To estimate the average education level, in years, of residents in a certain county, a marketing team draws a random sample of 100 residents. Their education level averages to about 12.5 years; the standard deviation is about 2.8 years. The team estimates the average education level of all the residents of that county to be 12.5 years with what margin of error, for approximately 90% confidence?

Answer:

The margin of error is  [tex]E = 0.4606 [/tex]      

Step-by-step explanation:

From the question we are told that

   The sample size is  n = 100

    The sample mean is  [tex]\= x = 12.5 \ years[/tex]

    The standard deviation is  [tex]\sigma = 2.8 \ years[/tex]

From the question we are told the confidence level is  90% , hence the level of significance is    

      [tex]\alpha = (100 - 90 ) \%[/tex]

=>   [tex]\alpha = 0.10[/tex]

Generally from the normal distribution table the critical value  of  [tex]\frac{\alpha }{2}[/tex] is  

   [tex]Z_{\frac{\alpha }{2} } =  1.645[/tex]

Generally the margin of error is mathematically represented as  

      [tex]E = Z_{\frac{\alpha }{2} } *  \frac{\sigma }{\sqrt{n} }[/tex]

=>   [tex]E =1.645*  \frac{ 2.8 }{\sqrt{100} }[/tex]

=>   [tex]E = 0.4606 [/tex]      

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