Ordinal and nominal I want to test whether men or women are higher risk takers. In the data, I have a likert scale for risk (strongly agree to take risk, agree to take risk, uncertain, disagree to take risk and strongly disagree to take risk). The data has a gender variable but I want to know whether men or women are high risk takers. which test should I use?
 A: You might want to see the answers to a recent Cross Validated questions here: CV Question.
A couple of potential solutions would be the Cochran-Armitage test or to use ordinal regression.
I'll use the sample data from @BruceET, and present these in R.
Input =(
"RiskTaking  1   2   3   4   5
Sex
Men          3   9  12  14  12
Women        7  23  12   2   6 
")

Tabla = as.table(read.ftable(textConnection(Input)))

Tabla

As @BruceET mentioned, a permutation test could be used.  In R, the coin package has an implementation for a table with one ordinal variable and one categorical variable. This is a different test than those described by @BruceET, and I think is supposed to be used like the Cochran-Armitage test.  Note that the ordinal categories (RiskTaking) are specified as being equidistant with the scores option.
### Adapted from:
###  http://rcompanion.org/handbook/H_09.html

library(coin)

Test = chisq_test(Tabla,
                  scores = list("RiskTaking" = c(-2, -1, 0, 1, 2)))

Test

   #### Asymptotic Linear-by-Linear Association Test
   #### 
   #### data:  RiskTaking (ordered) by Sex (Men, Women)
   #### Z = 3.6366, p-value = 0.0002762
   #### alternative hypothesis: two.sided

As @Gijs mentioned, a more EDIT: flexible approach is to use ordinal regression.  In R, the ordinal package is a great tool for this.
I'll  recreate the data to keep things simple.  Note that the dependent variable must be specified as an ordered factor variable.
Sex        = c(rep("Men", 5), rep("Women", 5))
RiskTaking = rep(1:5,2)
Count      = c(3,9,12,14,12,7,23,12,2,6)

Data = data.frame(Sex, RiskTaking, Count)

Data$RiskTaking = factor(Data$RiskTaking, ordered=TRUE)

str(Data)

The following conducts ordinal regression with the ordinal package.  Here I'll use the anova function between two models.  The summary function could be used instead.  
### Adapted from:
###  http://rcompanion.org/handbook/G_02.html

if(!require(ordinal)){install.packages("ordinal")}

library(ordinal)

model = clm(RiskTaking ~ Sex, data = Data, weight = Data$Count)

model.null = clm(RiskTaking ~ 1, data = Data, weight = Data$Count)

anova(model, model.null)

   #### Likelihood ratio tests of cumulative link models:
   ####  
   ####            formula:         link: threshold:
   #### model.null RiskTaking ~ 1   logit flexible  
   #### model      RiskTaking ~ Sex logit flexible  
   #### 
   ####            no.par    AIC  logLik LR.stat df Pr(>Chisq)    
   #### model.null      4 315.85 -153.93                          
   #### model           5 302.93 -146.46  14.924  1  0.0001119 ***

A: Let's use a scale with values 1 through 5, where 1 is least inclined toward
risk and 5 is most inclined. If you are willing to take the scale as
numerical, you might try to use a two-sample t test, even though such
a scale hardly produces normal data.
A Mann-Whitney-Wilcoxon rank sum test can deal with ordinal data, but would
probably not be useful for your data because of it will have many ties.
A one-sided two-sample permutation test might be the best choice. It
can compare responses of men and women using either differences in means
(numerical) or differences in medians (ordinal). Theoretically, it is
possible to find the exact permutation distribution under the null
hypothesis by using combinatorial methods, but this can be tedious. And a good
approximation of the permutation distribution can be found using simulation.
Example. Suppose that the proportional choices of responses 1 through 5
are $(1, 1, 2, 3, 2)$ for men and $(1, 3, 2, 1, 1)$ for women. Randomly
generated data for 50 men and 50 women according to these proportions
might be tabled as follows:
m (men)
 1  2  3  4  5 
 3  9 12 14 12 

w (women)
 1  2  3  4  5 
 7 23 12  2  6 

Thus the observed sample means for men and women are $\bar M = 3.46$ and
$\bar W = 2.54,$ respectively. ]A one-sided, pooled 2-sample t test has
P-value 0.0001; if it is valid, this P-value provides strong evidence that men tend to have higher scores.]
A permutation test begins by finding $\bar D_{obs} = \bar M - \bar W = 0.92.$
Then the 100 observations for men and women are randomly permuted.
Taking the first 50 observations to be for 'men' and the last 50 to be for 'women', we get an average difference $\bar D_{prm}.$ (If the null hypothesis
is true then the permutation should not make an important difference.)
If this permutation procedure is repeated many times, we can simulate the permutation
distribution of $\bar D$ and then see whether the observed value 0.92 is
an unusual value for the permutation distribution. The P-value is the
proportion of $\bar D_{prm}$-values that exceed 0.92.
For the fake data shown above, the P-value is about 0.0001. 
If data are taken to be ordinal (not numerical) one can use medians throughout, and the P-value is 0.03, so the null hypothesis
is still rejected at the 5% level of significance. 
You are not the first person to consider testing such data and several
other tests have been proposed for ordinal data. You can google around to
see various proprietary software packages for such analyses. Textbooks in
psychology and sociology provide additional sources. I prefer the
permutation test. (Perhaps other Answers will have different suggestions.)

Note: A short program using R statistical software for a permutation test using the difference in means as the
'metric' is shown below. Remove set.seed(1801) for different fake data;
remove set.seed(1118) for a different approximation of the P-value.
Change mean to median on lines with #* to use the difference in medians
as the metric. (In R, more elegant programs can be written for this procedure, but
the one below should be easy to understand, even for those not familiar with R.)
set.seed(1801)  # generate fake data
mp = c(1,1,2,3,2);  mw=c(1,3,2,1,1)
m = sample(1:5, 50, rep=T, prob=mp)  # R scales `mp` so elements sum to 1
w = sample(1:5, 50, rep=T, prob=wp)

set.seed(1118)   # one-sided permutation test: matric=means
mw = c(m,w);  df.obs = mean(m)-mean(w)                 #*
B = 10^5;  df.prm = numeric(B)
for(i in 1:B) {
  mw.prm=sample(mw,100)
  df.prm[i] = mean(mw.prm[1:50])-mean(mw.prm[51:100])  #*
  }
mean(df.prm > df.obs)   # mean of logical vector is its proportion of TRUE's
[1] 9e-05

summary(df.prm)
      Min.    1st Qu.     Median       Mean    3rd Qu.       Max. 
-1.2000000 -0.1600000  0.0000000  0.0001532  0.1600000  1.0800000 


