Confidence interval of the mean.
We have to calculate a 95% confidence interval for the mean.
The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.
The sample mean is M=512.
The sample size is N=51.
When σ is not known, s divided by the square root of N is used as an estimate of σM:
[tex]s_M=\dfrac{s}{\sqrt{N}}=\dfrac{109}{\sqrt{51}}=\dfrac{109}{7.141}=15.263[/tex]The degrees of freedom for this sample size are:
[tex]df=n-1=51-1=50[/tex]The t-value for a 95% confidence interval and 50 degrees of freedom is t=2.
The margin of error (MOE) can be calculated as:
[tex]MOE=t\cdot s_M=2\cdot15.263=30.526\approx30.5[/tex]Then, the lower and upper bounds of the confidence interval are:
[tex]\begin{gathered} LL=M-t\cdot s_M=512-30.5=481.5 \\ UL=M+t\cdot s_M=512+30.5=542.5 \end{gathered}[/tex]Answer: The 95% confidence interval for the mean is (481.5, 542.5).
