Question

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, 20% of the people said they were in favor of the plan. Determine a 95% confidence interval for the percentage of people who favor the tax plan, rounding values to the nearest tenth.

Answer 1

Answer: (0.5415 ; 0.6785)

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)