Question

An opinion poll surveyed 900 people and reported that 36% believe a certain governor broke campaign financing laws in his election campaign.a. Calculate a 95% confidence interval for the population proportion of people who believe the governor broke campaign financing laws. (Round your answers to three decimal places.)

Answer 1

Answer:

The confidence Interval = (0.329, 0.391)

Step-by-step explanation:

Formula for the Confidence Interval for proportion =

p ± z × √p(1 - p)/n

where

p = x/n

From the question

n = 900 people

x = 36% of 900 people

= 36/100 × 900

= 324

z = z score of 95% confidence Interval

z = 1.98

p = x/n

= 324/900

= 0.36

Confidence Interval = p ± z × √p(1 - p)/n

= 0.36 ± 1.96 × √0.36 (1 - 0.36)/900

= 0.36 ± 1.96 ×√0.36 (0.64)/900

= 0.36 ± 1.96 × √0.000256

= 0.36 ± 1.96 × 0.016

= 0.36 ± 0.03136

Confidence Interval = 0.36 ± 0.03136

= 0.36 - 0.03136

= 0.32864

Approximately to 3 decimal places = 0.329

= 0.36 + 0.03136

= 0.39136

Approximately to 3 decimal places = 0.391

Therefore, the 95% confidence interval for the population proportion of people who believe the governor broke campaign financing laws = (0.329, 0.391)