A survey was given to a random sample of 195 residents of a town to determine whether they support a new plan to raise taxes in order to increase education spending. Of those surveyed, 117 respondents said they were in favor of the plan. Determine a 95% confidence interval for the proportion of people who favor the tax plan.
Question
Answer:
(0.5415 ; 0.6785)
Step-by-step explanation:
Sample size, n = 195
Number of samples who werw in favor, x = 117
Phat = 117 / 195 = 0.609 = 0.61
The confidence interval :
Phat ± Margin of error
Margin of Error = Zcritical * √(phat(1-phat)/n)
Zcritical at 95% = 1.96
1 – phat = 1 – 0.61 = 0.39
Margin of Error = 1.96 * √((0.61*0.39)/195) = 0.0685
Confidence interval :
Lower boundary : 0.61 – 0.0685 = 0.5415
Upper boundary : 0.61 + 0.0685 = 0.6785
(0.5415 ; 0.6785)