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%

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