In late June 2012, Survey USA published results of a survey stating that 55% of the 579 randomly sampled Kansas residents planned to set off fireworks on July 4th. Determine the margin of error for the 55% point estimate using a 95% confidence level. The margin of error is: % (please round to the nearest percent)

The Margin of Error E 4.1 %

Step-by-step explanation:

Given that:

The sample size = 579

The sample proportion = 0.55

From the confidence interval of 95%

The level of significance ∝ = 1 - C.I = 1 - 0.95 = 0.05

The critical value of

Thus;

The Margin of Error E =

The Margin of Error E =

The Margin of Error E =

The Margin of Error E =

The Margin of Error E =

The Margin of Error E =

The Margin of Error E = 0.040523

The Margin of Error E 0.041

The Margin of Error E 4.1%

12/11c

or

12

c

11

i have no clue at all sorry

step-by-step explanation: