Answer:
The answer is "[tex]0.02235 \ and \ 0.02501[/tex]".
Step-by-step explanation:
Given:
[tex]x= 1184\\\\n = 50000\\\\c = 95\% = 0.95[/tex]
The sample proportion is computed by subtracting that sample size by both the successes:
[tex]\hat{p}=\frac{x}{n} =\frac{1184}{50000}= 0.02368[/tex]
Determine [tex]z_{\frac{\alpha}{2}} = Z_{0.025}[/tex] using table Z in Table Z for trust level [tex]1 -\alpha = 0.95[/tex]
add using appendix F ((0.025) inside the table, therefore the z-score is the found z-score with negative sign):
So, the boundaries of the confidence interval:
[tex]\to \hat{p}-z_{\frac{\alpha}{2}} \cdot \sqrt{\frac{\hat{p} (1-\hat{p})}{n}} = 0.02368 - 1.96 \cdot \sqrt{\frac{0.02368(1-0.02368)}{50000}}\approx 0.02235 \\\\\to \hat{p}-z_{\frac{\alpha}{2}} \cdot \sqrt{\frac{\hat{p} (1-\hat{p})}{n}} = 0.02368 + 1.96 \cdot \sqrt{\frac{0.02368(1-0.02368)}{50000}}\approx 0.02501\\\\[/tex]
