Question

A poll was taken of 588 residents in a county. The residents sampled were asked whether they think their local government did a good job overall. 490 responded ""yes"". Let p denote the proportion of all residents in that county who think their local government did a good job. Construct a 95% confidence interval for p: Round off to two decimal places.

Answer 1

Answer:

[472.29, 507.71]

Step-by-step explanation:

The proportion is  

490/588 = 0.8333

The mean can be estimated as  

$\bf \bar x$ = 490  

and the standard deviation can be approximated with

$\bf s = \sqrt{588*0.8333*(1-0.8333)}$ = 9.037

and we just need to find two values a and b such that the area under the Normal curve of mean 490 and standard deviation 9.037 between a and b equals 95% = 0.95

(See picture)

With the help of a spreadsheet we can easily find these values which are

a = 472.288

b = 507.712

and the interval we are looking for, rounded to 2 decimals, is [472.29, 507.71]