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]
