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Chapter 13: Problem 8

The article "Psychiatric and Alcoholic Admissions Do Not Occur Disproportionately Close to Patients' Birthdays"' (Psych. Rep., 1992: 944鈥�946) focuses on the existence of any relationship between date of patient admission for treatment of alcoholism and patient's birthday. Assuming a 365day year (i.e., excluding leap year), in the absence of any relation, a patient's admission date is equally likely to be any one of the 365 possible days. The investigators established four different admission categories: (1) within 7 days of birthday, (2) between 8 and 30 days, inclusive, from the birthday, (3) between 31 and 90 days, inclusive, from the birthday, and (4) more than 90 days from the birthday. A sample of 200 patients gave observed frequencies of \(11,24,69\), and 96 for categories 1,2 , 3 , and 4, respectively. State and test the relevant hypotheses using a significance level of \(.01\).

Short Answer

Expert verified
Reject the null hypothesis; the admission dates are not uniformly distributed, indicating a relationship with birthdays.

Step by step solution

01

State the Hypotheses

To determine if there is a relationship between admission date and birthdays, we formulate the null and alternative hypotheses. The null hypothesis ( H_0 ) is that the admission dates are uniformly distributed across the categories. The alternative hypothesis ( H_1 ) is that the admission dates are not uniformly distributed, implying a relationship between admission dates and birthdays.
02

Determine Expected Frequencies

Since each category should occur with equal probability in the absence of a relationship, we need to calculate the expected frequency for each category. There are 365 total days, so the proportion of days for each category is:- Category 1: 14 days (7 days before and after birthdays)- Category 2: 23 days- Category 3: 60 days- Category 4: 268 days The expected frequencies for each category using a sample of 200 patients are calculated as:- Expected Frequency 1: \( \frac{14}{365} \times 200 = 7.67 \)- Expected Frequency 2: \( \frac{23}{365} \times 200 = 12.60 \)- Expected Frequency 3: \( \frac{60}{365} \times 200 = 32.88 \)- Expected Frequency 4: \( \frac{268}{365} \times 200 = 146.9 \).
03

Compute the Chi-square Test Statistic

The Chi-square test statistic compares the observed frequencies with the expected frequencies. It is calculated as \[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \]for each category:- Category 1: \( \frac{(11-7.67)^2}{7.67} = 1.440 \)- Category 2: \( \frac{(24-12.60)^2}{12.60} = 10.414 \)- Category 3: \( \frac{(69-32.88)^2}{32.88} = 38.146 \)- Category 4: \( \frac{(96-146.9)^2}{146.9} = 17.417 \)Summing these gives the total Chi-square statistic: \( \chi^2 = 67.417 \).
04

Determine the Degrees of Freedom

The degrees of freedom (df) for this test are calculated as the number of categories minus 1. With 4 categories, \( df = 4 - 1 = 3 \).
05

Find the Critical Chi-square Value

Using a Chi-square distribution table and significance level \( \alpha = 0.01 \), we find the critical value for 3 degrees of freedom is approximately 11.345.
06

Compare Test Statistic to Critical Value

Compare the calculated Chi-square statistic (67.417) to the critical value (11.345). Since 67.417 > 11.345, the null hypothesis is rejected at the \(0.01\) significance level.
07

Draw Conclusion

Rejecting the null hypothesis suggests that there is a significant relationship between the admission dates and the patients' birthdays.

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Key Concepts

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

Hypothesis Testing
In hypothesis testing, we aim to evaluate two competing statements about a particular attribute of a population. In this exercise, we're looking at the distribution of admission dates to determine if they correlate with patients' birthdays. The first step in hypothesis testing is to determine the hypotheses.

**Null Hypothesis (H鈧�):** This is the default assumption that there is no effect or no difference. In our scenario, it would suggest that admissions happen randomly and uniformly across the 365-day calendar year, showing no relation to the birthdates.

**Alternative Hypothesis (H鈧�):** This is what researchers typically wish to prove. It asserts that there is a statistically significant effect or difference. For our case, this means patient admissions are not evenly distributed throughout the year but instead show some pattern or relationship to the patients' birthdays.

By testing these hypotheses using data, we determine if there is enough evidence to support the alternative over the null hypothesis.
Expected Frequencies
The concept of expected frequencies is central to carrying out a chi-square test, which helps decide if the observed frequencies in our sample differ from what we would expect if the null hypothesis were true. Calculating expected frequencies involves proportionally splitting our data into categories based on the likelihood of each category.

For our exercise with four categories of days relative to a birthday:
  • Category 1: 14 days total (7 before and 7 after the birthday).
  • Category 2: 23 days total.
  • Category 3: 60 days total.
  • Category 4: 268 days total.
Since the total number of days is 365, we can calculate the expected frequency for each category as a fraction of 200, based on the number of days in each category. This helps us understand what the data should look like if there is no pattern.
Degrees of Freedom
Degrees of freedom refer to the number of values that are free to vary in a statistical analysis, helping determine the distribution to assess significance. For a chi-square test, the degrees of freedom are calculated by taking the number of categories and subtracting one.

In our problem, there are four categories (days relative to the birthday), giving us a degree of freedom equal to 3.
  • Degrees of Freedom = Number of Categories - 1
  • So, Degrees of Freedom = 4 - 1 = 3
This concept is crucial because it affects the critical value against which we compare our chi-square statistic. Knowing the degrees of freedom, we can reference a chi-square distribution table to find our critical value for a given significance level.
Significance Level
The significance level (\(\alpha\)) in statistical hypothesis testing is a threshold that determines when we can reject the null hypothesis. It reflects the probability of incorrectly rejecting a true null hypothesis 鈥� a type I error.

In our exercise, the significance level is set at 0.01, or 1%. This means we accept a 1% chance of making an error by saying there is a relationship between admissions and birthdays when there really isn't one.

Significance level plays a vital role in determining the critical value from the chi-square distribution table. If our test statistic exceeds this critical value, the results are considered statistically significant. In this case, our calculated chi-square statistic was 67.417, which was significantly higher than our critical value at 3 degrees of freedom and a 0.01 significance level, leading to the rejection of the null hypothesis.

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Most popular questions from this chapter

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