Question

conds. Question Completion Status: A ne spaper article reported that ap ? based on a sample of 8 voters sho ed the President's ob approval rating stood at 62, 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. T TTT Paragraph Arial3 (12pt)E E T-

Answer 1

Given , n = 800, $\hat{p}$ = 62% = 0.62 , margin of error = 3% = 0.03

Formula for margin of error for proportion is given by,

$E = Z_{\alpha /2}*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$0.03 = Z_{\alpha /2}*\sqrt{\frac{0.62(1-0.62)}{800}}$

$0.03 = Z_{\alpha /2}*0.01716$

$Z_{\alpha /2}= 0.03/0.01716=1.7483$

This is two tailed critical value.

So confidence level = P( -1.7483 < z < 1.7483 )

P( -1.7483 < z < 1.7483 ) = P( z < 1.7483 ) - P( z < -1.7483 )

Using Excel,   =NORMSDIST( Z )    This function returns area less then z

P( -1.7483 < z < 1.7483 ) = NORMSDIST(1.7483) - NORMSDIST(-1.7483)

P( -1.7483 < z < 1.7483 ) =0.9195 $\approx$ 0.92 = 92%

So, confidence level is 92%