Question

Construct a confidence interval of the population proportion at the given level of confidence.

x equals=860​,

n equals=1200​,

90​% confidence

What are the upper and lower bounds

Answer 1

Solution :

Given that,

n = 1200

x = 860​

= x / n = 860​, /1200 = 0.717

1 - = 1 - 0.717 = 0.283

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

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

= 1.645 * (((0.717 * 0.283) /1200)

                           = 0.021

A 90 % confidence interval for population proportion p is ,

- E < P < + E

0.717 - 0.021 < p < 0.717 + 0.021

0.696 < p < 0.738

Upper bound = 0.738

lower bound = 0.696