Answer:
1 [tex]\sigma_{\= x } = 0.0130[/tex]
2 [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
[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
[tex]n = [ \frac{2.58 ^2 * 0.623 ( 1 -0.623 )}{0.02^2} ][/tex]
[tex]n = 3908.5[/tex]
