Answer:
The 99% confidence interval is  [tex] 165.776  <  \mu <  180.224  [/tex] Â
Step-by-step explanation:
From the question we are told that
   The sample size is  n  = 100
   The sample mean is  [tex]\= x = 173 \ days[/tex]
   The population standard deviation is  [tex]\sigma = 28 \ days[/tex] Â
From the question we are told the confidence level is  99% , hence the level of significance is  Â
   [tex]\alpha = (100 - 99 ) \%[/tex]
=> Â [tex]\alpha = 0.01[/tex]
Generally from the normal distribution table the critical value  of  [tex]\frac{\alpha }{2}[/tex] is Â
  [tex]Z_{\frac{\alpha }{2} } = 2.58 [/tex]
Generally the margin of error is mathematically represented as Â
   [tex]E = Z_{\frac{\alpha }{2} } *  \frac{\sigma }{\sqrt{n} }[/tex]
   [tex]E = 2.58  *  \frac{28 }{\sqrt{100} }[/tex]
=> Â Â [tex]E = 7.224 [/tex]
Generally 95% confidence interval is mathematically represented as Â
   [tex]\= x -E <  \mu <  \=x  +E[/tex]
=> Â Â [tex] 173 -7.224 Â < Â \mu < 173 + 7.224 Â [/tex] Â Â
=> Â Â [tex] 165.776 Â < Â \mu < 180.224 Â [/tex] Â
  Â
