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[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

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A certain crop can grow in tidal flats and saline-alkali land, and its irrigation is done by diluting seawater for irrigation. In order to study the effect of seawater concentration $x\left(\%\right)$ on yield per mu $y(\text{tons})$, the yield per mu of the crop with seawater concentration data was measured through planting experiments in the experimental field. The data is shown in the table below.

| Seawater Concentration $x/\%$ | $3$ | $4$ | $5$ | $6$ | $7$ |
|--------------------------------|-----|-----|-----|-----|-----|
| Yield per mu $y/\text{tons}$    | $0.57$ | $0.53$ | $0.44$ | $0.36$ | $0.30$ |
| Residual $\widehat {{e_i}}$    | $-0.01$ | $0.02$ | $m$ | $n$ | $0$ |

It was found by plotting the scatter plot that the yield per mu $y(\text{tons})$ and seawater concentration $x\left(\%\right)$ can be fitted with a linear regression model. By calculating with the least squares method, the linear regression equation between $y$ and $x$ is $y=-0.07x+\hat{a}$.

$(1)$ Find the values of $m$ and $n$;  
$(2)$ In statistics, the commonly used correlation coefficient $R^{2}$ is used to describe the regression effect. The larger $R^{2}$ is, the better the regression effect. For example, assuming $R^{2}=0.85$, it means that $85\%$ of the differences in the predicted variable $y$ are caused by the explanatory variable $x$. Calculate the correlation coefficient $R^{2}$ (rounded to $0.01$) and indicate to what extent the change in yield per mu is caused by irrigating seawater concentration.

Given: Residual ${\hat{e}_i}={y_i}-{y_i}$, correlation coefficient ${R^2}=1-\frac{{\sum_{i=1}^n{{{({{y_i}-{{\hat{y}}_i}})}^2}}}}{{\sum_{i=1}^n{{{({{y_i}-\overline{y}})}^2}}}}$, where $\sum_{i=1}^5{{{({{y_i}-\overline{y}})}^2}}=0.051$.

Present the answer in LaTex format: \boxed{Your answer}
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