Question

In the study of 10 patients treated for a kidney infection, the following data refer to the use of urinary analysis 1, 2, 5, 7, 3, 1, 4, 3, 6, and 4 Use the given data to conduct a 99% confidence interval for the mean use of this procedure.

Answer 1

For the given data sample mean is $\overline{x}=\frac{\sum x}{n}=3.6$

Create the following table.

data	data-mean	(data - mean)2
1	-2.6	6.76
2	-1.6	2.56
5	1.4	1.96
7	3.4	11.56
3	-0.6	0.36
1	-2.6	6.76
4	0.4	0.16
3	-0.6	0.36
6	2.4	5.76
4	0.4	0.16

Find the sum of numbers in the last column to get. $\sum (xi-\overline{x})^2=36.4$

So standard deviation is $\sigma=\sqrt{\frac{\sum (xi-\overline{x})^2}{n-1}}=2.0111$

Now t table value for 9 df is 2.262

So Margin of Error is $E=t*\frac{s}{\sqrt{n}}= 3.250*\frac{2.0111}{\sqrt{10}}=2.067$

Hence CI is $CI=\overline{x} \pm E=3.6 \pm 2.067=(1.533,5.667)$