Answer :
- Calculate the expected counts: $n_1 \hat{p}_c = 50 \times 0.592 = 29.6$, $n_1(1-\hat{p}_c) = 50 \times 0.408 = 20.4$, $n_2 \hat{p}_c = 75 \times 0.592 = 44.4$, $n_2(1-\hat{p}_c) = 75 \times 0.408 = 30.6$.
- Check if all expected counts are greater than or equal to 10.
- Since all counts are greater than or equal to 10, the large counts condition is met.
- Final Answer: $n_1 \hat{p}_c = 29.6$, $n_1(1-\hat{p}_c) = 20.4$, $n_2 \hat{p}_c = 44.4$, $n_2(1-\hat{p}_c) = 30.6$, Yes
### Explanation
1. Understand the problem and provided data
We are given two independent random samples: 50 adults who exercise regularly and 75 adults who do not exercise regularly. We are also given that the pooled proportion is $\hat{p}_{c} = 0.592$. We need to calculate the expected counts $n_1 \hat{p}_c$, $n_1(1-\hat{p}_c)$, $n_2 \hat{p}_c$, and $n_2(1-\hat{p}_c)$ and check if the large counts condition is met.
2. Calculate $n_1 \hat{p}_c$
First, we calculate $n_1 \hat{p}_c = 50 \times 0.592 = 29.6$.
3. Calculate $n_1(1-\hat{p}_c)$
Next, we calculate $n_1(1-\hat{p}_c) = 50 \times (1-0.592) = 50 \times 0.408 = 20.4$.
4. Calculate $n_2 \hat{p}_c$
Then, we calculate $n_2 \hat{p}_c = 75 \times 0.592 = 44.4$.
5. Calculate $n_2(1-\hat{p}_c)$
Finally, we calculate $n_2(1-\hat{p}_c) = 75 \times (1-0.592) = 75 \times 0.408 = 30.6$.
6. Check if the large counts condition is met
The large counts condition is met if all four expected counts are greater than or equal to 10. Since $29.6 \geq 10$, $20.4 \geq 10$, $44.4 \geq 10$, and $30.6 \geq 10$, the large counts condition is met.
7. State the final answer
Therefore, $n_1 \hat{p}_c = 29.6$, $n_1(1-\hat{p}_c) = 20.4$, $n_2 \hat{p}_c = 44.4$, $n_2(1-\hat{p}_c) = 30.6$, and the large counts condition is met.
### Examples
In medical research, checking the large counts condition is crucial when comparing the effectiveness of a treatment (e.g., a new drug) versus a placebo. Researchers need to ensure they have enough data in both groups (treated and placebo) to make reliable conclusions about whether the treatment truly has an effect. If the counts are too small, the results might be misleading due to random chance.
- Check if all expected counts are greater than or equal to 10.
- Since all counts are greater than or equal to 10, the large counts condition is met.
- Final Answer: $n_1 \hat{p}_c = 29.6$, $n_1(1-\hat{p}_c) = 20.4$, $n_2 \hat{p}_c = 44.4$, $n_2(1-\hat{p}_c) = 30.6$, Yes
### Explanation
1. Understand the problem and provided data
We are given two independent random samples: 50 adults who exercise regularly and 75 adults who do not exercise regularly. We are also given that the pooled proportion is $\hat{p}_{c} = 0.592$. We need to calculate the expected counts $n_1 \hat{p}_c$, $n_1(1-\hat{p}_c)$, $n_2 \hat{p}_c$, and $n_2(1-\hat{p}_c)$ and check if the large counts condition is met.
2. Calculate $n_1 \hat{p}_c$
First, we calculate $n_1 \hat{p}_c = 50 \times 0.592 = 29.6$.
3. Calculate $n_1(1-\hat{p}_c)$
Next, we calculate $n_1(1-\hat{p}_c) = 50 \times (1-0.592) = 50 \times 0.408 = 20.4$.
4. Calculate $n_2 \hat{p}_c$
Then, we calculate $n_2 \hat{p}_c = 75 \times 0.592 = 44.4$.
5. Calculate $n_2(1-\hat{p}_c)$
Finally, we calculate $n_2(1-\hat{p}_c) = 75 \times (1-0.592) = 75 \times 0.408 = 30.6$.
6. Check if the large counts condition is met
The large counts condition is met if all four expected counts are greater than or equal to 10. Since $29.6 \geq 10$, $20.4 \geq 10$, $44.4 \geq 10$, and $30.6 \geq 10$, the large counts condition is met.
7. State the final answer
Therefore, $n_1 \hat{p}_c = 29.6$, $n_1(1-\hat{p}_c) = 20.4$, $n_2 \hat{p}_c = 44.4$, $n_2(1-\hat{p}_c) = 30.6$, and the large counts condition is met.
### Examples
In medical research, checking the large counts condition is crucial when comparing the effectiveness of a treatment (e.g., a new drug) versus a placebo. Researchers need to ensure they have enough data in both groups (treated and placebo) to make reliable conclusions about whether the treatment truly has an effect. If the counts are too small, the results might be misleading due to random chance.
