High School

Counts Condition for a Two-Proportion Test

Next, let's prepare to check the large counts condition. To do so, identify the following values:

1. [tex]\hat{p}_C = \frac{\text{out of } 100}{100} = 0.58[/tex]
2. [tex]\hat{p}_A = \frac{\text{out of } 100}{100} = 0.18[/tex]

Now determine:

- [tex]n_1 = \square[/tex]
- [tex]n_2 = \square[/tex]
- Combined proportion: [tex]\hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \square[/tex]

Answer :

We are given two groups with the following information:

- Group C has an observed proportion of successes
[tex]$$\hat{p}_C = 0.58$$[/tex]
out of 100 individuals.
- Group A has an observed proportion of successes
[tex]$$\hat{p}_A = 0.18$$[/tex]
out of 100 individuals.

We first set the sample sizes for each group:

[tex]$$
n_1 = 100 \quad \text{(for Group C)}, \qquad n_2 = 100 \quad \text{(for Group A)}
$$[/tex]

Next, we compute the number of successes in each group. For Group C, the number of successes is calculated by multiplying the proportion by the total number in the group:

[tex]$$
x_1 = n_1 \times \hat{p}_C = 100 \times 0.58 = 58.
$$[/tex]

Similarly, for Group A, the number of successes is:

[tex]$$
x_2 = n_2 \times \hat{p}_A = 100 \times 0.18 = 18.
$$[/tex]

To check the large counts condition, we combine the results from both groups. The overall pooled proportion, which is the total number of successes in both groups divided by the total number of observations, is given by:

[tex]$$
\hat{p}_c = \frac{x_1 + x_2}{n_1 + n_2} = \frac{58 + 18}{100 + 100} = \frac{76}{200} = 0.38.
$$[/tex]

Thus, the following values have been determined:

- Proportion for Group C: [tex]$\hat{p}_C = 0.58$[/tex]
- Proportion for Group A: [tex]$\hat{p}_A = 0.18$[/tex]
- Sample sizes: [tex]$n_1 = 100$[/tex], [tex]$n_2 = 100$[/tex]
- Combined proportion:
[tex]$$\hat{p}_c = 0.38.$$[/tex]

These values satisfy the counts condition for performing a two-proportion analysis.