Answer:
1 Â [tex]\sigma_{\= x } = 0.0130[/tex]
2 Â [tex]n = 3908.5[/tex]
Step-by-step explanation:
From the question we are told that
   The  sample size is  [tex]n_p = 1400[/tex]
   The  number of those that said the would use internet is [tex]k = 872[/tex]
    The margin of error is  [tex]E = 0.02[/tex]
Generally the sample proportion is mathematically evaluated as
      [tex]\r p = \frac{k}{n_p}[/tex]
substituting values
      [tex]\r p = \frac{ 872}{1400}[/tex]
substituting values
      [tex]\r p = 0.623[/tex]
Generally the standard error of  [tex]\r p[/tex] is mathematically evaluated as
      [tex]\sigma_{\= x } = \sqrt{\frac{\r p (1- \r p)}{n} }[/tex]
substituting values
       [tex]\sigma_{\= x } = \sqrt{\frac{0.623 (1- 0.623)}{1400} }[/tex]
       [tex]\sigma_{\= x } = 0.0130[/tex]
For  a  95% confidence interval the confidence level is  95%
Given that the confidence interval is 95% the we can evaluated the level of confidence as
          [tex]\alpha = 100 - 99[/tex]
          [tex]\alpha = 1\%[/tex]
          [tex]\alpha = 0.01[/tex]
Next we obtain the critical value of  [tex]\frac{\alpha }{2}[/tex] from normal distribution table (reference math dot  armstrong dot edu) , the value is Â
     [tex]Z_{\frac{\alpha }{2} } = 2.58[/tex]
Give that the population size is very large  the sample size is mathematically represented as
      [tex]n = [ \frac{Z_{\frac{\alpha }{2} ^2 * \r p ( 1 - \r p )}}{E^2} ][/tex]
substituting values Â
       [tex]n = [ \frac{2.58 ^2 * 0.623 ( 1 -0.623 )}{0.02^2} ][/tex]
       [tex]n = 3908.5[/tex]
