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Zorluk: OrtaStatistical Inference and Hypothesis Testing

A team of epidemiologists is investigating a potential association between the regular consumption of a novel energy drink and the incidence of new-onset atrial fibrillation in young adults. A cross-sectional survey is administered to a random sample of 1,0001,000 university students. The survey reveals that 200200 students consume the energy drink regularly. Additionally, clinical records indicate that 5050 students in the total sample have experienced at least one episode of atrial fibrillation.

To determine if there is a statistically significant association between energy drink consumption and atrial fibrillation, the researchers plan to use a Chi-square test of independence. Assuming the null hypothesis is true, what is the expected number of students who both consume the energy drink regularly and have experienced atrial fibrillation? (Enter your answer as a whole number).

Cevap: 10 students

Cevap

10
The Chi-square test of independence evaluates whether two categorical variables are associated. The null hypothesis states that the two variables (energy drink consumption and atrial fibrillation) are completely independent. If they are independent, the probability of a student having both characteristics is simply the product of their individual probabilities. The probability of consuming the drink is 200/1,000=0.2200/1,000 = 0.2. The probability of having atrial fibrillation is 50/1,000=0.0550/1,000 = 0.05. Therefore, the expected joint probability under the null hypothesis is 0.2×0.05=0.010.2 \times 0.05 = 0.01. To find the expected count, this probability is multiplied by the total sample size: 0.01×1,000=100.01 \times 1,000 = 10. Alternatively, this is calculated directly using the standard formula: Expected count = Row Total×Column TotalGrand Total=200×501,000=10\frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}} = \frac{200 \times 50}{1,000} = 10.

Adım Adım Çözüm

1
Identify the total sample size.
The total sample size (NN) is 1,0001,000.
The total number of observations is required for the denominator in the expected frequency formula.
2
Identify the marginal totals for both variables of interest.
The total number of energy drink consumers (row total) is 200200. The total number of students with atrial fibrillation (column total) is 5050.
Marginal totals represent the overall frequency of each independent characteristic.
3
Apply the expected frequency formula for a Chi-square test.
Expected frequency = 200×501,000\frac{200 \times 50}{1,000}.
Under the null hypothesis of independence, the expected frequency of a cell is the product of its marginal totals divided by the grand total.
4
Calculate the final expected value.
10,0001,000=10\frac{10,000}{1,000} = 10.
This yields the expected number of students who would have both characteristics if the variables were completely independent.

Anahtar Kavram

Calculating expected frequencies for a Chi-square test of independence.
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Statistical Inference and Hypothesis Testing

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