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diamong [38]
3 years ago
14

The 2010 General Social Survey reported a sample where about 48% of US residents thought marijuana should be made legal. If we w

anted to limit the margin of error of a 95% confidence interval to 4%, about how many Americans would we need to survey
Mathematics
1 answer:
Sati [7]3 years ago
5 0

Answer:

The sample size is  n  = 600

Step-by-step explanation:

From the question we are told that

    The  sample proportion is  \r  p = 0.48

     The  margin of error is  MOE  =  0.04

Given that the confidence level is 95%  the level of significance is mathematically represented as

        \alpha  =  100 - 95

        \alpha  =  5 \%

        \alpha  =  0.05

Next  we obtain the critical value of  \frac{\alpha }{2} from the normal distribution table , the values is

                   Z_{\frac{\alpha }{2}  } =  1.96

The reason we are obtaining critical value of    \frac{\alpha }{2} instead of    \alpha is because  

\alpha represents the area under the normal curve where the confidence level interval (  1-\alpha) did not cover which include both the left and right tail while  

\frac{\alpha }{2} is just the area of one tail which what we required to calculate the margin of error

Generally the margin of error is mathematically represented as

      MOE  =  Z_{\frac{\alpha }{2} } *  \sqrt{ \frac{\r p(1-  \r p )}{n} }

substituting values

          0.04= 1.96*  \sqrt{ \frac{0.48(1-  0.48 )}{n} }

         0.02041 = \sqrt{ \frac{0.48(52 )}{n} }

         0.02041 = \sqrt{ \frac{ 0.2496}{n} }

          0.02041^2  = \frac{ 0.2496}{n}

           0.0004166 = \frac{ 0.2496}{n}

=>       n  = 600

   

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