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. \({ }^{1}\) In an attempt to verify their claim, 200 working people who had recently had heart attacks were surveyed and the day on which their heart attacks occurred was recorded: $$ \begin{array}{lc} \hline \text { Day } & \text { Observed Count } \\ \hline \text { Sunday } & 24 \\ \text { Monday } & 36 \\ \text { Tuesday } & 27 \\ \text { Wednesday } & 26 \\ \text { Thursday } & 32 \\ \text { Friday } & 26 \\ \text { Saturday } & 29 \\ \hline \end{array} $$ 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 order to determine if there is a statistically significant difference in the incidence of heart attacks depending on the day of the week, a Chi-Square test was conducted. Based on the data and calculations, the calculated Chi-Square statistic (5.14) is less than the critical value (12.59) at a significance level of 0.05. As a result, we cannot reject the null hypothesis, which means there is no statistically significant evidence to suggest any difference in the incidence of heart attacks depending on the day of the week.

Step-by-step solution

Define the null and alternative hypotheses

The null hypothesis (H0) states there is no significant difference in the heart attack distribution across different days of the week, while the alternative hypothesis (H1) states there is a significant difference in the heart attack distribution across different days of the week. H0: Heart attack distribution is equal across all days of the week. H1: Heart attack distribution is not equal across all days of the week.

Calculate expected frequencies

To determine the expected frequencies for each day of the week, we first need to find the total number of observed heart attacks (200). If the distribution is equal across the days, we would expect that each day would have the same number of heart attacks. Expected frequency for each day = Total observed heart attacks / 7 = 200 / 7 ≈ 28.57

Calculate the Chi-Square statistic

For each day of the week, we will calculate the Chi-Square statistic using the following formula: \(\chi^{2}=\sum \frac{\left( O-E\right)^{2}}{E}\) where O represents the observed frequency, E represents the expected frequency, and the sum is taken over all days of the week. Based on the data given: \(\chi^{2}=\frac{(24-28.57)^{2}}{28.57}+\frac{(36-28.57)^{2}}{28.57}+\frac{(27-28.57)^{2}}{28.57}+\frac{(26-28.57)^{2}}{28.57}+\frac{(32-28.57)^{2}}{28.57}+\frac{(26-28.57)^{2}}{28.57}+\frac{(29-28.57)^{2}}{28.57}\approx 5.14\)

Find the critical value for the Chi-Square distribution

To determine the critical value, we first need to find the degrees of freedom (df), which can be calculated as: df = number of categories - 1 = 7 (days) - 1 = 6 Now, using a Chi-Square distribution table and given the significance level of \(\alpha = 0.05\), we find the critical value corresponding to 6 degrees of freedom: Critical value at 0.05 significance level and 6 df ≈ 12.59

Compare the calculated Chi-Square statistic to the critical value

Now we will compare the calculated Chi-Square statistic (5.14) to the critical value (12.59). Since 5.14 is less than 12.59, we cannot reject the null hypothesis. This means there is no statistically significant evidence to suggest any difference in the incidence of heart attacks depending on the day of the week.

Key concepts

Null and Alternative Hypotheses

Understanding the null and alternative hypotheses is crucial when conducting a statistical test. In our heart attack incidence example, the null hypothesis (H0) posits that there is no difference in heart attack frequency among different days of the week, indicating uniform distribution across all days. In contrast, the alternative hypothesis (H1) suggests that there is an uneven distribution, implying that the frequency of heart attacks may indeed vary by day.

Formulating these hypotheses lays the foundation for statistical testing. The null hypothesis represents a default position, reflecting no effect or no change, while the alternative embodies what researchers are set out to prove or observe, typically suggesting a new effect or change in the population. It's vital to define these hypotheses correctly because they determine the direction and framework of the statistical analysis.

Expected Frequency Calculation

The next step in a Chi-Square Test involves the calculation of expected frequencies. These frequencies represent the number of observations that we would anticipate for each category if the null hypothesis were true. To calculate expected frequencies, as illustrated in the heart attack incidence on different days of the week, we divide the total number of observations by the number of categories. Here, we would expect an equal number of heart attacks across the seven days of the week if the null hypothesis were valid.

Therefore, with 200 total heart attacks, the expected frequency is \( 200 / 7 \) or approximately 28.57 for each day. By comparing observed frequencies to these expected values, we can assess whether the data significantly deviates from what we would expect under the null hypothesis. This step is vital because the Chi-Square statistic is based on the differences between observed and expected frequencies.

Chi-Square Distribution

Once the expected frequencies and the Chi-Square statistic have been calculated, understanding the Chi-Square distribution is key to determining statistical significance. The Chi-Square distribution is a theoretical distribution used in hypothesis testing that is applicable when the test statistic follows a Chi-Square distribution under the null hypothesis.

The distribution is specified by degrees of freedom (df), which, in the case of the Chi-Square Test, are determined by the number of categories minus one. For instance, there are seven days a week, so with seven categories, the degrees of freedom would be six (7 - 1 = 6). This value directly influences the shape of the Chi-Square distribution and is used to ascertain the likelihood of observing a test statistic as extreme as, or more than, the one calculated from the sample data, assuming the null hypothesis is correct.

Significance Level

Lastly, it is important to consider the significance level (\( \alpha \)) when performing a hypothesis test. The significance level is the criterion used to decide whether to reject the null hypothesis and is often set at 0.05, meaning a 5% risk of concluding that a difference exists when there is no actual difference.

When the calculated Chi-Square value is compared to the critical value from the Chi-Square distribution at the chosen significance level and degrees of freedom, the decision to reject or not reject the null hypothesis is made. If the Chi-Square statistic is greater than the critical value, the null hypothesis is rejected, suggesting the data does not fit the expected distribution. In our example, with a significance level of 0.05 and 6 df, the critical value is approximately 12.59. Because the calculated Chi-Square statistic (\( 5.14 \) is less than this critical value, we fail to reject the null hypothesis, indicating that the observed data do not provide sufficient evidence to suggest a variation in heart attack occurrences based on the day of the week.