3 A full descriptive

To look at a full descriptive example including the mean, standard deviation, standard error, 95% Confidence Intervals, and z-scores, we will use the following data from 10 participants:

Participants	X
1	8
2	11
3	6
4	3
5	1
6	6
7	1
8	0
9	1
10	4

3.0.1 The Mean

We know the formula for the mean is:

\[\overline{X} = \frac{\sum_i^n{x_i}}{n}\]

If we start to fill in the information from above we get:

\[\overline{X} = \frac{8 + 11 + 6 +3 +1 +6 +1 +0 +1 +4}{10}\]

And if we sum all the values on the top half together we get:

\[\overline{X} = \frac{41}{10}\]

And finally divide the top half by the bottom half, leaving:

\[\overline{X} = 4.1\]

We find that the mean, rounded to two decimal places, is \(\overline{X} = 4.1\)

3.0.2 Standard Deviation

Next we want an interval estimate, most commonly the standard deviation. The formula for the standard deviation is:

\[s = \sqrt\frac{\sum_i^n(x_{i} - \overline{x})^2}{n-1}\] So let's again start by filling in the numbers to the formula:

\[s = \sqrt\frac{(8 - 4.1)^2 + (11 - 4.1)^2 + (6 - 4.1)^2 + (3 - 4.1)^2 + (1 - 4.1)^2 + \\(6 - 4.1)^2 + (1 - 4.1)^2 + (0 - 4.1)^2 + (1 - 4.1)^2 + (4 - 4.1)^2}{10 - 1}\]

And we do the subtractions in all those brackets of the top half and the one on the bottom:

\[s = \sqrt\frac{(3.9)^2 + (6.9)^2 + (1.9)^2 + (-1.1)^2 + (-3.1)^2 + \\(1.9)^2 + (-3.1)^2 + (-4.1)^2 + (-3.1)^2 + (-0.1)^2}{9}\]

Then we square all those values on the top half:

\[s = \sqrt\frac{15.21 + 47.61+ 3.61+ 1.21+ 9.61+ 3.61+ 9.61+ 16.81+ 9.61+ 0.01}{9}\]

Sum those values together:

\[s = \sqrt\frac{116.9}{9}\] Divide the top half by the bottom half:

\[s = \sqrt{12.9888889}\]

And finally take the square root:

\[s = 3.6040101\]

We find that the standard deviation, rounded to two decimal places, is \(s = 3.6\)

3.0.3 Standard Error of the Mean

We also want to include a measure of how our sample relates to the population of interest and to do that we can use the Standard Error of the Mean and 95% Confidence Intervals. The formula for the standard error can be written as:

\[SE = \frac{SD}{\sqrt{n}}\]

We know that our \(SD = 3.6040101\) and that we have \(n = 10\) participants, so:

\[SE = \frac{3.6040101}{\sqrt{10}}\]

And if we do the square root of the bottom half (denominator):

\[SE = \frac{3.6040101}{3.1622777}\]

And divide the top by the bottom:

\[SE = 1.1396881\]

We find that the standard error, rounded to two decimal places, is \(SE = 1.14\)

3.0.4 Confidence Intervals

And we can use the standard error not to calculate the Upper and Lower 95% Confidence Intervals using the cut-off value (assuming \(\alpha = .05\) and two-tailed) of \(z = 1.96\). The key formulas are:

\[Upper \space 95\%CI = \overline{x} + (z \times SE)\]

And

\[Lower \space 95\%CI = \overline{X} - (z \times SE)\]

We know that:

• \(\overline{x} = 4.1\)
• \(SE = 1.1396881\)
• \(z = 1.96\)

Upper 95% CI

Dealing with the Upper 95% CI we get:

\[Upper \space 95\%CI = 4.1 + (1.96 \times 1.1396881)\]

And if we sort out the bracket first:

\[Upper \space 95\%CI = 4.1 + 2.2337886\]

Which leaves us with:

\[Upper \space 95\%CI = 6.3337886\]

Lower 95% CI

And now the Lower 95% CI we get:

\[Lower \space 95\%CI = 4.1 - (1.96 \times 1.1396881)\]

And if we sort out the bracket first:

\[Lower \space 95\%CI = 4.1 - 2.2337886\]

Which leaves us with:

\[Lower \space 95\%CI = 1.8662114\]

And if we round both those values to two decimal places, then you get Lower 95%CI = 1.87 and Upper 95%CI = 6.33

3.0.5 The Write-Up

And if we were to do a decent presentation of this data, including a measure of central tendency (the mean), a measure of spread relating to the sample (the standard deviation), and a measure of spread relating to the population (the 95% CI), then it would start as something like: "We ran 10 participants (M = 4.1, SD = 3.6, 95%CI = [1.87, 6.33])....."