Which field are you? are these tools they are marking standardized tools developed through science or technology or were they designed by you?

Anyway, lets talk statistics. Usually in Pure or physical sciences, researchers set their probability level to 1 percent (0.01 level of significance). However, 5% (0.05 LS) is also in wide use among social and behavioural scientists. The choice is however, depending on you, or the conference guidelines or your discipline.

You may wish to use independent t-test if you are comparing two groups. For comparison involving three or more groups, you may use ANOVA. This answer is only tentative in the interim, since i do not have complete knowledge of your study's objective and/or design.

If you explain your project yet further, someone may help better.

Best regards

Owan

Which field are you? are these tools they are marking standardized tools developed through science or technology or were they designed by you?

Anyway, lets talk statistics. Usually in Pure or physical sciences, researchers set their probability level to 1 percent (0.01 level of significance). However, 5% (0.05 LS) is also in wide use among social and behavioural scientists. The choice is however, depending on you, or the conference guidelines or your discipline.

You may wish to use independent t-test if you are comparing two groups. For comparison involving three or more groups, you may use ANOVA. This answer is only tentative in the interim, since i do not have complete knowledge of your study's objective and/or design.

If you explain your project yet further, someone may help better.

Best regards

Owan

Hello Tereza,

It seems to me that your best bet would be a chi-square goodness of fit test. I'd suggest summing the total number of tool endorsements given by each respondent across all respondents and then dividing that number by the number of tools from which they could select. That value would be the expected cell frequency for each tool under the null hypothesis that there were no differences in the rates with which any of the tools were used.

The observed cell frequency would be the actual number (of your 20 respondents) who endorsed a given tool.

For each cell (tool): compute (Fo - Fe)^2 / Fe

where: Fo = Observed Frequency

Fe = Expected Frequency

Then sum the result of all similar cell computations.

Refer to a critical chi-square based on cells (tools) - 1 df (degrees of freedom).

However, with a sample size of 20 cases, you might find that this won't be a very powerful statistical test. Only a bigger sample size could improve that situation.

Good luck with your work!

It appears you have "frequency" data. That is, if n=15 respondents mark that they use "interactive tables", then the frequency is (15 / 20 = ) 0.75. For "polygraph", it would be (11 / 20 = ) 0.55; and so on.

In some circumstances, you could do a *Chi square goodness of fit* test for this type of data, although in your case I don't think you can assume *independence* and therefore it would not be appropriate. (Your sample size is also very small, and I do not know how many options the respondents were given to select from.)

At the least, you can report the frequency/percent of each answer (tool), you could stratify by respondent gender, age, etc. You should research "multiple response analysis", "frequency analysis", etc.

Chi Square, as our colleagues above said!