A survey is done to estimate the proportion of US Adults who think that texting while...

A survey is done to estimate the proportion of US Adults who think that texting while driving should be illegal. In a survey, 82% of a randomly selected sample of 1025 individuals said that this should be illegal.

q1: A 95% confidence interval is construct to estimate the proportion, the STANDARD ERROR is 0.012. What is the margin of error?

Solutions

Expert Solution

Solution :

Given that,

n = 1025

Point estimate = sample proportion = = 82%=0.82

1 - = 1-0.82 =0.18

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 *STANDARD ERROR

= 1.96*0.012

margin of error =0.02352

orchestra answered 1 year ago

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