In the following table,likely voters' preferences of two candidates are cross-classified by gender. For the chi-square test of independence,the value of the test statistic is _______.
A) 2.34
B) 1.62
C) 3.25
D) 4

Question

In the following table,likely voters' preferences of two candidates are cross-classified by gender.  For the chi-square test of independence,the value of the test statistic is _______.
A) 2.34 
B) 1.62 
C) 3.25 
D) 4

Quizplus · Accepted Answer

To find the chi-square test statistic, we need to calculate the expected frequencies for each cell. To do this, we first calculate the row and column totals:

- Row 1 total: 216 + 130 = 346 
- Row 2 total: 197 + 174 = 371 
- Column 1 total: 216 + 197 = 413 
- Column 2 total: 130 + 174 = 304

Next, we calculate the expected frequency for each cell using the formula: 
(expected frequency) = (row total * column total) / (sample size)

For example, the expected frequency for the first cell is: 
(expected frequency) = (346 * 413) / 917 = 155.34

We repeat this calculation for each cell and get the following table of expected frequencies:

| Male     | Female   | Total
----------|----------|----------|----------
Candidate A| 155.34   | 114.66   | 270
Candidate B| 128.66   | 94.34    | 223
----------|----------|----------|----------
Total     | 284      | 209      | 493

Now we can calculate the chi-square test statistic using the formula:

X^2 = Σ ( (O-E)^2 / E )

where O is the observed frequency and E is the expected frequency. We sum this expression over all cells.

For example, in the first cell: 
(O-E)^2 / E = (216-155.34)^2 / 155.34 = 9.66

We repeat this calculation for each cell and get:

| Male     | Female   | Total
----------|----------|----------|----------
Candidate A| 9.66     | 2.26     | 11.92
Candidate B| 11.74    | 2.74     | 14.48
----------|----------|----------|----------
Total     | 21.4     | 5        | 26.4

Finally, we sum up all of these values to get the chi-square test statistic: 
X^2 = 11.92 + 14.48 + 9.66 + 2.26 = 38.32

Using a significance level of 0.05 and 1 degree of freedom (df = (2-1)*(2-1) = 1), we can compare this test statistic to the critical value from the chi-square distribution table. We find that the critical value is 3.84. Since our test statistic (38.32) is greater than the critical value, we reject the null hypothesis of independence and conclude that there is a significant difference between the preferences of male and female likely voters. Therefore, the best choice based on the given options is C.

In the Following Table,likely Voters' Preferences of Two Candidates Are