To obtain an estimate of the proportion , p . of New YorkCity residents who feel

Question

To obtain an estimate of the proportion , p. of New YorkCity residents who feel that the quality of life in NYC has becomeworse in the past few years, a telephone poll by Time/CNN in 1990revealed that 686 out of 1009 NYC residents said that the qualityof life in NYC had become worse. a) Give a 90% confidence level for p based on thepoll mentioned above. b) To update the above estimate, how large a sample isrequired to be 98% confident that the maximum error of the estimateof p is 2.5%?
Please show all work-thanks
a) Give a 90% confidence level for p based on thepoll mentioned above. b) To update the above estimate, how large a sample isrequired to be 98% confident that the maximum error of the estimateof p is 2.5%?
Please show all work-thanks

Explanation / Answer

p=686/1009=0.68 (a) =0.1 , Z(0.05)=1.645 (check normal table) The 90% CI is p±Z*p*(1-p)/n --> 0.68±1.645*sqrt(0.68*(1-0.68)/1009) --> (0.6558,0.7042) (b)=0.02, Z(0.01)=2.326 E=0.025 n=(Z/E)^2*p*(1-p) = (2.326/0.025)^2*0.68*(1-0.68) = 1883.642 Take n=1884

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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