Answer:
The margin of error is [tex]E = 6.9 [/tex]
The 95% confidence interval is [tex] 26.5 < \mu < 40.3 [/tex]
Step-by-step explanation:
From the question we are told that
The sample size is n = 31
The mean is [tex]\mu = 33.4 \ ng/ml[/tex]
The standard deviation is [tex]\sigma = 19.6 \ ng/ml[/tex]
From the question we are told the confidence level is 95% , hence the level of significance is
[tex]\alpha = (100 - 95 ) \%[/tex]
=> [tex]\alpha = 0.05[/tex]
Generally from the normal distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] is
[tex]Z_{\frac{\alpha }{2} } = 1.96[/tex]
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }[/tex]
=> [tex]E = 1.96 * \frac{19.6 }{\sqrt{31 } }[/tex]
=> [tex]E = 6.9 [/tex]
Generally 95% confidence interval is mathematically represented as
[tex]\= x -E < \mu < \=x +E[/tex]
=> [tex] 33.4 - 6.9 < \mu < 33.4 + 6.9 [/tex]
=> [tex] 26.5 < \mu < 40.3 [/tex]
