likert scale standard deviation

eoins23456

Hey Guys , I have a survey with a likert scale question with 4 options (c1,c2,c3,c4). They have 24,24,25 and 28 responses respectively and there weights would be 1,2,3,4 . The software i am using is throwing out a standard deviation of 1.14. Anybody know what formula would have been used to get this ?

MathsManiac

The regular formula for standard deviation of a sample gives approx 1.13504 for this data set, which rounds to 1.14. (Find the mean = 2.564356...; sum the 101 squared deviations from it; divide by n-1 = 100.)

Some people would question, by the way, whether it is appropriate to use a measure such as standard deviation (or mean) when analysing data that is only ordinal rather than interval, but people often go ahead and use such parametric statistics on Likert scale items anyway. What do you hope to infer from the standard deviation?

eoins23456

MathsManiac wrote: »

The regular formula for standard deviation of a sample gives approx 1.13504 for this data set, which rounds to 1.14. (Find the mean = 2.564356...; sum the 101 squared deviations from it; divide by n-1 = 100.)

Some people would question, by the way, whether it is appropriate to use a measure such as standard deviation (or mean) when analysing data that is only ordinal rather than interval, but people often go ahead and use such parametric statistics on Likert scale items anyway. What do you hope to infer from the standard deviation?

I must be missing something here. I get the below.

sqrt ( ( (24-2.56 ) + (24-2.56 ) + (25-2.56 ) + (28-2.56 ) ) /100 ) = 4.5

I'm not really try to infer anything but just wanted to know where the number came from

MathsManiac

sqrt{[(1-2.56)^2 + (1-2.56)^2 + ... + (1-2.56)^2 + (2-2.56)^2 + (2-2.56)^2 + ... + (2-2.56)^2 + (3-2.56)^2 + (3-2.56)^2 + ... + (3-2.56)^2 + (4-2.56)^2 + (4-2.56)^2 + ... + (4-2.56)^2]/100}

= sqrt{[24*(1-2.56)^2 + 24*(2-2.56)^2 + 25*(3-2.56)^2 + 28*(4-2.56)^2]/100}

= 1.135