Question

In: Statistics and Probability

We have 30 cross-validation results as below: 0.81, 0.20, 0.92, 0.99, 0.75, 0.88, 0.98, 0.42, 0.92,...

We have 30 cross-validation results as below: 0.81, 0.20, 0.92, 0.99, 0.75, 0.88, 0.98, 0.42, 0.92, 0.90, 0.88, 0.72, 0.94, 0.93, 0.77, 0.78, 0.79, 0.69, 0.91, 0.92, 0.91, 0.62, 0.82, 0.93, 0.85, 0.83, 0.95, 0.70, 0.80, 0.90 Calculate the 95% confidence interval of the mean.

Solutions

Expert Solution

Let us first find mean and standard deviation for the given data

Sample mean is

And standard deviation is

Create the following table.

data data-mean (data - mean)2
0.81 -0.0036999999999999 1.3689999999999E-5
0.20 -0.6137 0.37662769
0.92 0.1063 0.01129969
0.99 0.1763 0.03108169
0.75 -0.0637 0.00405769
0.88 0.0663 0.00439569
0.98 0.1663 0.02765569
0.42 -0.3937 0.15499969
0.92 0.1063 0.01129969
0.90 0.0863 0.00744769
0.88 0.0663 0.00439569
0.72 -0.0937 0.00877969
0.94 0.1263 0.01595169
0.93 0.1163 0.01352569
0.77 -0.0437 0.00190969
0.78 -0.0337 0.00113569
0.79 -0.0237 0.00056169
0.69 -0.1237 0.01530169
0.91 0.0963 0.00927369
0.92 0.1063 0.01129969
0.91 0.0963 0.00927369
0.62 -0.1937 0.03751969
0.82 0.0063 3.969E-5
0.93 0.1163 0.01352569
0.85 0.0363 0.00131769
0.83 0.0163 0.00026569
0.95 0.1363 0.01857769
0.70 -0.1137 0.01292769
0.80 -0.0137 0.00018769
0.90 0.0863 0.00744769

Find the sum of numbers in the last column to get.

Hence standard deviation is

Now t table value 29 df for 95% CI is TINV(0.05,29)=2.045

So Margin of Error is

Hence CI is


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