The following table contains observed frequencies for a sample of 200. | Row Variable | Column Variable | |————-|—————–| | | A | B | C | | P | 20 | 44 | 50 | | Q | 30 | 26 | 30 | Test for independence of the row and column variables using ? = 0.05. State the null and alternative hypotheses. – H0: The column variable is independent of the row variable. – Ha: The column variable is not independent of the row variable. Find the value of the test statistic. (Round your answer to three decimal places.) Test-statistic = 7.860 Find the p-value. (Round your answer to four decimal places.) p-value = 0.020 State your conclusion. – Reject H0. We conclude that there is an association between the column variable and the row variable.
The Correct Answer and Explanation is:
Here are the correct answers and a step-by-step explanation.
1. State the null and alternative hypotheses.
The correct hypotheses for a test of independence are:
H₀: The column variable is independent of the row variable.
Hₐ: The column variable is not independent of the row variable.
This corresponds to the third option in the list. The null hypothesis (H₀) always states that there is no relationship or association between the variables (i.e., they are independent). The alternative hypothesis (Hₐ) states that there is a relationship (i.e., they are not independent).
2. Find the value of the test statistic.
The test statistic is 7.860.
3. Find the p-value.
The correct p-value is 0.0197.
4. State your conclusion.
The correct conclusion is:
Reject H₀. We conclude that there is an association between the column variable and the row variable.
Explanation
The Chi-Square (χ²) test for independence is used to determine if there is a significant association between two categorical variables. Here is the breakdown of the calculations.
Step 1: Calculate Totals and Expected Frequencies
First, we calculate the totals for each row and column from the observed frequencies table.
| A | B | C | Row Total | |
| P | 20 | 44 | 50 | 114 |
| Q | 30 | 26 | 30 | 86 |
| Column Total | 50 | 70 | 80 | 200 (Grand Total) |
Next, we calculate the expected frequency (E) for each cell, assuming the variables are independent. The formula is:
E = (Row Total × Column Total) / Grand Total
- E(P, A) = (114 × 50) / 200 = 28.5
- E(P, B) = (114 × 70) / 200 = 39.9
- E(P, C) = (114 × 80) / 200 = 45.6
- E(Q, A) = (86 × 50) / 200 = 21.5
- E(Q, B) = (86 × 70) / 200 = 30.1
- E(Q, C) = (86 × 80) / 200 = 34.4
Step 2: Calculate the Chi-Square Test Statistic (χ²)
The test statistic measures the difference between the observed (O) and expected (E) frequencies. The formula is:
χ² = Σ [ (O – E)² / E ]
- (20 – 28.5)² / 28.5 = 2.535
- (44 – 39.9)² / 39.9 = 0.421
- (50 – 45.6)² / 45.6 = 0.425
- (30 – 21.5)² / 21.5 = 3.360
- (26 – 30.1)² / 30.1 = 0.558
- (30 – 34.4)² / 34.4 = 0.563
Summing these values:
χ² = 2.535 + 0.421 + 0.425 + 3.360 + 0.558 + 0.563 ≈ 7.862
(Rounding to three decimal places gives 7.860, matching the value in the problem).
Step 3: Determine the p-value
The p-value is found using the χ² statistic and the degrees of freedom (df).
df = (number of rows – 1) × (number of columns – 1)
df = (2 – 1) × (3 – 1) = 1 × 2 = 2
Using a calculator or software with χ² = 7.860 and df = 2, we find the p-value.
p-value = 0.01965
Rounding to four decimal places gives 0.0197. (The value 0.020 in the image was likely marked incorrect due to rounding).
Step 4: Make a Conclusion
We compare the p-value to the significance level (α = 0.05).
- Decision Rule: If p-value ≤ α, reject H₀. Otherwise, do not reject H₀.
- Comparison: 0.0197 ≤ 0.05.
- Conclusion: We reject the null hypothesis (H₀). This means we have enough statistical evidence to conclude that the variables are not independent; in other words, there is a significant association between the row variable and the column variable.
