Question

Rasmussen reported that 1000 taxpayers were surveyed recently and 610 said they believe that tax cuts help the economy. What is the 99% confidence interval for the true proportion of voters saying that tax cuts will help the economy?

Answer 1

Solution :

Given that,

n = 1000

x = 610

Point estimate = sample proportion = = x / n = 610/1000=0.61

1 -   = 1- 0.61 =0.39

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* (((0.61*0.39) /1000 )

E = 0.0489

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.61-0.0489 < p <  0.61+0.0489

0.5611< p < 0.6589

The 99% confidence interval for the population proportion p is : 0.5611, 0.6589