Question

Gender of Patients at a Medical Center The gender of the patients at a medical center is recorded. Test the claim at \(\alpha=0.05\) that they are admitted at random. $$ \begin{array}{llllllllll} \mathrm{F} & \mathrm{F} & \mathrm{M} & \mathrm{M} & \mathrm{M} & \mathrm{M} & \mathrm{M} & \mathrm{F} & \mathrm{F} & \mathrm{F} \\ \mathrm{M} & \mathrm{M} & \mathrm{M} & \mathrm{M} & \mathrm{M} & \mathrm{M} & \mathrm{F} & \mathrm{M} & \mathrm{M} & \mathrm{F} \\ \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{M} & \mathrm{M} & \mathrm{M} & \mathrm{F} & \mathrm{M} & \mathrm{F} & \mathrm{M} \\ \mathrm{M} & \mathrm{M} & \mathrm{M} & \mathrm{M} & \mathrm{M} & \mathrm{F} & \mathrm{M} & \mathrm{M} & \mathrm{F} & \mathrm{M} \\ \mathrm{F} & \mathrm{F} & \mathrm{M} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{M} \end{array} $$

Short answer

The claim is not rejected; admissions seem random regarding gender.

Step-by-step solution

Define Hypotheses

In this step, we define the null and alternative hypotheses. The null hypothesis \(H_0\) is that the genders are admitted at random, meaning there is no preference for either gender. The alternative hypothesis \(H_1\) is that the admission is not random with respect to gender.

Count the Frequency of Each Gender

Count the occurrences of males \(M\) and females \(F\) in the dataset provided. There are 50 total patients. Count of \(M = 28\) and count of \(F = 22\).

Determine Expected Frequencies

If admissions are random, each gender should represent half of the patient population. For 50 patients, both \(M\) and \(F\) should ideally have an expected frequency of 25.

Conduct a Chi-Square Test

Use the chi-square test formula: \[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \] where \(O_i\) is the observed frequency and \(E_i\) is the expected frequency. For males: \( \frac{(28 - 25)^2}{25} = 0.36 \) and for females: \( \frac{(22 - 25)^2}{25} = 0.36 \). Thus, \( \chi^2 = 0.36 + 0.36 = 0.72 \).

Compare to Critical Value

Find the critical value from the chi-square distribution table for \(\alpha = 0.05\) with \(df = 1\) (one degree of freedom for two categories). The critical value is 3.841. Compare the calculated \( \chi^2 = 0.72 \) to this critical value.

Decision

Since \( 0.72 < 3.841 \), we fail to reject the null hypothesis. This suggests that there is no statistical evidence of a gender-based preference in admissions at the medical center.

Key concepts

Chi-Square Test

The Chi-Square Test is a statistical method for examining the distribution of categorical data. It helps us determine if there is a significant difference between the observed frequencies in a dataset and the frequencies expected under a random distribution. This test is great for checking if there are patterns or preferences in data that should be random. In our exercise, the genders of the patients at a medical center are analyzed to see if admissions are truly random. The test compares the observed counts of male and female patients to what we'd expect if there were no preference for either gender.

• Observed frequencies: The actual counts we recorded, 28 males and 22 females.
• Expected frequencies: The counts we expect when assuming random admission, which are 25 males and 25 females.

We use a formula to calculate the Chi-Square statistic, \[\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \]Where \(O_i\) is the observed frequency, and \(E_i\) is the expected frequency. The result tells us if there's enough evidence to claim that admissions are not random.

Null and Alternative Hypotheses

Understanding the null and alternative hypotheses is essential in any hypothesis testing. The null hypothesis, denoted as \(H_0\), is the assumption that any observed difference from expectations is due to random chance. For our medical center study, the null hypothesis is that patient admissions are random with respect to gender. This means there's no preference for either males or females.
On the other hand, the alternative hypothesis, \(H_1\), is the assertion that there is a meaningful difference that is not attributable to chance. In this context, the alternative hypothesis suggests that there is a preference or a systematic error in how genders are admitted. Formulating these hypotheses sets the stage for statistical testing.
Such formulation helps statisticians and researchers focus on the testing purpose—either to support the null hypothesis or to find sufficient evidence to reject it in favor of the alternative hypothesis.

Significance Level Alpha

The significance level, denoted as \(\alpha\), is a threshold used to determine whether a result is statistically significant. It represents the probability of rejecting the null hypothesis when it is actually true, commonly known as a Type I error. In our problem, the significance level is set at \(\alpha = 0.05\), which means we're willing to accept a 5% chance of incorrectly claiming that admissions are not random.
Choosing this level involves a trade-off: a lower \(\alpha\) can reduce the chance of a Type I error but might increase the possibility of a Type II error, failing to reject a false null hypothesis. Thus, \(\alpha = 0.05\) is a balance that is widely accepted in scientific studies.This significance level is also the critical value threshold in our chi-square test. We compare our calculated chi-square statistic to a critical value, determined by \(\alpha\) and the degrees of freedom, to decide whether to reject the null hypothesis.

Degrees of Freedom

Degrees of freedom refer to the number of independent values or quantities which can be assigned to a statistical distribution. They are crucial for determining the critical value in hypothesis testing, including chi-square tests. In our exercise, we have two categories (male and female), making our degrees of freedom equal to one, calculated as the number of categories minus one \[ df = (n - 1) \].
This calculation helps us find the critical value in the chi-square distribution table corresponding to \(\alpha = 0.05\).
In practice, the degrees of freedom can affect how stringent the test is. With more categories, degrees of freedom increase, and critical values get harsher, requiring more substantial statistical evidence to make a claim. For this exercise with only one degree of freedom, the critical value is 3.841. Thus, the calculated chi-square value of 0.72 does not exceed this threshold, indicating no significant difference from random admission.