Question

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)

Answer 1

The Margin of Error E
`\simeq` 4.1 %

Explanation:

Given that:

The sample size = 579

The sample proportion
`\hat p = 55\%` = 0.55

From the confidence interval of 95%

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

The critical value of
`Z_(\alpha/2 ) = Z_(0.025) = 1.96`

Thus;

The Margin of Error E =
`Z_(\alpha/2 ) * \sqrt{{\frac {\hat p ( 1 - \hat p }{n}}`

The Margin of Error E =
`1.96 * \sqrt{{\frac {0.55 ( 1 - 0.55) }{579}}`

The Margin of Error E =
`1.96 * \sqrt{{\frac {0.55 ( 0.45 )}{579}}`

The Margin of Error E =
`1.96 * \sqrt{{\frac {0.2475}{579}}`

The Margin of Error E =
`1.96 * \sqrt{{4.2746114 * 10^(-4)}`

The Margin of Error E =
`1.96 * .020675`

The Margin of Error E = 0.040523

The Margin of Error E
`\simeq` 0.041

The Margin of Error E
`\simeq` 4.1%