Question

As discussed in Exercise 6.10, the General Social Survey reported a sample where about 61% of US residents thought marijuana should be made legal. If we wanted to limit the margin of error of a 95% confidence interval to 2%, about how many Americans would we need to survey?

Answer 1

Answer: 2285

Step-by-step explanation:

Formula to find the sample size is given by :-

$n=p(1-p)(\dfrac{z_c}{E})^2$

, where p = prior estimate of population proportion.

E= Margin of error.

$z_c$ = z-value for confidence interval of c.

Given : Confidence interval : 95%

From the z-value table , the z-value for 95% confidence interval = $z_c=1.96$

The General Social Survey reported a sample where about 61% of US residents thought marijuana should be made legal.

i.e. the prior estimate of population proportion of US residents thought marijuana should be made legal : p=0.61  

 [∵ sample proportion is the best estimate for population proportion.]

Margin of error : E= 2%=0.02

Now, the required minimum sample size would be :-

$n=0.61(1-0.61)(\dfrac{1.96}{0.02})^2$

Simplify ,

$n=0.2379\times9604=2284.7916\approx2285$

Thus , we need to survey 2285 Americans .