Question

NOTE VOTE Out of 300 people sampled, 249 preferred Candidate A. Based on this, estimate what proportion of the voting population (TT) prefers Candidate A. Use a 95% confidence level, and give your answers as decimals, to three places. <t<

Answer 1

Solution-1: Sample size = n = 300 Sample number of people that preferred candidate A = x = 249 Sample proportion of people that preferred candidate A: X 249 P = = = 200 = 0.830 Standard Error of the Point estimate: 0.830 x (1 – 0.830) Op = ✓ n ✓ 300 = 0.0217 For 95% confidence interval, a = 1-0.95 = 0.05 Hence, a/2=0.05/2=0.025 From normal distribution table; 20.025 = 1.96 Margin of error: E = 20.025 X OD E = 1.96 x 0.0217 E = 0.0425 95% confidence interval for proportion of the voting population that prefers candidate A: =p+E = 0.830 + 0.0425 Lower limit of 95% confidence interval = 0.830 - 0.0425 = 0.787 Upper limit of 95% confidence interval = 0.830 + 0.0425 = 0.873 Hence, 0.787 < < 0.873

Solution-2: Sample size = n = 560 Sample proportion = p = 0.23 Standard Error of the Point estimate: PC P) Op = 0.23 x (1 - 0.23 560 = 0.0178 For 90% confidence interval, a = 1-0.90 = 0.10 Hence, a/2=0.10/2=0.05 From normal distribution table; 20.05 = 1.645 Margin of error: E = 20.05 X Op E = 1.645 x 0.0178 E = 0.029