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Answer :
We are given two independent samples with sample sizes and sample proportions as follows:
[tex]\[
n_1 = 325,\quad \hat{p}_1 = 0.71; \qquad n_2 = 455,\quad \hat{p}_2 = 0.64.
\][/tex]
For the large-counts condition, we need to check that the counts of successes and failures are at least 10. This requires calculating:
1. [tex]$$ n_1 \hat{p}_1, $$[/tex]
2. [tex]$$ n_1 (1 - \hat{p}_1), $$[/tex]
3. [tex]$$ n_2 \hat{p}_2, $$[/tex]
4. [tex]$$ n_2 (1 - \hat{p}_2). $$[/tex]
Step 1: Calculation for the first sample ([tex]$n_1 = 325$[/tex], [tex]$\hat{p}_1 = 0.71$[/tex])
- The number of successes is:
[tex]$$
n_1 \hat{p}_1 = 325 \times 0.71 = 230.75.
$$[/tex]
- The number of failures is:
[tex]$$
n_1 (1 - \hat{p}_1) = 325 \times (1 - 0.71) = 325 \times 0.29 = 94.25.
$$[/tex]
Step 2: Calculation for the second sample ([tex]$n_2 = 455$[/tex], [tex]$\hat{p}_2 = 0.64$[/tex])
- The number of successes is:
[tex]$$
n_2 \hat{p}_2 = 455 \times 0.64 = 291.2.
$$[/tex]
- The number of failures is:
[tex]$$
n_2 (1 - \hat{p}_2) = 455 \times (1 - 0.64) = 455 \times 0.36 = 163.8.
$$[/tex]
Conclusion:
The four calculated quantities for the large-counts condition are:
[tex]\[
\begin{aligned}
n_1 \hat{p}_1 & = 230.75, \\
n_1 (1-\hat{p}_1) & = 94.25, \\
n_2 \hat{p}_2 & = 291.2, \\
n_2 (1-\hat{p}_2) & = 163.8.
\end{aligned}
\][/tex]
Since all these values are greater than 10, the large-counts condition is satisfied.
[tex]\[
n_1 = 325,\quad \hat{p}_1 = 0.71; \qquad n_2 = 455,\quad \hat{p}_2 = 0.64.
\][/tex]
For the large-counts condition, we need to check that the counts of successes and failures are at least 10. This requires calculating:
1. [tex]$$ n_1 \hat{p}_1, $$[/tex]
2. [tex]$$ n_1 (1 - \hat{p}_1), $$[/tex]
3. [tex]$$ n_2 \hat{p}_2, $$[/tex]
4. [tex]$$ n_2 (1 - \hat{p}_2). $$[/tex]
Step 1: Calculation for the first sample ([tex]$n_1 = 325$[/tex], [tex]$\hat{p}_1 = 0.71$[/tex])
- The number of successes is:
[tex]$$
n_1 \hat{p}_1 = 325 \times 0.71 = 230.75.
$$[/tex]
- The number of failures is:
[tex]$$
n_1 (1 - \hat{p}_1) = 325 \times (1 - 0.71) = 325 \times 0.29 = 94.25.
$$[/tex]
Step 2: Calculation for the second sample ([tex]$n_2 = 455$[/tex], [tex]$\hat{p}_2 = 0.64$[/tex])
- The number of successes is:
[tex]$$
n_2 \hat{p}_2 = 455 \times 0.64 = 291.2.
$$[/tex]
- The number of failures is:
[tex]$$
n_2 (1 - \hat{p}_2) = 455 \times (1 - 0.64) = 455 \times 0.36 = 163.8.
$$[/tex]
Conclusion:
The four calculated quantities for the large-counts condition are:
[tex]\[
\begin{aligned}
n_1 \hat{p}_1 & = 230.75, \\
n_1 (1-\hat{p}_1) & = 94.25, \\
n_2 \hat{p}_2 & = 291.2, \\
n_2 (1-\hat{p}_2) & = 163.8.
\end{aligned}
\][/tex]
Since all these values are greater than 10, the large-counts condition is satisfied.
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