Question

A survey of 492 citizens found that 360 of them favor a new bill introduced by the city. We want to find a 95% confidence interval for the true proportion of the population who favor the bill. What is the lower limit of the interval? (Round to 3 decimal digits)

Answer 1

Given that,

n = 492

x = 360

= x / n = 360 /492 = 0.732

1 - = 1 - 0.732 = 0.268

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

=0.05

Z = Z_0.05 = 1.645

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

= 1.645 * (((0.732 * 0.268) / 492) = 0.033

A 95 % confidence interval for population proportion p is ,

- E

0.732 - 0.033

0.699

The 95% confidence interval for the population proportion p is : lower bound = 0.699