Question

In: Statistics and Probability

Assume that a sample is used to estimate a population proportion p. Find the margin of...

Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places. 95% confidence; the sample size is 9900, of which 30% are successes.

Solutions

Expert Solution

Solution :

Given that,

n = 9900

Point estimate = sample proportion = = 0.30

1 -   = 1- 0.30 =0.70

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E = Z/2   * $\sqrt$ ((( * (1 - )) / n)

= 1.96 ($\sqrt$((0.30*0.70) / 9900)

=0.009027

Margin of error = E =0.0090

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