Question

Amelia wants to estimate what proportion of her community members favor a tax increase for more local
school funding. She wants her margin of error to be no more than 2% at the 95% condence level. What is
the smallest sample size required to obtain the desired margin of error?

(2) Suppose a preliminary poll reveals that 45% of 50 voters support a presidential candidate. At 95% condence,
how large a sample is needed to obtain a margin of error of 4 percentage points?

Answer 1

Solution :

Given that,

1) = 0.50

1 - = 0.50

margin of error = E = 0.02

At 95% confidence level

= 1-0.95% =1-0.95 =0.05

/2 =0.05/ 2= 0.025

Z/2 = Z_{0.025 = 1.960}

Z_/2 =1.960

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.96/0.02)² * 0.50*0.50

= 2401

sample size = n = 2401

2)

= 0.45

1 - =1 -0.45 = 0.55

margin of error = E = 0.04

At 95% confidence level

= 1-0.95% =1-0.95 =0.05

/2 =0.05/ 2= 0.025

Z/2 = Z_{0.025 = 1.960}

Z_/2 =1.960

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.96/0.04)² * 0.45*0.55

= 594

sample size = n =594