why is oxidation number of H in NaBH_{4 =}-1?

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Solution

H have the characters of both alkali metals and halogens, the electronegativity of hydrogen is higher than the boron.
B have 2.0 while H have 2.2 so hydrogen show negative oxidation state rather then positive.
Therefore it will show a negative oxidation state and boron will show a positive charge. These types of compounds where Hydrogen exist in negative oxidation are known as hydride.

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