Answer:
95% confident the population mean is between (9.1,9.9)
Step-by-step explanation:
We are given the following data set:
9.3, 9.7, 9.2, 9.7
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]Mean =\displaystyle\frac{37.9}{4} = 9.475[/tex]
Sum of squares of differences = 0.2075
[tex]S.D = \sqrt{\frac{0.2075}{3}} = 0.262[/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 3 and}~\alpha_{0.05} = \pm 3.182[/tex]
[tex]9.475 \pm 3.182(\frac{0.262}{\sqrt{4}} ) = 9.475 \pm 0.416842 =(9.058158,9.891842) \approx (9.1,9.9)[/tex]