Question

In North Carolina, they survey 400 residents and asked if they like crab cakes, 343 responded yes to the question. Find a 98% confidence interval estimate of the proportion of North Carolina residents who like crab cakes.

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 33 / 400 = 0.8575

Z_/2 = 2.326

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

= 2.326 * (((0.8575 * 0.1425) / 400)

= 0.041

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.8575 - 0.041 < p < 0.8575 + 0.041

0.817 < p < 0.898

(0.817 , 0.898)