Respuesta :
Answer:
Confidence Interval: (21596,46428)
Step-by-step explanation:
We are given the following data set:
10520, 56910, 52454, 17902, 25914, 56607, 21861, 25039, 25983, 46929
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{340119}{10} = 34011.9[/tex]
Sum of squares of differences = 551869365.6 + 524322983.6 + 340111052.4 + 259528878 + 65575984.41 + 510538544 + 147644370.8 + 80512934.41 + 64463235.21 + 166851472.4 = 2711418821
[tex]S.D = \sqrt{\frac{2711418821}{9}} = 17357.09[/tex]
Confidence interval:
[tex]\mu \pm t_{critical}\frac{\sigma}{\sqrt{n}}[/tex]
Putting the values, we get,
[tex]t_{critical}\text{ at degree of freedom 9 and}~\alpha_{0.05} = \pm 2.2621[/tex]
[tex]34011.9 \pm 2.2621(\frac{17357.09}{\sqrt{10}} ) = 34011.9 \pm 12416.20 = (21595.7,46428.1) \approx (21596,46428)[/tex]
