Question

A survey of 571 citizens found that 334 of them favor anew 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

Solution :

Given that,

n = 571

x = 334

Point estimate = sample proportion = = x / n = 334/571 = 0.585

1 - = 1 - 0.585 = 0.415

At 95% confidence level

= 1-0.95% =1-0.95 =0.05

/2 =0.05/ 2= 0.025

Z/2 = Z_{0.025 = 1.960}

Z_/2 = 1.960

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

= 1.960 * ((0.585*(0.415) /571 )

= 0.040

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.585 - 0.040 < p < 0.585 + 0.040

0.545< p < 0.625

( 0.545,0.625)

The lower limit of the interval = 0.545