Question

4. During April 5-12, 2010, A CBS News/New York Times poll surveyed a sample of 1,580 adult Americans. 40% of those surveyed believed current gun control laws should be made more strict. a. Construct a 95% confidence interval for the proportion of all adult Americans who believe gun control laws should be made more strict. State the width and the margin of error for this confidence interval. [6]

Answer 1

1580 14 Here we have: ma.1580, P = 0.40 ☺ sf gs y confidence level L=1-0.95 -0.05 Zorit= 20.05/2=41.96 margin it enor( 6) = zerit 18.11-P) = + 1.36 V-4c1-a 4s) 717 0.0242 width=2€ =240.0242 = 10.0484] Now 95% cat ê ŽE = 0.40 +0.0242 = 10.3758, 0.49427 12 Ad 90% confidence level a )-0.90 - 0.10 Zerit: 20.10/2 = 10 645 New.margin of error [€ ) = zouty = 10695 Joie x 0.6 -10-0203) o At go% confidence level the margin of error is smaller because of smaller critical value at goy confidence level as compared to 95% confidence level ral at n=3000 Af 95% confidence level Margin of error(e) – 1096 X 10.6x004 3000

I 10.01751 width = 2XE - 2 x 0.0175 - 1000350 New 95% cat 0.40 0.0175 0.3825, 0.4175 10 After increasing the sample sire the margin of error decreased the width of the interval decreased The new contidence interval is negrower - Two ways to decrease the width of the contidence interval: o- decrease the confidence level - increase the sample size. o when width of the confidence interval is decreased, it means there is no smaller margin of error and hence we will have more accuracy. -