*URGENT* Unsure of which statistical test to use

Question

I am assessing the effect of premature birth on cognitive development. I conducted a survey with quantitative answers on a scale (e.g., how easy to you find mathematics? Very hard, moderately hard, neutral, etc).

I want to conduct a statistical analysis between gestational age (which is nominal) vs the difficulty level of tasks (ordinal). Some of the answers are also "yes/or" (which is nominal).

I wanted to do Pearsons but that only looks at numerical data. I thought of Kendalls Tau and Spearmans but they look at ordinal data only.

What statistical test is best suited?

Answer 1

Hi Michelle,

You could try a chi-squared analysis with a two-way table. Put gestational age (nominal) in column and perceived difficulty (ordinal) in row. It will look something like this, although I can't really draw a table here in Wyzant:

Infant 3-6 Months 7-9 Months ....

Perceived Difficulty

Very Hard

Moderate

Neutral

..

Then, you can compute the chi-squared test statistic to see if the variables are related. This is called the chi-squared test of independence. Test statistic is given by:

X² = Sum [(Observed – Expected)²/Expected]

Observed are your actual counts per cell in the table. To compute expected, compute (row total * column total)/ grand total for each cell. Do the addition and division, and you will have your test statistic.

You will also need a degrees of freedom, which is:

df= (r-1) * (c-1),

r= number of rows

c=number of columns

In the example above, we would do df= (3-1) * (3-1) = 4

You would then check your test statistic against the corresponding chi-squared value, which you can find readily enough in a chi-squared table (Google this; Wyzant reviews posts when I link) or with statistical software. An example from the table. Above, we have 4 degrees of freedom, so you look for that on the right, then look for (1-alpha) on top, where alpha is your significance level. Most common one is 0.05, but that's up to you for your study.

Anyway, if your chi-squared test statistic exceeds your critical value, you can conclude that a relationship exists. If not, you cannot draw that conclusion. If you need p-value, as you probably do, that will likely require statistical software. I hope this at least gives you an option.