Use the normal distribution to find a confidence interval for a proportion P giv

Question

Use the normal distribution to find a confidence interval for a proportion P given the relevant sample results. Give the best point estimate for p, the margin of error, and the confidence interval. Assume the results come from a random sample. A 90% confidence interval for p given that p = 0.33 and n = 450 Round your answer for the point estimate to two decimal places, and your answers for the margin of error and the confidence interval to three decimal places. point estimate = margin of error = plusminus The 90% confidence interval is to

Explanation / Answer

point estimate = 0.33

for 90% confidence interval, z=1.64
Margin of error = z*sqrt(p*(1-p)/n)
Margin of error = 1.64*sqrt(0.33*(1-0.33)/450)
Margin of error = 0.0364

The 90% Confidence interval (pcap - ME, pcap + ME) = (0.33 - 0.0364, 0.33 + 0.0364) = (0.2936, 0.3664)

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.