How to test statistical significance of differences between before and after nominal data I'm wondering what the best statistical test to run on my dissertation survey would be. 
I'm investigating letting agents attitudes towards the recently introduced tenant fee ban in the UK. A survey conducted before the ban came into force (with a sample size of 1,008) found that 87% of agents believed the ban would lead to an increase in rents; my survey conducted after the ban came into force (with a sample size of 31) found that 91% of agents believe it will lead to an increase in rents. 
The question was worded 'How will the tenant fee ban impact on rents in the long-term?', and the available answers were 'increase', 'decrease', 'stay the same' and 'don't know' (which I believe is nominal data?)
I'm investigating whether or not there has been a change in agent's attitudes since the ban came into force, so I guess the null hypothesis is 'there has been no change in attitudes since the ban came into force'. 
Is it possible to run a test to see if there is a statistically significant difference between attitudes before the ban came into force and attitudes after?
Thanks!
 A: Classic tests for changes in distributions of nominal data (each individual in exactly 1 category, comparing 2 sets of data from different conditions) would be a Chi-squared test or Fisher's exact test. Your data, however, are more like ordinal data, where there is a natural ordering among the 3 categories "decrease," "no change," "increase" (excluding for now the "don't know" category). The link above to ordinal data suggests some ways to examine such data.
Your study, however, is just not adequately powered to detect a difference from the situation before the ban came into force. Of the 31 cases in your sample, you evidently found 28 in the "increase" category to get the 91% value. Note that 87% of 31 cases would have been 27 in the "increase" category, only 1 different from the 28 that you found. With errors in count values typically on the order of the square root of the number of counts, you simply don't have enough cases to detect a difference. You need a much larger sample, with careful attention paid to matching the characteristics of your sample to that of the pre-ban study, to accomplish your goal. You should try to find a competent local statistician, and get advice on study design first.
A: One difficulty with your proposed approach is that you are looking
only at "Increase" vs. "Not Increase" instead of "Increase, Decrease, No Change."
You say that 87% of 1109 in the published survey, which I take to mean 961 respondents said "Increase". And among those in your sample of 31 subjects,
28 (about 91%) said "Increase." Even if you chose your subjects at random from the same population used in the published survey, 31 subjects is not enough to establish a difference between 87% and 91%. (In a survey of 1109 respondents, the margin of error is for a 95% confidence interval
is about $\pm$ 3%.)  Here is a printout from Minitab of results of a test of two proportions. Results from a 2-sample (approximate) normal test and from a hypergeometric Fisher exact test are shown, neither significant: [When performing such a test it is important to enter counts, not proportions; the software will compute proportions.]
Test and CI for Two Proportions 

Sample    X     N  Sample p
1       961  1108  0.867329
2        28    31  0.903226

Difference = p (1) - p (2)
Estimate for difference:  -0.0358973
95% CI for difference:  (-0.141871, 0.0700767)
Test for difference = 0 (vs ≠ 0):  
  Z = -0.58  P-Value = 0.560

* NOTE * The normal approximation may be inaccurate for small samples.

Fisher’s exact test: P-Value = 0.788

If you have proportions about .90 saying "Increase," and want to
have probability 90% of rejecting the null hypothesis (no difference) when the actual difference is about $\pm$ 3%, then you
need sample sizes around 1280 (around  1590 for 95% rejection probability). So, as @EdM suggests (+1), you would need a massively larger study than currently
contemplated.
Power and Sample Size 

Test for Two Proportions

Testing comparison p = baseline p (versus ≠)
Calculating power for baseline p = 0.91
α = 0.05

              Sample  Target
Comparison p    Size   Power  Actual Power
        0.87    1284    0.90      0.900052
        0.87    1588    0.95      0.950087

The sample size is for each group.

Here is the accompanying power curve:

