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:
where
are data points,
is the mean and n is the number of observations.


Sum of squares of differences = 551869365.6 + 524322983.6 + 340111052.4 + 259528878 + 65575984.41 + 510538544 + 147644370.8 + 80512934.41 + 64463235.21 + 166851472.4 = 2711418821

Confidence interval:

Putting the values, we get,

