Question

In one year, there were 17,737 fatal injuries in California, and 11,149 of them were unintentional. Using the data from this year, construct a 98% confidence interval estimate of the percentage of California fatal injuries that are unintentional. Interpret the interval as well.

Answer 1

Solution :

Given that,

n = 17737

x = 11149

= x / n = 11149 /17737 = 0.629

1 - = 1 - 0.629 = 0.371

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z_/2 = Z_0.01 = 2.326

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

= 2.326 * (((0.629 * 0.371) / 17737)

= 0.008

A 98 % confidence interval for population proportion p is ,

- E < P < + E

0.629 - 0.008< p < 0.629 + 0.008

0.621 < p < 0.637

The 98% confidence interval for the population proportion p is : ( 0.621 , 0.637)