Question

If n=230 and X = 184, construct a 99% confidence interval.

1. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

   Confidence interval = answer _____

2. 
   
3. Express the same answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

   answer _____ < p < answer _____

4. 
   
5. Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places.

   p = answer _____ ±± answer ____

Answer 1

Sample proportion = 184 / 230 = 0.8

99% confidence interval for p is

- Z/2 * sqrt [ ( 1 - ) / n ] < p < + Z/2 * sqrt [ ( 1 - ) / n ]

0.8 - 2.5758 * sqrt [ 0.8 * ( 1 - 0.8) / 230] < p < 0.8 + 2.5758 * sqrt [ 0.8 * ( 1 - 0.8) / 230]

0.732 < p < 0.0.868

99% C I is ( 0.732 , 0.868 )

b)

Sample proportion = 184 / 230 = 0.8

99% confidence interval for p is

- Z/2 * sqrt [ ( 1 - ) / n ] < p < + Z/2 * sqrt [ ( 1 - ) / n ]

0.8 - 2.5758 * sqrt [ 0.8 * ( 1 - 0.8) / 230] < p < 0.8 + 2.5758 * sqrt [ 0.8 * ( 1 - 0.8) / 230]

0.732 < p < 0.0.868

c)

E = Z/2 * sqrt [ ( 1 - ) / n ]

= 0.8 2.5758 * sqrt [ 0.8 * ( 1 - 0.8) / 230]

= 0.8 0.068