Question

For the given confidence level and values of x and n, find the following.x=48, n=97, confidence level 99%

The point estimate for the given data is .4948.

The standard error for the given data is .0502.

(c) Find the margin of error. Round the answers to at least four decimal places, if necessary.

Answer 1

Given that,

Point estimate = sample proportion = = x / n = 0.4948

1 -   = 1- 0.4948 =0.5052

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

  Margin of error = E = Z / 2    * (((( * (1 - )) / n)

= 2.576* ((( 0.4948*0.5052) /97)

E = 0.1333