Question

1. In a sample, we have n=140 and p=0.80. Using the normal distribution to find 99% confidence interval for the population proportion p. Round all answers to 3 decimal places.

1. What is the standard error?

2. What is the margin of error?

3. Construct a 99% confidence interval for the population proportion p.

Answer 1

Solution :

Given that,

n = 140

Point estimate = sample proportion = = 0.80

1 -   = 1- 0.80 =20

standard error = (((( * (1 - )) / n) =(((0.80*0.20) /140 )=0.034

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* (((0.80*0.20) /140 )

E = 0.087

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.80-0.087 < p < 0.80+0.087

0.713< p < 0.887