Answer:
95% confidence interval: (2.339,3.521)
Step-by-step explanation:
We are given the following in the question:
3.32, 2.49, 3.45, 2.38, 3.01
Formula:
[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]
where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]\bar{x} =\displaystyle\frac{14.65}{5} = 2.93[/tex]
Sum of squares of differences = 0.925
[tex]\sigma = \sqrt{\frac{0.925}{4}} = 0.48[/tex]
95% confidence interval:
[tex]\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}[/tex]
Putting the values, we get,
[tex]t_{critical}\text{ at degree of freedom 4 and}~\alpha_{0.05} = \pm 2.77[/tex]
[tex]2.93 \pm 2.77(\frac{0.48}{\sqrt{5}} ) = 2.93 \pm 0.591 = (2.339,3.521)[/tex]
