Question

In a survey of 2844 adults , 1435 says they have started paying bills online in the last year.

Construct a 99% confidence interval for the population proportion. Interpret the results.

Answer 1

Solution :

Given that,

n = 2844

x = 1435

Point estimate = sample proportion = = x / n = 1435/2844=0.505

1 -   = 1-0.505 =0.495

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* (((0.505*0.495) /2844 )

E = 0.024

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.505-0.024 < p < 0.505+0.024

0.481< p < 0.529