# Desperate Dissertation help: Likert scale analysis 1-5 and Standard Deviation

• Apr 10th 2008, 09:25 AM
Desperate Dissertation help: Likert scale analysis 1-5 and Standard Deviation
I've completely forgot how this works.

Here's my table, could anyone please start me off with the std dev of some of these:

http://img170.imageshack.us/img170/5...viationsm4.jpg

Thanks in advance! In a pickle.
• Apr 10th 2008, 02:25 PM
Aryth
I'll do the Standard Deviation for Item 2:

1: Find arithmetic mean:

$\displaystyle \frac{(1 + 6 + 5 + 8 + 4 + 7)}{6} = \frac{31}{6} = 5.1666667$

2. Find the deviation of each number from the mean:

$\displaystyle 1 - 5.1666667 = -4.1666667 = -\frac{25}{6}$

$\displaystyle 6 - 5.1666667 = 1.1666667 = \frac{7}{6}$

$\displaystyle 5 - 5.1666667 = 0.1666667 = \frac{1}{6}$

$\displaystyle 8 - 5.1666667 = 3.1666667 = \frac{19}{6}$

$\displaystyle 4 - 5.1666667 = -1.1666667 = -\frac{7}{6}$

$\displaystyle 7 - 5.1666667 = 2.1666667 = \frac{13}{6}$

3. Square each Deviation:

$\displaystyle \left(-\frac{25}{6}\right)^2 = \frac{625}{36}$

$\displaystyle \left(\frac{7}{6}\right)^2 = \frac{49}{36}$

$\displaystyle \left(\frac{1}{6}\right)^2 = \frac{1}{36}$

$\displaystyle \left(\frac{19}{6}\right)^2 = \frac{361}{36}$

$\displaystyle \left(-\frac{7}{6}\right)^2 = \frac{49}{36}$

$\displaystyle \left(\frac{13}{6}\right)^2 = \frac{169}{36}$

$\displaystyle 625 + 49 + 1 + 361 + 49 + 169 = 1254$

So, we have: $\displaystyle \frac{1254}{36}$

5. Now we divide by the number of values:

$\displaystyle \frac{\frac{1254}{36}}{6} = \frac{1254}{216}$

6. Now we take the positive square root of what we have so far:

$\displaystyle \sqrt{\frac{1254}{216}} = 2.4095$

And there you go.

A formula would be:

$\displaystyle \sigma = \sqrt{\frac{1}{N}\sum^N_{i=1} (x_i - \bar{x})^2}$

Where:

$\displaystyle \sigma = \text{Standard Deviation}$

$\displaystyle N = \text{Number of data points}$

$\displaystyle x_i = \text{i-th data point}$

$\displaystyle \bar{x} = \text{Arithmetic Mean of data points}$
• Apr 10th 2008, 02:38 PM
Quote:

Originally Posted by Aryth
I'll do the Standard Deviation for Item 2:

1: Find arithmetic mean:

$\displaystyle \frac{(1 + 6 + 5 + 8 + 4 + 7)}{6} = \frac{31}{6} = 5.1666667$

2. Find the deviation of each number from the mean:

$\displaystyle 1 - 5.1666667 = -4.1666667 = -\frac{25}{6}$

$\displaystyle 6 - 5.1666667 = 1.1666667 = \frac{7}{6}$

$\displaystyle 5 - 5.1666667 = 0.1666667 = \frac{1}{6}$

$\displaystyle 8 - 5.1666667 = 3.1666667 = \frac{19}{6}$

$\displaystyle 4 - 5.1666667 = -1.1666667 = -\frac{7}{6}$

$\displaystyle 7 - 5.1666667 = 2.1666667 = \frac{13}{6}$

3. Square each Deviation:

$\displaystyle \left(-\frac{25}{6}\right)^2 = \frac{625}{36}$

$\displaystyle \left(\frac{7}{6}\right)^2 = \frac{49}{36}$

$\displaystyle \left(\frac{1}{6}\right)^2 = \frac{1}{36}$

$\displaystyle \left(\frac{19}{6}\right)^2 = \frac{361}{36}$

$\displaystyle \left(-\frac{7}{6}\right)^2 = \frac{49}{36}$

$\displaystyle \left(\frac{13}{6}\right)^2 = \frac{169}{36}$

$\displaystyle 625 + 49 + 1 + 361 + 49 + 169 = 1254$

So, we have: $\displaystyle \frac{1254}{36}$

5. Now we divide by the number of values:

$\displaystyle \frac{\frac{1254}{36}}{6} = \frac{1254}{216}$

6. Now we take the positive square root of what we have so far:

$\displaystyle \sqrt{\frac{1254}{216}} = 2.4095$

And there you go.

A formula would be:

$\displaystyle \sigma = \sqrt{\frac{1}{N}\sum^N_{i=1} (x_i - \bar{x})^2}$

Where:

$\displaystyle \sigma = \text{Standard Deviation}$

$\displaystyle N = \text{Number of data points}$

$\displaystyle x_i = \text{i-th data point}$

$\displaystyle \bar{x} = \text{Arithmetic Mean of data points}$

Thanks! I think I've figured it out now.
• Nov 2nd 2009, 06:14 PM
Lee
st deviation query
Can you explain how you got top numbers in step 2......25/6, 1/6 etc?

Thanks
Lee