Question

Rasmussen reported that 1000 taxpayers were surveyed recently and 630 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?

Select one:

(0.6001, 0.6599)

(0.5907, 0.6693)

(0.6049, 0.6551)

(0.6294, 0.6306)

none of these are correct

Answer 1

Solution :

Given that,

n = 1000

x = 630

Point estimate = sample proportion = = x / n = 630/1000=0.63

1 -   = 1-0.63 =0.37

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* (((0.63*0.37) / 1000)

E = 0.0393

A 99% confidence interval for proportion p is ,

- E < p < + E

0.63 - 0.0393< p < 0.63+ 0.0393

0.5907< p < 0.6693

(0.5907, 0.6693)

The 99% confidence interval for the proportion p is : (0.5907, 0.6693)