Question

(1 point) (For the following question, use the methods described in lecture - do not use R prop.test(). You must use at least four decimal places for all intermediate calculations and your final answer) A poll is taken in which 316 out of 525 randomly selected voters indicated their preference for a certain candidate. (a) Find the "Large-Sample" 90% confidence interval for p. sps (b) Find the margin of error for this 90% confidence interval for p. (c) Without doing any calculations, indicate whether the margin of error is larger or smaller or the same for an 80% confidence interval. A. larger B. smaller C. same

Answer 1

a. $p=\frac{316}{525}=0.602$

z value for 90% CI is 1.645 as P(-1.645<z<1.645)=0.90

So Margin of Error is

pg E= * 1.645 * 0.602 * 0.398 525 = 0.021 n

Hence CI is $p \pm E=0.602\pm 0.021=(0.581,0.623 )$

b. Margin of Error is

pg E= * 1.645 * 0.602 * 0.398 525 = 0.021 n

c. If we will use 80% CI, z value will decrease and so the margin of error will be smaller

Hence answer here is smaller