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.
[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.
