Question

A study of the career paths of hotel general managers sent questionnaires to an SRS of 210 hotels belonging to major U.S. hotel chains. There were 108 responses. The average time these 108 general managers had spent with their current company was 8.8 years. (Take it as known that the standard deviation of time with the company for all general managers is 3.5 years.)
Required:
a. Find the margin of error for an 85% confidence interval to estimate the mean time a general manager had spent with their current company.
b. Find the margin of error for a 99% confidence interval to estimate the mean time a general manager had spent with their current company.

Answer 1

Answer:

a

  $E = 0.4850$  

b

 $E = 0.8689$  

Step-by-step explanation:

From the question we are told that

   The sample size is  n  =  108

    The sample mean is  $\= x = 8.8 \ years$  

     The standard deviation is  $\sigma = 3.5$

From the question we are told the confidence level is  85% , hence the level of significance is    

      $\alpha = (100 - 85 ) \%$

=>   $\alpha = 0.15$

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

   $Z_{\frac{\alpha }{2} } = 1.44$

Generally the margin of error is mathematically represented as  

      $E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=>    $E = 1.44 * \frac{ 3.5 }{\sqrt{108} }$

=>    $E = 0.4850$  

From the question we are told the confidence level is  99% , hence the level of significance is    

      $\alpha = (100 - 99 ) \%$

=>   $\alpha = 0.01$

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

   $Z_{\frac{\alpha }{2} } = 2.58$

Generally the margin of error is mathematically represented as  

      $E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=>    $E = 2.58 * \frac{ 3.5 }{\sqrt{108} }$

=>    $E = 0.8689$