Question

a study involves 612 randomly selected deaths, with 29 caused by accidents. Construct a 98% confidence interval for the percentage of all deaths that are caused by accidents.

Answer 1

Solution :

Given that,

n = 612

x = 29

Point estimate = sample proportion = = x / n = 29/612=0.047

1 - = 1-0.047=0.953

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table    )

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 2.326 (((0.047*0.953) / 612)

E= 0.0199

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.047 -0.0199 < p <  0.047+0.0199

0.0271< p < 0.0669

The 98% confidence interval for the population proportion p is : 0.0271, 0.0669