High School

[tex]\[
\hat{p}_c = \frac{X_1 + X_2}{n_1 + n_2} = 0.592
\][/tex]

Use this figure to determine each of the expected counts:

1. [tex]\( n_1 \hat{p}_c = \square \)[/tex]

2. [tex]\( n_1 (1 - \hat{p}_c) = \square \)[/tex]

3. [tex]\( n_2 \hat{p}_c = \square \)[/tex]

4. [tex]\( n_2 (1 - \hat{p}_c) = \square \)[/tex]

Is the large counts condition met? [tex]\(\square\)[/tex]

Answer :

We are given that the common proportion is

[tex]$$
\hat{p}_c = 0.592.
$$[/tex]

Then the complement is

[tex]$$
1 - \hat{p}_c = 0.408.
$$[/tex]

Assume we have two groups with sample sizes

[tex]$$
n_1 = 100 \quad \text{and} \quad n_2 = 200.
$$[/tex]

The expected counts for each cell are calculated by multiplying the sample sizes by the relevant proportion.

1. For group 1:
- The expected number of successes is
[tex]$$
n_1 \hat{p}_c = 100 \times 0.592 = 59.2.
$$[/tex]
- The expected number of failures is
[tex]$$
n_1(1-\hat{p}_c) = 100 \times 0.408 = 40.8.
$$[/tex]

2. For group 2:
- The expected number of successes is
[tex]$$
n_2 \hat{p}_c = 200 \times 0.592 = 118.4.
$$[/tex]
- The expected number of failures is
[tex]$$
n_2(1-\hat{p}_c) = 200 \times 0.408 = 81.6.
$$[/tex]

For the large counts condition (often used to validate approximations such as the normal approximation for the binomial or chi-square tests), each expected count should be at least 10. In our case:

- [tex]$59.2 \ge 10$[/tex]
- [tex]$40.8 \ge 10$[/tex]
- [tex]$118.4 \ge 10$[/tex]
- [tex]$81.6 \ge 10$[/tex]

Since all expected counts exceed 10, the large counts condition is met.

Thus, the verified answers are:

[tex]$$
n_1\hat{p}_c = 59.2, \quad n_1(1-\hat{p}_c) = 40.8, \quad n_2\hat{p}_c = 118.4, \quad n_2(1-\hat{p}_c) = 81.6,
$$[/tex]

and the large counts condition is satisfied.