Words:0 Path: p QUESTION 17 A newspaper article reported that a poll based on a

Question

Words:0 Path: p QUESTION 17 A newspaper article reported that a poll based on a sample of 1150 residents of a state showed that the state's Governor's job approval rating stood at 58%. They claimed a margin of error of ±3%, what level of confidence were the pollsters using? Show all work using the equation editor beginning with the margin of error formula. Round your solution to the nearest whole percent. Write a sentence that gives your solution. 3(12pt) ·-·T·e TTT Paragraph ' Arial HTHL ESS Words:0 A Path: p

Explanation / Answer

Margin of error = = +- 0.03

The (1-alpha) *100% confidence interval for population propertion is

( p - margin of error ,p + margin of error)

i.e. ( p - Z alpha/2 *sqrt(p*q/n) , p + Z alpha/2 *sqrt(p*q/n)

hence Z alpha/2 * sqrt ( p*q /n) =0.03

n = 1150 = total number of sample

p = sample propertion = 0.58 q = 1-p =0.42

E(p) = 0.58 and Var(p) = sqrt ( p*q/n)

Z alpha/2 * ( 0.58 *.42 /1150) =0.03

Zalpha/2 = 2.061

i.e. P ( -Zalpha/2 < Z < Zalpha/2 ) = 1-alpha where Z = ( p-E(p)) / SD(p) ~ N(0,1)

P( -2.061 < Z < 2.061) = 1-alpha ---------------(I)

From normal probability table

P( -2.061 < Z < 2.061) = 0.9607 ------------(II)

From (I) and(II)

1-alpha = 0.9607

Hence level of confidence = 0.9607 *100% = 96.07 %

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.