Equation of the ellipse is $\frac{x^2}{4}+\frac{y^2}{3}=1$ eccentricity is  ${\mathrm{e}}^2=1-\frac{3}{4}=\frac{1}{4}\Rightarrow \mathrm{e}=\frac{1}{2}$

Coordinates of the foci of the ellipse are  $(\pm 1,0)$ Hyperbola has same foci, if  ${\mathrm{e}}_1$ is the eccentricity of the hyperbola then

$\left.2{\mathrm{ae}}_1=2\text{ and }2\mathrm{a}=\frac{1}{2}\text{ (given }\right)$

$\Rightarrow {\mathrm{e}}_1=4\text{ and }{\mathrm{b}}^2={\mathrm{a}}^2\left({\mathrm{e}}_1^2-1\right)=\frac{1}{16}(16-1)=\frac{15}{16}$

So the equation of hyperbola is $\text{ or }\frac{x^2}{1}-\frac{y^2}{15}=\frac{1}{16}$ Thus Statement-I is true

Statement II is false as the equation of the hyperbola conjugate to 

$\frac{x^2}{15}-\frac{y^2}{1}=16\text{ is }\frac{x^2}{15}-\frac{y^2}{1}=-16$