Question

Construct a

99​%

confidence interval of the population proportion using the given information.

x=120, n=200

The lower bound is

-----

The upper bound is

------

​(Round to three decimal places as​ needed.)

Answer 1

Solution :

Given that,

n = 200

x = 120

= x / n = 120 /200 = 0.60

1 - = 1 - 0.60 = 0.40

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 2.576 * (((0.60 * 0.40) / 200) = 0.089

At 99 % confidence interval for population proportion p is ,

- E < P < + E

0.60 - 0.089< p < 0.60 + 0.089

0.511 < p < 0.689

The 99% confidence interval for the population proportion p is : (lower bound = 0.511 upper bound = 0.689)