Question

(1 point) A random sample of 1700 voters in a particular city found 544 voters who voted yes on proposition 200. Find a 95% confidence interval for the true percent of voters in this city who voted yes on proposition 200. Express your results to the nearest hundredth of a percent. Your answer should be expressed as a percentage in the form XX.XX Answer:

Answer 1

Let denotes the sample proportion and n denotes the sample size. Then 100 (1-0)% confidence interval for population proportion is CB-E, þ + E), where E margin of Errore za JB (16) n . 1700 for a :0.05 , 100(1-0)% = 95%. whence we get 95% confidence interval. Q:0.05 -> Q12 = .025 .2012 = Z0025 = 1.960 (obtained from a table) Given that n= 1700 Ø : 544 = 0:32 - 1-P = 1-0.32 0.68 E = 1.960x) 0.32 x068 = 0.022175 1700 i lower limit - ÅE=0.32-0.022175= 0.297825 = 29.7825 y. 29.79 y. Ubber limit pt = 0.32 +0.022175 -0.342175 = 34.2175 y. 34.22 y. : 95% confidence interval is ([29.78] % to [34:22 %)