Comparing the rate of admitting to not quitting smoking between two groups? One group of patients was interviewed by doctors and 2% patients admitted to smoking. Another group of patients was interviewed by nurses and their admittance rate to smoking was 10%. 
How would I show that the difference in rates is statistically significant?
 A: Assuming the patients are random, independent samples for each group, consider a 2-proportion z-test where:
$$z=(\hat{p}_1-\hat{p}_2)/SE$$
$$SE = \sqrt{\hat{p}(1-\hat{p})(1/n_1)+(1/n_2)}$$
$$\hat{p}=(\hat{p}_1*n_1+\hat{p}_2*n_2)/(n_1+n_2)$$
Or is your sample paired?
A potential solution for paired samples is McNemar's test.
A: @sairahsharif  You have 6 more questions. I think what you have is the following design:A: said yes to doctor, yes to nurse
B: said yes to doctor, no to nurse
C: said no to doctor, yes to nurse
D: said no to doctor, no to nurse
and for question 1 "1-How many admit to second hand smoking" the answers can be either "admit" or "don't admit".  
I'd approach this as a 4x2 chi-square.  Depending on the size of the cells, you may not have enough (too many expected values less than 5 can be a problem). But let's not deal with that problem until we know we have it.
I think for questions 2-5 the same thing works. Question 6 (do babies do better) you probably have a continuous measure rather than a yes/no measurement so a dummy variable regression would seem appropriate.  That may also make sense if people get multiple interventions and comply with some, but not all of them -- but I'd start with the question by question analysis first since it will be easier to understand.
Isn't there also a question 0: of those referred, how many actuallly show up?
I also have a question unrelated to statistical issues, just curiosity: what do you mean by "quitting secondary smoking"? 
