A survey was taken of 850 people in a neighborhood, and the results were that 230 people were planning on voting for a particular candidate. Determine the margin of error for a 99% confidence interval of population proportion of people in a neighborhood who were planning on voting for this candidate. Round it to the nearest hundredth and interpret its meaning in context.(2 points) The margin of error=0.03. With 99% confidence, the population proportion of people in a neighborhood who are planning on voting for a particular candidate falls between blank* and blank*.
The margin of error is 0.03. With 99% confidence, the population proportion of people in a neighborhood who are planning on voting for a particular candidate falls between 0.21 and 0.27. This means that we are 99% confident that the true proportion of people in the neighborhood planning to vote for the candidate is between 21% and 27%.
This is wrong.
To calculate the margin of error for a 99% confidence interval, we will use the formula:
Margin of error = Z * sqrt((p*(1-p))/n)
Where:
Z = Z-value for 99% confidence level = 2.576
p = proportion of people planning on voting for the candidate = 230/850 = 0.27
n = total number of people surveyed = 850
Margin of error = 2.576 * sqrt((0.27*(1-0.27))/850)
Margin of error = 2.576 * sqrt((0.1962)/850)
Margin of error = 2.576 * sqrt(0.000231]
Margin of error = 2.576 * 0.0152
Margin of error = 0.04
So, the margin of error for a 99% confidence interval of the population proportion of people in the neighborhood who were planning on voting for this candidate is 0.04.
With 99% confidence, the population proportion of people in the neighborhood who are planning on voting for a particular candidate falls between 0.23 and 0.31.