Answer:
Explanation:
| Male | Female | Total
------|------|--------|-------
Overweight | 15 | 36 | 51
Not Overweight | 85 | 64 | 149
------|------|--------|-------
Total | 100 | 100 | 200
Now, we can test the hypothesis that the two genders are homogeneous with respect to being overweight using a chi-square test. The null hypothesis is that there is no association between gender and overweight status.
The formula for the chi-square test is:
χ² = Σ [(O - E)² / E]
Where:
χ² = chi-square value
O = observed frequency
E = expected frequency
To calculate the expected frequency, we can use the formula:
E = (row total x column total) / grand total
Using this formula, we can calculate the expected frequencies for each cell of the contingency table:
| Male | Female | Total
------|------|--------|-------
Overweight | 12.75 | 38.25 | 51
Not Overweight | 87.25 | 61.75 | 149
------|------|--------|-------
Total | 100 | 100 | 200
Now we can use the formula for the chi-square test to calculate the chi-square value:
χ² = [(15 - 12.75)² / 12.75] + [(36 - 38.25)² / 38.25] + [(85 - 87.25)² / 87.25] + [(64 - 61.75)² / 61.75]
χ² = 1.06 + 0.76 + 0.68 + 0.56
χ² = 3.06
Next, we need to determine the degrees of freedom (df) for our test. For a contingency table with r rows and c columns, the degrees of freedom are (r-1) x (c-1). In this case, we have 2 rows and 2 columns, so our degrees of freedom are:
df = (2-1) x (2-1) = 1
We can then use a chi-square distribution table or a statistical software program to determine the critical value of chi-square for our chosen level of significance and degrees of freedom. For a significance level of 0.05 and 1 degree of freedom, the critical value of chi-square is 3.84.
Since our calculated chi-square value of 3.06 is less than the critical value of 3.84, we fail to reject the null hypothesis. This means that there is not enough evidence to conclude that there is a significant association between gender and overweight status among diabetic patients.