Assume that a sample is used to estimate a population proportion p. Find the 98% confidence interval for a sample of size 322 with 270 successes. Enter your answer using decimals (not percents) accurate to three decimal places.

Answer with explanation:Sample Size=322Success=270Probability of Success(p)              [tex]=\frac{270}{322}\\\\=\frac{135}{161}[/tex]At 98% confidence interval , population proportion will be [tex]=z_{98 \text{Percent}}\times \sqrt{\frac{p(1-p)}{n}}\\\\=0.8365 \times\sqrt{\frac{\frac{270}{322}(1-\frac{270}{322})}{322}}\\\\=0.8365 \times\sqrt{\frac{\frac{270}{322}\times(\frac{52}{322})}{322}}\\\\=\frac{0.8365}{322} \times \sqrt{270 \times 52}\\\\=0.00259782 \times \sqrt{14040}\\\\=0.00259782 \times 118.490\\\\=0.307817[/tex]          = 0.308( approx to three decimal places.)