Question

Given a random sample of size 322. Find a 99% confidence interval for the population proportion if the number of successes was 168.
(Use 3 decimal places.)

lower limit
upper limit

Answer 1

Solution :

Given that,

n = 322

x = 168

Point estimate = sample proportion = = x / n = 168/322=0.522

1 -   = 1- 0.522 =0.478

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

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

= 2.576* (((0.522*0.478) /322 )

E = 0.072

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.522-0.072 < p < 0.522+ 0.072

0.450< p < 0.594

The 99% confidence interval for the population proportion p is : lower limit=0.450,upper limit=0.594