Question

"Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places. ​98% confidence; the sample size is​ 800, of which​ 40% are successes"

Answer 1

Answer:

The error E = ± 4.04 %

Step-by-step explanation:

Solution:-

- The sample data is used to estimate the population proportion ( p ).

- The success p^ = success percentage = 40 %

- The confidence interval CI = 98%

- The sample size n = 800

- The margin of error E:

- The margin of error "E" for estimation of population proportion ( p ) is given by:

                          $E = z-critical*\sqrt{\frac{p~ ( 1 - p~ )}{n} }$

Where, Z-critical value is defined by the significance level:

                         P ( Z < Z-critical ) = α / 2

Where, α : Significance level

                          α = 1 - CI

                         P ( Z < Z-critical ) = (1 - 0.98) / 2

                         P ( Z < Z-critical ) = 0.01  

                         Z-critical = 2.33

- The error E of estimation is:

                        $E = 2.33*\sqrt{\frac{0.4 ( 1 - 0.4 )}{800} }\\\\E = 0.04035$

- The error E = ± 4.04 %