi^{2}=\sum\frac{(O-E)^{2}}{E}\) where O is the observed frequency for each day of the week and E is the expected frequency for each day of the week. Calculate the test statistic for the given data and use a chi-squared distribution table with 6 degrees of freedom (\(df = 7 - 1\)) to find the p-value.

Determine whether there is enough evidence to reject the null hypothesis

With a significance level of \(\alpha=0.05\), we will reject the null hypothesis if the p-value is less than 0.05. If the p-value is less than 0.05: - There is enough evidence to reject the null hypothesis and support the claim that there is a difference in the incidence of heart attacks depending on the day of the week. If the p-value is greater than or equal to 0.05: - There is not enough evidence to reject the null hypothesis, and we cannot support the claim that there is a difference in the incidence of heart attacks depending on the day of the week. Compare the calculated p-value to the significance level \(\alpha=0.05\) and make a conclusion accordingly.

Key concepts

Statistical Hypothesis Testing

Statistical hypothesis testing is a powerful method used to determine if there is enough statistical evidence in a sample of data to infer a particular condition for the entire population. In our context, this means analyzing the data collected from working individuals who experienced heart attacks. By using a specific test, such as the Chi-Squared Test, researchers can either accept or reject a hypothesis related to the occurrence of heart attacks on different weekdays.

Here's how it works:

• We first make an assumption that reflects the lack of a specific effect or relationship (the null hypothesis).
• An alternative hypothesis is set to capture the presence of an effect or a relationship.
• We gather and analyze data to evaluate these propositions and make a decision based on calculated probabilities.

Statistical hypothesis testing allows us to make decisions or predictions about a population from a sample with a quantified level of confidence.

Null Hypothesis

The null hypothesis, often labeled as \(H_0\), is a statement that there is no effect or no difference. In statistical hypothesis tests like the one used in this heart attack frequency study, it acts as a reference point. Here, it suggests that heart attacks occur randomly across different days of the week, following no particular pattern.

Formulating this hypothesis involves assuming an even distribution of heart attacks every day. So, for 200 observed heart attacks, we would expect each day to account for about \( \frac{200}{7} \) or approximately 28.57 attacks.

In practice, the null hypothesis is what the test aims to assess. If the collected data doesn't present significant deviations from what's expected under \(H_0\), the hypothesis remains upheld. If, however, the evidence strongly contradicts \(H_0\), we might consider rejecting it in favor of the alternative hypothesis.

Alternative Hypothesis

Contrasting the null hypothesis, the alternative hypothesis, denoted as \(H_a\), posits that there is indeed a difference in the phenomenon being studied. For our exercise, \(H_a\) claims that the occurrence of heart attacks does vary depending on the weekday.

The alternative hypothesis challenges the assumption of uniformity. Rather than heart attacks being equally likely on any given day, \(H_a\) suggests an imbalance or a pattern in the data, such as a spike in occurrences on specific days like Monday. By introducing this hypothesis, researchers are examining if their gathered evidence sufficiently supports the idea of non-uniform distribution.

Testing \(H_a\) involves showing that the real-world data significantly diverge from the outcomes predicted by the null hypothesis. When the alternative hypothesis is substantiated, it opens pathways for further investigation into the factor or reasons causing the variation.

P-Value

The p-value is a critical measure in statistical hypothesis testing. It represents the probability of observing results at least as extreme as those recorded, assuming the null hypothesis is true. In simple terms, the p-value helps us decide whether our observed data are unusual under the assumption of \(H_0\).

For the heart attack data, once the Chi-Squared Test statistic is calculated, we use it to find a corresponding p-value. If this value is very low, it indicates that observing such data (or more extreme) is unlikely if the null hypothesis were true. Thus, a low p-value favors the alternative hypothesis.

The interpretation of the p-value depends on our chosen significance level, typically set at 0.05. If the p-value is below 0.05, it suggests strong evidence against \(H_0\), prompting researchers to consider rejecting the null hypothesis.

Significance Level

The significance level, denoted as \( \alpha \), determines the threshold for deciding whether to reject the null hypothesis. It quantifies the risk we are willing to take in rejecting \(H_0\) when it actually is true—a "false positive."

In this exercise, the significance level is set at 0.05. This means there is a 5% risk of concluding that there is a difference in heart attack frequencies across weekdays when, in truth, there is none.

Choosing a significance level is crucial as it balances the types of errors one might make (Type I errors if \(H_0\) is wrongly rejected, and Type II errors if \(H_0\) is wrongly accepted). Therefore, a predetermined \( \alpha \) value ensures that the results of the hypothesis test are interpreted consistently and meaningfully.