ISSUE: Comparison study

DISCRETE DATA: If the answer you collected is in the form of (Yes | No), the data set is discrete. Select a reference group, say "married." Where married people answer "happier" score yes = 1, all else no = 0. Each score 1 = s for success. Use Laplace Rule of Success to determine the probability:

(1)   p = (s + 1) / (n + 2)

The probability of "not happier" is q = 1 - p. Assume now that you have X number of married people answered happier, what is the probability of that in the group? i.e. what is P(X)?

(2)   P(X) = [n! / (n - X)!X!] pX qn - X

Is it statistically significant? The test statistic for discrete distribution is given by:

(3)   Zbin = [(X / n) - p] / sqrt(pq /n)

Here assuming that the sample is large enough to affect normal distribution in order to use Z. Set you confidence interval, say CI = 95%; then the null hypothesis is H0: Z(obs) < 1.65 otherwise significant is Z(obs) > 1.65. Double check you work by follow the same steps but using the unmarried people as the reference group.

CONTINUOUS DATA: If the data comes in a form of continuous set, say scalar data, i.e. [1,2,3,4,5], here you can examine both married and unmarried people in one set and select out those with significant outliers by using the standard score analysis. Say both groups gave score individually as Xi: (x1, x2, ..., xn); the observed mean is X^; standard deviation is S, the Z score is given by:

(4)  Z = (Xi - X^) / S

This Z(obs) is the standard score. With this standard score, look in the Z-Table for the corresponding % probability. Compare that percent probability to your CI = 95%, if less than 95% = 1.65, then not significant if more than 95% then significant.

Next step, collect all significant counts (both married and unmarried), then compare then which have more frequency. The conclusion would be "The ..... group shows more statistically significant counts for being happier than ..... group." where group = married or unmarried.

Now taken only the one with Z score higher than 1.65 or with % larger than 95%, doe a group comparison using the following formula:

(5)   Z = (A - B) / C

(5.1)   A = (X1^ - X2^)               .... X1^ & X2^ = mean of groups 1 & 2

(5.2)   B = (µ1 - µ2)

(5.3)   C = S(sqrt(1/n1 + 1/n2)  .... n1 & n2 = sample size of groups 1 & 2

(5.4)   S = sqrt(S2)

(5.5)   S2 = (SS1 + SS2) / n1 + n2 - 2

(5.6)   SS1 = Σ(X1i - X1^)2

(5.7)   SS2 = Σ(X2i - X2^)2

This last process tells you whether the unmarried and married group are statistically different (among the one who already score Z > 1.65.

REFERENCES: See attached. I hope this has been helpful. Cheers.

http://www.xlstat.com/en/products-solutions/feature/two-sample-t-and-z-tests.html

Paul explains it comprehensively. However,

Sir if you use STATA or Eviews then u can test such mean differences without making these adjustments.

I think you will need to take into account also other variables, for example health, age, education... They all can influence the answers to whether one is happy/unhappy, in addition to being/not being married.

Dear Paul Louangrath, thank you very much for such a comprehensive and argumentative answer.I feel obliged and empowered.

Dear Jiri AMbros thanks for adding more dimensions which influence happiness. Dear Umar Farooq regards for you as well.