Question

Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $30,000 and $45,000 with a population standard deviation of $3750. Assume that a 95% confidence interval estimate of the population mean annual starting salary is desired. How large a sample should be taken if the desired margin of error is $200?

Answer 1

Answer:

The value is   $n =1351$

Step-by-step explanation:

From the question we are told that

  The standard deviation is  $\sigma = \ 3750$

   The margin of error is  $E = \ 200$

From the question we are told the confidence level is  95% , hence the level of significance is    

      $\alpha = (100 - 95 ) \%$

=>   $\alpha = 0.05$

Generally from the normal distribution table the critical value  of  $\frac{\alpha }{2}$ is  

   $Z_{\frac{\alpha }{2} } = 1.96$

Generally the sample size is mathematically represented as

         $n = [\frac{Z_{\frac{\alpha }{2} } * \sigma }{E} ] ^2$

=>      $n = [1.96 * 3750 }{200} ] ^2$

=>      $n =1351$