Question

In: Statistics and Probability

In a survey of 2901 adults, 1495 say they have started paying bills online in the...

In a survey of 2901 adults, 1495 say they have started paying bills online in the last year.

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

A 99% confidence interval for the population proportion is

Expert Solution

Solution :

Given that,

n = 2901

x = 1495

Point estimate = sample proportion = = x / n = 1495/2901=0.515

1 - = 1-0.515=0.485

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 * $\sqrt$ (( * (1 - )) / n)

= 2.576 ( $\sqrt$ ((0.515*0.485) / 2901)

= 0.024

A 99% confidence interval is ,

- E < p < + E

0.515-0.024 < p < 0.515+0.024

(0.491 , 0.539)

orchestra answered 1 year ago

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