Question

A survey was given to a random sample of 1100 residents of a town to determine whether they support a new plan to raise taxes in order to increase education spending. Of those surveyed, 34% of the people said they were in favor of the plan. At the 95% confidence level, what is the margin of error for this survey expressed as a percentage to the nearest tenth? (Do not write ±±).

Answer 1

Solution :

Given that,

n = 1100

Point estimate = sample proportion = =0.34=0.34%

1 - = 1-0.34=0.66

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

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

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

= 1.96 (((0.34*0.66) /1100 )

= 0.028

=2.8%