Answer:
The 95% confidence interval is [tex]10.5 < \mu <13.3[/tex]
Step-by-step explanation:
From the question we are told that
The sample size is [tex]n = 41[/tex]
The sample mean is [tex]\= x = 11.9 \ hr[/tex]
The standard deviation is [tex]\sigma = 4.5[/tex]
For a 95% confidence interval the confidence level is 95%
Given that the confidence level is 95% then the level of significance can be mathematically represented as
[tex]\alpha = 100 - 95[/tex]
[tex]\alpha = 5 \%[/tex]
[tex]\alpha = 0.05[/tex]
Next we obtain the critical values of [tex]\frac{\alpha }{2}[/tex] from the normal distribution table
The values 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]
substituting values
[tex]E = 1.96 * \frac{ 4.5 }{ \sqrt{41} }[/tex]
[tex]E = 1.377[/tex]
The 95% confidence interval is mathematically represented as
[tex]\= x - E < \mu < \= x - E[/tex]
substituting values
[tex]11.9 - 1.377 < \mu <11.9 + 1.377[/tex]
[tex]10.5 < \mu <13.3[/tex]
