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=23.
The sample size is N=19.
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{11}{\sqrt{19}}=\dfrac{11}{4.359}=2.524[/tex]The degrees of freedom for this sample size are:
[tex]df=n-1=19-1=18[/tex]The t-value for a 95% confidence interval and 18 degrees of freedom is t=2.101.
The margin of error (MOE) can be calculated as:
[tex]MOE=t\cdot s_M=2.101\cdot2.524=5.302[/tex]Then, the lower and upper bounds of the confidence interval are:
[tex]\begin{gathered} LL=M-t\cdot s_M=23-5.302=17.698 \\ UL=M+t\cdot s_M=23+5.302=28.302 \end{gathered}[/tex]The 95% confidence interval for the mean is (17.698, 28.302).
The interval is a T-interval, as the sampleis small and we don't know the standard deviation of the population.
