Question

A survey was given to a random sample of 140 residents of a town to determine whether they support a new plan to raise taxes in order to increase education spending. Of those surveyed, 105 respondents said they were in favor of the plan Determine a 95% confidence interval for the proportion of people who favor the tax plan, rounding values to the nearest thousandth.

Answer 1

95% confidence interval for the proportion of people who favor the tax plan is (71.35%, 78.65%)

Step-by-step explanation:

According to the given data we have the following:

survey was given to a random sample of 140 residents, hence n=140

Of those surveyed, 105 respondents said they were in favor of the plan, therefore, P=75%

Therefore, to calculate a 95% confidence interval for the proportion of people who favor the tax plan we would have to make the following calculation:

95% confidence interval for the proportion of people who favor the tax plan=P±Z*√P(1-P)/n

=0.75±1.95√0.75(1-0.75)/140

=(0.75±0.0365)

=(0.7135,0.7865)

95% confidence interval for the proportion of people who favor the tax plan=(71.35%, 78.65%)