Question

(1 point) A poll is taken in which 350 out of 600 randomly selected voters indicated their preference for a certain candidate.

(a) Find a 90% confidence interval for p.

I keep getting 58.2 and 58.4 which is wrong

(b) Find the margin of error for this 90% confidence interval for p.

Answer 1

Solution :

Given that,

n = 600

x = 350

Point estimate = sample proportion = = x / n = 350/600=0.583

1 -   = 1- 0.583 =0.417

At 90% confidence level the z is

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

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

= 1.645 *((0.583*0.417) /600 )

E = 0.0331

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.583- 0.0331< p < 0.583+0.0331

0.5499< p < 0.6161

The 90% confidence interval for the population proportion p is : 0.5499, 0.6161