Answer:
The value is [tex]E = 0.196[/tex]
Step-by-step explanation:
From the question we are told that
The sample size is n = 119
The sample mean is [tex]\= x = 11.63 \ years[/tex]
The standard deviation is [tex]s = 1. 3 \ years[/tex]
Given that the confidence level is 90% then the level of significance is mathematically represented as
[tex]\alpha = (100 - 90) \%[/tex]
=> [tex]\alpha = 0.10[/tex]
The critical value of [tex]\frac{\alpha }{2}[/tex] obtained from the normal distribution table is
[tex]Z_{\frac{\alpha }{2} } = 1.645[/tex]
Generally th margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{s}{\sqrt{n} }[/tex]
=> [tex]E = 1.645* \frac{1.3}{\sqrt{119} }[/tex]
=> [tex]E = 0.196[/tex]
