How to interpret data from Mann-Whitney U Test So I am doing a research project and I was told to d

U = 341978.5
This is the value of the Mann-Whitney U statistic, which is defined as
$$\begin{gathered} U=\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}\underset{j=1}{\overset{m}{\sum <!--\; \sum \; -->}}S(X_i,Y_j), \\ S(X,Y)=\{\begin{array}{ll}1,& Y<X\\ 1/2,& Y=X\\ 0,& Y>X.\end{array} \end{gathered}$$
Here, $X_i$ represents the response of the $i^{\mathrm{t}\mathrm{h}}$ person in the first group, which are women, and $Y_j$ is the response of the $j^{\mathrm{t}\mathrm{h}}$ person in the second group, which are men. However, this definition could have X and Y reversed, so you might want to double check.
U (standardized) 0.532
This is the standardized value of the U-statistic, equal to
$\frac{U-<!--\; -\; -->{\mathrm{\mu <!--\; \mu \; -->}}_U}{{\mathrm{\sigma <!--\; \sigma \; -->}}_U},$,
where ${\mathrm{\mu <!--\; \mu \; -->}}_U$ and ${\mathrm{\sigma <!--\; \sigma \; -->}}_U$ are given below. This standardized value is approximately normally distributed with mean 0 and variance 1, so 0.532 is a z-score; the more extreme it is, the more evidence there is to support the hypothesis that the two groups differ in their responses.
Expected value 336960
This is ${\mathrm{\mu <!--\; \mu \; -->}}_U=mn/2$, where m=936 and n=720.
Variance (U) 89023289.87
This is the sample variance of the U statistic,
${\mathrm{\sigma <!--\; \sigma \; -->}}_U^2=\frac{mn}{12}((m+n-<!--\; -\; -->1)-<!--\; -\; -->\underset{i=1}{\overset{k}{\sum <!--\; \sum \; -->}}\frac{t_i^3-<!--\; -\; -->t_i}{(m+n)(m+n-<!--\; -\; -->1)})$
where $t_i$ is the number of people sharing rank i, where your ranks range from 1 to k=7.
p-value (Two-tailed) 0.595
This is the conditional probability that, given there is no difference between the two groups, you would obtain a sample that is at least as extreme as the one you observed. It is a measure of the plausibility of the data you observed, assuming the null is true. Therefore, the smaller this value, the more evidence there is to favor rejecting the null hypothesis.
alpha 0.05
This is the predefined significance level of the test and is the maximum Type I error you are willing to accept--i.e., you wish to limit the probability of incorrectly rejecting the null hypothesis to be at most 5%. Since the p-value exceeds $\mathrm{\alpha <!--\; \alpha \; -->}$, you do not reject $H_0$ and your conclusion is that the data furnishes insufficient evidence to suggest the two groups responded differently to the survey.