In June 2012, a CBS/New York Times Poll asked 906 randomly selected Americans wh

Question

In June 2012, a CBS/New York Times Poll asked 906 randomly selected Americans whether they approve of the job the US Supreme Court is doing. 451 approved and 455 disapproved. We wish to estimate the proportion p of Americans that approve of the job the Supreme Court is doing. Find the sample proportion p that serves as a point estimate for p. Give your answer to two decimal places. Find the Standard Error of p Give your answer to four decimal places. Standard Error of p =.0166 What is the Margin of Error for the 95% confidence interval? Give your answer to three decimal places. Find the 95% confidence interval for p. Give the endpoints to three decimal places. Find a 99% confidence interval for p. Give the endpoints to three decimal places.

Explanation / Answer

p = 451 / 906= 0.50

Serror = srqt ( 0.50*0.50 / 906 ) = 0.0166

alpha / 2 = 0.025 Z= 1.96

MOE 1.96 * 0.0166 = 0.033

confidence interval

0.5 - 0.033 < p < 0.5 +0.033

0.465 < p < 0.530

99% confidence

alpha / 2 = 0.005 Z=2.57

MOE = 2.57 * 0.0166 = 0.042662

0.5 - 0.042662 < p < 0.5 + 0.042662

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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