A survey was given to a random sample of 1450 residents of a town to determine whether they support a new plan to raise taxes in order to increase education spending. Of those surveyed, 48% 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.

(0.4543 ; 0.5057)

Step-by-step explanation:

This is a confidence interval for one sample proportion :

Phat , = 0.48

Sample size = 1450

1 - phat = 1 - 0.48 = 0.52

Zcritical = 95% ; Zcritical at 95% = 1.96

Using the relation :

Phat ± Zcritical *√[phat(1-phat) / n]

√[phat(1-phat) / n] = √(0.48*0.52) /1450) = 0.0131201

Lower boundary :

0.48 - (1.96 * 0.0131201) = 0.4543

Upper boundary :

0.48 + (1.96 * 0.0131201) = 0.5057

(0.4543 ; 0.5057)

(c, 0) i might be wrong so don't put full faith on my answer

-9

step-by-step explanation: