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Old April 10th, 2008, 10:25 AM
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Default 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:



Thanks in advance! In a pickle.
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Old April 10th, 2008, 03:25 PM
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I'll do the Standard Deviation for Item 2:

1: Find arithmetic mean:

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

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

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

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

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

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

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

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

3. Square each Deviation:

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

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

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

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

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

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

4. Add the squares:

625 + 49 + 1 + 361 + 49 + 169 = 1254

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

5. Now we divide by the number of values:

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

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

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

And there you go.

A formula would be:

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

Where:

\sigma = \text{Standard Deviation}

N = \text{Number of data points}

x_i = \text{i-th data point}

\bar{x} = \text{Arithmetic Mean of data points}
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Old April 10th, 2008, 03:38 PM
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Quote:
Originally Posted by Aryth View Post
I'll do the Standard Deviation for Item 2:

1: Find arithmetic mean:

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

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

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

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

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

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

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

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

3. Square each Deviation:

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

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

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

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

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

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

4. Add the squares:

625 + 49 + 1 + 361 + 49 + 169 = 1254

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

5. Now we divide by the number of values:

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

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

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

And there you go.

A formula would be:

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

Where:

\sigma = \text{Standard Deviation}

N = \text{Number of data points}

x_i = \text{i-th data point}

\bar{x} = \text{Arithmetic Mean of data points}
Thanks! I think I've figured it out now.
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Old November 2nd, 2009, 07:14 PM
Lee Lee is offline
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Default st deviation query

Can you explain how you got top numbers in step 2......25/6, 1/6 etc?

Thanks
Lee
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