Answer:
The 95% confidence interval is [tex] 101.13 < \mu < 104.87 [/tex]
The 99% confidence interval is [tex] 100.54 < \mu < 105.46 [/tex]
Step-by-step explanation:
From the question we are told that
The sample size is n = 110
The sample mean is [tex]\= x = 103 \ mg[/tex]
The population standard deviation is [tex]\sigma = 10 \ mg[/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{ 10 }{\sqrt{110} }[/tex]
=> [tex]E =1.8688 [/tex]
Generally 95% confidence interval is mathematically represented as
[tex]\= x -E < \mu < \=x +E[/tex]
=> [tex]103 - 1.8688 < \mu < 103 + 1.8688 [/tex]
=> [tex] 101.13 < \mu < 104.87 [/tex]
Considering question b
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{ 10 }{\sqrt{110} }[/tex]
=> [tex]E =2.4599 [/tex]
Generally 99% confidence interval is mathematically represented as
[tex]\= x -E < \mu < \=x +E[/tex]
=> [tex]103 - 2.4599 < \mu <103 + 2.4599 [/tex]
=> [tex] 100.54 < \mu < 105.46 [/tex]
