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