Question

A survey was given to a random sample of 1950 residents of a town to determine whether they support a new plan to raise taxes in order to increase education spending. Of those surveyed, 80% 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

[0.8,0.8]

[0.7822,0.8178]

Explanation:

Confidence interval=p+/-z*(√p(1-p)/n)

p :sample proportion

z : critical

n: sample proportion

alpha =0.05

z critical=1.96

Cl(proportion)=(0.8- or +1.96×√(0.8(1-0.8)/1950

=(0.782,0.818)

=(0.8,0.8)